HackerRank How Many Substrings? Solution

Yashwant Parihar,

In this post, we will solve HackerRank How Many Substrings? Problem Solution.

Consider a string of n characters, s, of where each character is indexed from 0 ton – 1. You are given q queries in the form of two integer indices: left and right. For each query, count and print the number of different substrings of $ in the inclusive range between left and right.
Note: Two substrings are different if their sequence of characters differs by at least one. For example, given the string s = aab, substrings 8[0,0]=a and 8[1,1] = a are the same but substrings $[0,1] = aa and $[1,2] = ab are different. | |
Input Format
The first line contains two space-separated integers describing the respective values of n
and q.
The second line contains a single string denoting s.
Each of the q subsequent lines contains two space-separated integers describing the respective values of left and right for a query.

Output Format
For each query, print the number of different substrings in the inclusive range between
index left and index right on a new line.

Sample Input 0

5 5
aabaa
1 1
1 4
1 1
1 4
0 2

Sample Output 0

1
8
1
8
5

Explanation 0
Given s = aabaa, we perform the following q = 5 queries:
1.1 1: The only substring of a is itself, so we print 1 on a new line.

  1. 1 4: The substrings of abaa are a, b, ab, ba, aa, aba, baa, and abaa, so we print 8 on a new line.
  2. 1 1: The only substring of a is itself, so we print 1 on a new line.
  3. 1 4: The substrings of abaa are a, b, ab, ba, aa, aba, baa, and abaa, so we print 8 on a new line.
    5.0 2: The substrings of aab are a, b, aa, ab, and aab, so we print 5 on a new line.
HackerRank How Many Substrings? Problem Solution
HackerRank How Many Substrings? Problem Solution

How Many Substrings? C++ Solution

#include<bits/stdc++.h>
#define pii pair<int,int>
#define fi first
#define se second
#define pb push_back
#define SZ(x) ((int)((x).size()))
#define rep(i,j,k) for(int i=(int)j;i<=(int)k;i++)
#define per(i,j,k) for(int i=(int)j;i>=(int)k;i--)
using namespace std;
typedef long long LL;
typedef double DB;
const DB pi=acos(-1.0);
const int N=100005;
int go[N<<1][26],fail[N<<1],len[N<<1],tot,last;
int n;
int cnt=0;
namespace seg{
    int tag[N<<2];
    LL sum[N<<2];
    inline void Tag(int me,int l,int r,int v){
        tag[me]+=v;
        sum[me]+=(r-l+1)*1ll*v;
    }
    inline void down(int me,int l,int r){
        if(tag[me]==0)return;
        int mid=(l+r)>>1;
        Tag(me<<1,l,mid,tag[me]);
        Tag(me<<1|1,mid+1,r,tag[me]);
        tag[me]=0;
    }
    void add(int me,int l,int r,int x,int y,int v){
        //if(l==1&&r==n)printf("add %d %d %d\n",x,y,v);
        if(l^r)down(me,l,r);
        if(x<=l&&r<=y){
            Tag(me,l,r,v);
            return;
        }
        int mid=(l+r)>>1;
        if(x<=mid)add(me<<1,l,mid,x,y,v);
        if(y>mid)add(me<<1|1,mid+1,r,x,y,v);
        sum[me]=sum[me<<1]+sum[me<<1|1];
    }
    LL ask(int me,int l,int r,int x,int y){
        if(l^r)down(me,l,r);
        if(x<=l&&r<=y)return sum[me];
        int mid=(l+r)>>1;
        LL ret=0;
        if(x<=mid)ret+=ask(me<<1,l,mid,x,y);
        if(y>mid)ret+=ask(me<<1|1,mid+1,r,x,y);
        return ret;
    }
    void Do(int pre,int now,int L,int R){
        if(L>R)return;
        ++cnt;
        //printf("_%d %d %d %d\n",pre,now,L,R);
        if(pre)add(1,1,n,pre-R+1,pre-L+1,-1);
        add(1,1,n,now-R+1,now-L+1,1);
    }
};
namespace lct{
    int l[N<<2],r[N<<2],fa[N<<2];
    int last[N<<2];
    inline bool top(int x){return (!fa[x])||(l[fa[x]]!=x&&r[fa[x]]!=x);}
    inline void left(int x){
        int y=fa[x];int z=fa[y];
        r[y]=l[x];if(l[x])fa[l[x]]=y;
        fa[x]=z;if(l[z]==y)l[z]=x;else if(r[z]==y)r[z]=x;
        l[x]=y;fa[y]=x;
    }
    inline void right(int x){
        int y=fa[x];int z=fa[y];
        l[y]=r[x];if(r[x])fa[r[x]]=y;
        fa[x]=z;if(l[z]==y)l[z]=x;else if(r[z]==y)r[z]=x;
        r[x]=y;fa[y]=x;
    }
    inline void down(int x){
        if(l[x])last[l[x]]=last[x];
        if(r[x])last[r[x]]=last[x];
    }
    int q[N<<2];
    inline void splay(int x){
        q[q[0]=1]=x;
        for(int k=x;!top(k);k=fa[k])q[++q[0]]=fa[k];
        per(i,q[0],1)down(q[i]);
        while(!top(x)){
            int y=fa[x];int z=fa[y];
            if(top(y)){
                if(l[y]==x)right(x);else left(x);
            }
            else{
                if(r[z]==y){
                    if(r[y]==x)left(y),left(x);
                    else right(x),left(x);
                }
                else{
                    if(l[y]==x)right(y),right(x);
                    else left(x),right(x);
                }
            }
        }
    }
    void Access(int x,int cov){
        int y=0;
        for(;x;y=x,x=fa[x]){
            splay(x);
            down(x);
            r[x]=0;

            int L,R;
            int z=x;
            while(l[z])z=l[z];
            L=len[fail[z]]+1;
            splay(z);splay(x);
            z=x;
            while(r[z])z=r[z];
            R=len[z];
            splay(z);splay(x);
            seg::Do(last[x],cov,L,R);
            r[x]=y;
            last[x]=cov;
        }
    }
    void SetFa(int x,int y,int po){
        fa[x]=y;
        Access(x,po);
    }
    void split(int x,int y,int d){
        splay(y);
        down(y);
        r[y]=0;
        fa[d]=y;
        splay(x);
        fa[x]=d;
        last[d]=last[x];
    }
};
namespace sam{
    void init(){
        tot=last=1;
    }
    void expended(int x,int po){
        int gt=++tot;len[gt]=len[last]+1;int p=last;last=tot;
        for(;p&&(!go[p][x]);p=fail[p])go[p][x]=gt;
        if(!p){
            fail[gt]=1;
            lct::SetFa(gt,1,po);
            return;
        }
        int xx=go[p][x];
        if(len[xx]==len[p]+1){
            fail[gt]=xx;
            lct::SetFa(gt,xx,po);
            return;
        }
        int tt=++tot;
        len[tt]=len[p]+1;
        fail[tt]=fail[xx];
        int dt=fail[xx];
        fail[xx]=fail[gt]=tt;
        lct::split(xx,dt,tt);
        lct::SetFa(gt,tt,po);
        rep(i,0,25)go[tt][i]=go[xx][i];
        for(;p&&(go[p][x]==xx);p=fail[p])go[p][x]=tt;
    }
};
int Q;
char str[N];
int qL[N];
vector<int>que[N];
LL ans[N];
void Main(){
    rep(i,1,n){
        sam::expended(str[i]-'a',i);
        rep(j,0,que[i].size()-1){
            int id=que[i][j];
            ans[id]=seg::ask(1,1,n,qL[id],n);
        }
    }
}
void init(){
    scanf("%d%d",&n,&Q);
    scanf("%s",str+1);
    rep(i,1,Q){
        int r;scanf("%d%d",&qL[i],&r);
        qL[i]++;r++;
        que[r].pb(i);
    }
    sam::init();
}
void Output(){
    rep(i,1,Q)printf("%lld\n",ans[i]);
}
int main(){
    init();
    Main();
    Output();
    return 0;
}

How Many Substrings? C Sharp Solution

using System;
using System.Collections.Generic;
using System.IO;
using System.Linq;
class Solution {

    static void Main(String[] args) {
        string[] tokens_n = Console.ReadLine().Split(' ');
        int n = Convert.ToInt32(tokens_n[0]);
        int q = Convert.ToInt32(tokens_n[1]);
        string s = Console.ReadLine();
        var queries = new Query[q];
        for(int a0 = 0; a0 < q; a0++){
            string[] tokens_left = Console.ReadLine().Split(' ');
            queries[a0] = new Query(Convert.ToInt32(tokens_left[0]), Convert.ToInt32(tokens_left[1]));
        }
        SubstringCounter.solve(s.ToCharArray(), queries);    
        foreach (Query e in queries)
            Console.WriteLine(e.Answer);
    }
}
public class SubstringCounter
{
    public static void solve(char[] str, Query[] queries)
    {

        int[] suffixArray = SuffixArray.BuildSuffixArray(str);
        int[] suffixArrayInverse = new int[str.Length];
        for (int i = 0; i < str.Length; i++)
        {
            suffixArrayInverse[suffixArray[i]] = i;
        }
        int[] lcp = SuffixArray.BuildLcpKasai(suffixArray, suffixArrayInverse, str);

        //defaultdict instead of List?
        var queriesIndexedByLeft = new List<Query>[str.Length];
        for (int i = 0; i < str.Length; i++)
            queriesIndexedByLeft[i] = new List<Query>();

        foreach (Query query in queries)
            queriesIndexedByLeft[query.Left].Add(query);

        var maximumPrefixContainingInterval = LcpIntervalHelper.BuildIntervals(lcp);

        SubstringSum substringSum = new SubstringSum(str.Length);
        for (int i = str.Length - 1; i >= 0; i--)
        {
            //No ValueTuples in hacker rank :o(
            Stack<Tuple<int, int>> tups = LcpIntervalHelper.GetLastSeen(i, 
                                                                        maximumPrefixContainingInterval,                                                                                            suffixArrayInverse);

            int matchedLcp = 0;
            while (tups.Count > 0)
            {
                var tup = tups.Pop();
                int j = tup.Item1;
                int newLcp = tup.Item2;
                substringSum.addRange(j + matchedLcp, j + newLcp - 1);
                matchedLcp = newLcp;
            }

            foreach (Query query in queriesIndexedByLeft[i])
            {
                query.Answer = substringSum.getSum(query.Left, query.Right);
            }
        }
    }
}
public class Query
{
    public int Left { get; }
    public int Right { get; }
    public long Answer { get; set; }

    public Query(int left, int right)
    {
        this.Left = left;
        this.Right = right;
    }
}


//When we want count substrings whilst we change the right value of a string for a prebuilt suffix array the standard
//Lcp array is less helpful as it is not ordered in terms of the original string and will become full of holes. With holes
//we would need to consider the minimum of all intervening missing LCP elements to calculate the LCP between two elements. 
//
//So instead we flip the idea on its side to have a set of LcpIntervals. Each lcp interval contains all Suffixes in
//the Suffix Array with a minimum lcp.
//
//At least the minimum LCP of all but the first element in the interval can be used to reduce the substring count.
//For each Suffix Array item we only store the LcpInterval with the maximum MinimumLcp. This ensures at most n
//intervals.
//So as the right value of a query changes the problem becomes how to spot if this is the first suffix in the
//interval or not.
//We do this by processing the string in reverse order and noting the leftmost time a string in the interval is seen.

// Each LcpInterval contains a link to its enclosing superinterval, each interval will be enclosed by one and only
// one Super-Interval, apart from the root.
// I suppose you could consider it to be a tree and perhaps I should have researched if an appropriate standard
// pattern / data structure existed.

// The curious thing is that when I look at the finished working result I notice that Left and Right of LcpInterval 
// don't actually appear to be needed. But if I remove them the motivating idea of the code would be impossible to
// understand. I will leave them in as comments, with the caveat of all comments that as they don't
// affect the correct working of the code they may be wrong.


public class LcpIntervalHelper
{
    public class LcpInterval
    {
        //public int Left { get; }
        //public int Right { get; set; }
        public LcpInterval SuperLcpInterval { get; set; }
        public int MinimumLcp { get; }
        public int LastSeen { get; set; }


        //public LcpInterval(int length, int left, int right, LcpInterval superLcpInterval)
        public LcpInterval(int length, LcpInterval superLcpInterval)
        {
            MinimumLcp = length;
            //Left = left;
            //Right = right;
            SuperLcpInterval = superLcpInterval;
            LastSeen = int.MaxValue;
        }

        public override string ToString()
        {
            return
                   //                      $"{nameof(Left)}: {Left}, " +
                   //                      $"{nameof(Right)}: {Right}" +
                   $"{nameof(MinimumLcp)}: {MinimumLcp} " +
                   $"{nameof(LastSeen)}: {LastSeen} ";
        }
    }

    public static Stack<Tuple<int, int>> GetLastSeen(int i, LcpInterval[] lcpIntervals, int[] saInverse)
    {
        var interval = lcpIntervals[saInverse[i]];
        var ret = new Stack<Tuple<int, int>>();
        int lastSeen = int.MaxValue;

        while (interval.SuperLcpInterval != null)
        {
            if (interval.LastSeen < lastSeen)
            {
                ret.Push(new Tuple<int, int>(interval.LastSeen, interval.MinimumLcp));
                lastSeen = interval.LastSeen;
            }
            interval.LastSeen = i;
            interval = interval.SuperLcpInterval;
        }

        return ret;
    }

    public static LcpInterval[] BuildIntervals(int[] lcp)
    {
        // the root interval should be the only one with parent == null
        //var interval = new LcpInterval(0, 0, lcp.Length , null);
        var interval = new LcpInterval(0, null);

        var suffixArrayIntervals = new LcpInterval[lcp.Length];

        //Step through lcp build interval assigning max count interval to each item of the initial string.
        //Each interval has a link to the containit interval.

        for (int i = 0; i < lcp.Length - 1; i++)
        {
            //start off we are in the current interval
            //We know we are in the same interval as indexes to the left. We might be in 
            //a sub interval if we have a higher matching prefex to the right.
            suffixArrayIntervals[i] = interval;
            LcpInterval oldLcpInterval = interval;
            while (interval.MinimumLcp > lcp[i + 1] && interval.SuperLcpInterval != null)
            {
                //close the current ineterval
                //                  interval.Right = i;
                oldLcpInterval = interval;
                interval = interval.SuperLcpInterval;
            }

            if (interval.MinimumLcp < lcp[i + 1])
            {
                if (interval != oldLcpInterval)
                {
                    // We have closed a higher LCP interval but are inserting a new one between it and its previous
                    // SuperInterval.
                    // e.g. oldInterval was MinimumLCP = 2 with SuperInterval MinimumLCP=0
                    // but we now whish to open an interval MinimumLCP=1

                    //interval = new LcpInterval(lcp[i + 1], oldLcpInterval.Left, lcp.Length - 1, interval);
                    interval = new LcpInterval(lcp[i + 1], interval);

                    oldLcpInterval.SuperLcpInterval = interval;
                }
                else
                {
                    // Starting a new interval from the previous interval with higher LCP
                    //interval = new LcpInterval(lcp[i + 1], i, lcp.Length - 1, interval);
                    interval = new LcpInterval(lcp[i + 1], interval);
                    suffixArrayIntervals[i] = interval;
                }
            }

        }
        suffixArrayIntervals[lcp.Length - 1] = interval;
        return suffixArrayIntervals;
    }
}
// For the Hackerrank problem we are interested in how the matching SuffixArray LCP values to reduce the number
// of substrings
//
// Without matching LCP prefixes the number of substrings is n=(right+1-left)(right+2-left)/2
// This is just math for SUM(i=left to right){i}. This is clearly Trivial to calculate.
//
// So we are only really interested in LCP values and when they should be used to reduce substring count.
// When we have a matching prefix in SuffixArray index j > i. The potential to use those matching prefixes 
// drops off as the j extends further to the right i.e j+lcp > right.
// So naively we want
//  for (int k = j; j < j+lcp; j++)
//  {
//      OffsetSum.UpdateValue(k, 1);
//  }
// And we can use OffsetSum to reduce the number of Substrings as we would normally for an LCP/SuffixArray
// However if lcp is > 4 potentially much > 4 this becomes slow.
// So we simulate this by using a math formula
// And calculate the sum using k * LinearSum(k) + OffsetSum(k)
//
// For instance for n = 13
// left =3 right =10 i.e Offset = [0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0 , 0, 0]
//
// Can be simulated with
//  Offset = [0, 0, 0, -2, 0, 0, 0, 0, 0, 0, 10, 0, 0, 0 ]
//  Linear = [0, 0, 0, 1, 0, 0, 0, 0, 0, 0, -1, 0, 0, 0 ]
//
// Sum( 1 ) = SumOffset(1) + SumLinear(1) = 0 + 1*0 = 0
// Sum( 3 ) = SumOffset(3) + SumLinear(3) = -2 + 3*1 = 1
// Sum( 4 ) = SumOffset(4) + SumLinear(4) = -2 + 4*1 = 2
// Sum( 5 ) = SumOffset(5) + SumLinear(5) = -2+ 5*1 = 3
//
// Sum( 8 ) = SumOffset(8) + SumLinear(8) = -2 + 8*1 = 6
// Sum( 9 ) = SumOffset(9) + SumLinear(9) = -2 + 9*1 = 7
// Sum( 10 ) = SumOffset(10) + SumLinear(10) = 8 + 10*0 = 8
// Sum( 11 ) = SumOffset(11) + SumLinear(11) = 8  + 11*0 = 8
// Sum( 12 ) = SumOffset(8) + SumLinear(8) = 8 + 12*0 = 8

public class SubstringSum
{
    class FenwickTree
    {
        private long[] fenwickArray;

        public FenwickTree(int n)
        {
            fenwickArray = new long[n];
        }

        public void UpdateValue(int index, long value)
        {
            while (index < fenwickArray.Length)
            {
                fenwickArray[index] += value;
                index += LSB(index + 1);
            }
        }

        private int LSB(int i)
        {
            return i & -i;
        }

        public long GetValue(int i)
        {
            long sum = 0;
            while (i != 0)
            {
                sum += fenwickArray[i - 1];
                i -= LSB(i);
            }
            return sum;
        }
    }

    private FenwickTree LinearSum;
    private FenwickTree OffsetSum;
    private int n;

    public SubstringSum(int n)
    {
        this.n = n;
        LinearSum = new FenwickTree(n);
        OffsetSum = new FenwickTree(n);
    }

    public void add(int i, long v)
    {
        if (i < n)
        {
            OffsetSum.UpdateValue(i, v);
        }
    }


    public void addRange(int left, int right)
    {
        LinearSum.UpdateValue(left, 1);
        OffsetSum.UpdateValue(left, 1 - left);

        LinearSum.UpdateValue(right, -1);
        OffsetSum.UpdateValue(right, right);
    }

    public long getSum(int i)
    {
        long li = i;
        return li * LinearSum.GetValue(i + 1) + OffsetSum.GetValue(i + 1);
    }

    //inclusive right
    public long getSum(int left, int right)
    {
        // I don't understand the rules of implicit type conversions in intermediate calculations.
        // so better safe than sorry. 100k^2 is too big for an int.
        long rightlong = right;
        long leftlong = left;

        return (rightlong + 1 - leftlong) * (rightlong + 2 - leftlong) / 2 - getSum(right);
    }
}

public static class SuffixArray
{

    //Naive c# sort LCP generation for test10 n=100,000 was 973 ms
    //This Kasai trick bought it down to 6 ms
    //The simple idea is that if a suffix matches k characters. The suffix generated by one character further along
    //in the string must match at least k-1 characters, i.e. just use the previous matching prefix one char
    //further along. 
    //So we don't need to check the first k-1 chars. This gives us an O(n) sort.
    public static int[] BuildLcpKasai(int[] suffixArray, int[] suffixArrayInverse, char[] str)
    {
        int[] lcp = new int[suffixArray.Length];

        int n = suffixArray.Length;
        int k = 0;
        for (int i = 0; i < n; i++)
        {
            int inversei = suffixArrayInverse[i];
            if (inversei == 0)
            {
                k = 0;
                continue;
            }
            int j = suffixArray[inversei - 1];
            while (i + k < n && j + k < n && str[i + k] == str[j + k])
                k++;

            lcp[inversei] = k;
            if (k > 0)
                k--;

        }
        return lcp;
    }
    // Naive SuffixArray creation using CSharp Collection.sort was too slow approx 6 seconds for 100k 
    // SO I used a more efficient algorithm
    // Code adapted from:
    // http://www.cs.cmu.edu/afs/cs/academic/class/15492-f07/www/handouts/ks-icalp03.pdf
    // Juha Karkkainen and Peter Sanders
    public static int[] BuildSuffixArray(char[] str)
    {

        int[] s = new int[str.Length + 3];
        int[] SA = new int[str.Length];

        for (int i = 0; i < str.Length; i++)
            s[i] = str[i];
        s[str.Length] = 0;
        s[str.Length + 1] = 0;
        s[str.Length + 2] = 0;

        suffixArray(s, SA, str.Length, 128);
        return SA;
    }

    static int GetI(int[] SA12, int n0, int t)
    {
        return (SA12[t] < n0 ? SA12[t] * 3 + 1 : (SA12[t] - n0) * 3 + 2);
    }

    static bool leq(int a1, int a2, int b1, int b2)
    {
        // lexic. order for pairs
        return (a1 < b1 || a1 == b1 && a2 <= b2);
    } // and triples

    static bool leq(int a1, int a2, int a3, int b1, int b2, int b3)
    {
        return (a1 < b1 || a1 == b1 && leq(a2, a3, b2, b3));
    }

    // stably sort a[0..n-1] to b[0..n-1] with keys in 0..K from r
    static void radixPass(int[] a, int[] b, int[] r, int rOffset, int n, int K)
    {
        // count occurrences
        int[] c = new int[K + 1]; // counter array
        for (int i = 0; i <= K; i++) c[i] = 0; // reset counters
        for (int i = 0; i < n; i++) c[r[rOffset + a[i]]]++; // count occurences
        for (int i = 0, sum = 0; i <= K; i++)
        {
            // exclusive prefix sums
            int t = c[i];
            c[i] = sum;
            sum += t;
        }

        for (int i = 0; i < n; i++) b[c[r[rOffset + a[i]]]++] = a[i]; // sort
    }

    // find the suffix array SA of s[0..n-1] in {1..K}^n
    // require s[n]=s[n+1]=s[n+2]=0, n>=2
    static void suffixArray(int[] s, int[] SA, int n, int K)
    {
        int n0 = (n + 2) / 3, n1 = (n + 1) / 3, n2 = n / 3, n02 = n0 + n2;
        int[] s12 = new int[n02 + 3];
        s12[n02] = s12[n02 + 1] = s12[n02 + 2] = 0;
        int[] SA12 = new int[n02 + 3];
        SA12[n02] = SA12[n02 + 1] = SA12[n02 + 2] = 0;
        int[] s0 = new int[n0];
        int[] SA0 = new int[n0];

        // generate positions of mod 1 and mod  2 suffixes
        // the "+(n0-n1)" adds a dummy mod 1 suffix if n%3 == 1
        for (int i = 0, j = 0; i < n + (n0 - n1); i++)
            if (i % 3 != 0)
                s12[j++] = i;

        // lsb radix sort the mod 1 and mod 2 triples
        radixPass(s12, SA12, s, 2, n02, K);
        radixPass(SA12, s12, s, 1, n02, K);
        radixPass(s12, SA12, s, 0, n02, K);

        int name = 0, c0 = -1, c1 = -1, c2 = -1;
        for (int i = 0; i < n02; i++)
        {
            if (s[SA12[i]] != c0 || s[SA12[i] + 1] != c1 || s[SA12[i] + 2] != c2)
            {
                name++;
                c0 = s[SA12[i]];
                c1 = s[SA12[i] + 1];
                c2 = s[SA12[i] + 2];
            }

            if (SA12[i] % 3 == 1)
            {
                s12[SA12[i] / 3] = name;
            } // left half
            else
            {
                s12[SA12[i] / 3 + n0] = name;
            } // right half
        }

        if (name < n02)
        {
            suffixArray(s12, SA12, n02, name);
            // store unique names in s12 using the suffix array 
            for (int i = 0; i < n02; i++) s12[SA12[i]] = i + 1;
        }
        else // generate the suffix array of s12 directly
            for (int i = 0; i < n02; i++)
                SA12[s12[i] - 1] = i;

        // stably sort the mod 0 suffixes from SA12 by their first character
        for (int i = 0, j = 0; i < n02; i++)
            if (SA12[i] < n0)
                s0[j++] = 3 * SA12[i];
        radixPass(s0, SA0, s, 0, n0, K);

        // merge sorted SA0 suffixes and sorted SA12 suffixes
        for (int p = 0, t = n0 - n1, k = 0; k < n; k++)
        {
            int i = GetI(SA12, n0, t); // pos of current offset 12 suffix
            int j = SA0[p]; // pos of current offset 0  suffix
            if (SA12[t] < n0
                ? leq(s[i], s12[SA12[t] + n0], s[j], s12[j / 3])
                : leq(s[i], s[i + 1], s12[SA12[t] - n0 + 1], s[j], s[j + 1], s12[j / 3 + n0]))
            {
                // suffix from SA12 is smaller
                SA[k] = i;
                t++;
                if (t == n02)
                {
                    // done --- only SA0 suffixes left
                    for (k++; p < n0; p++, k++) SA[k] = SA0[p];
                }
            }
            else
            {
                SA[k] = j;
                p++;
                if (p == n0)
                {
                    // done --- only SA12 suffixes left
                    for (k++; t < n02; t++, k++) SA[k] = GetI(SA12, n0, t);
                }
            }
        }
    }
}

How Many Substrings? Java Solution

import java.io.ByteArrayInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Comparator;
import java.util.InputMismatchException;
import java.util.List;

public class How_Many_Substrings {
    InputStream is;
    PrintWriter out;
    String INPUT = "";
    
    void solve()
    {
        int n = ni(), Q = ni();
        char[] s = ns(n);
        int[][] qs = new int[Q][];
        for(int z = 0;z < Q;z++){
            qs[z] = new int[]{ni(), ni(), z};
        }
        Arrays.sort(qs, new Comparator<int[]>() {
            public int compare(int[] a, int[] b) {
                return a[1] - b[1];
            }
        });

        SuffixAutomatonOfBit sa = SuffixAutomatonOfBit.build(s);
        sa.sortTopologically();
        SuffixAutomatonOfBit.Node[] nodes = sa.nodes;
        int[] from = new int[nodes.length-1];
        int[] to = new int[nodes.length-1];
        int p = 0;
        for(SuffixAutomatonOfBit.Node node : nodes){
            if(node.id >= 1){
                from[p] = node.link.id; to[p] = node.id; p++;
            }
        }
        assert p == nodes.length-1;
        int[][] g = packU(nodes.length, from, to);
        int[][] pars = parents3(g, 0);
        int[] par = pars[0], ord = pars[1], dep = pars[2];
        HeavyLightDecomposition hld = new HeavyLightDecomposition(g, par, ord, dep);
        int m = hld.cluspath.length;
        SegmentTreeOverwrite[] sts = new SegmentTreeOverwrite[m];
        for(int i = 0;i < m;i++){
            sts[i] = new SegmentTreeOverwrite(hld.cluspath[i].length);
        }
        
        int[] base = new int[n];
        int qp = 0;
        int np = 0;
        long[] ft0 = new long[n+3];
        long[] ft1 = new long[n+3];
        long[] ans = new long[Q];
        for(int i = 0;i < n;i++){
            while(!(nodes[np].len == i+1 && nodes[np].original == null))np++;
            base[i] = np;
//            tr("base", base[i]);
            
            // 5 3 1 0
            // 5 3 1 0
            // 8 6 3 1 0 ?
            // aaba
            
            // delete
            int cur = 0;
            int ppos = 0;
            while(true){
                int last = sts[hld.clus[cur]].get(hld.clusiind[cur]);
                if(last == -1)break;
                int lca = hld.lca(base[last], base[i]);
                // delete from lca to cur
    //            nodes[cur].len, nodes[lca].len;
                int inf = last-nodes[lca].len+1;
                int sup = last-ppos+1;
//                tr("del", last, ppos, nodes[lca].len, inf, sup);
                // _/
                addFenwick(ft0, 0, -(sup-inf));
                addFenwick(ft0, sup+1, +(sup-inf));
                addFenwick(ft0, inf+1, -(inf+1));
                addFenwick(ft0, sup+1, +inf+1);
                addFenwick(ft1, inf+1, 1);
                addFenwick(ft1, sup+1, -1);
//                tr(i, restoreRangeFenwick(ft0, ft1));
                ppos = nodes[lca].len;
                assert dep[base[i]]-dep[lca]-1 >= 0;
                cur = hld.ancestor(base[i], dep[base[i]]-dep[lca]-1);
            }
            // x
            //b a
            //   a
            
            // paint
            int cx = hld.clus[base[i]]; // cluster
            int ind = hld.clusiind[base[i]]; // pos in cluster
            while(true){
                sts[cx].update(0, ind+1, i);
                int con = par[hld.cluspath[cx][0]];
                if(con == -1)break;
                ind = hld.clusiind[con];
                cx = hld.clus[con];
            }
            
//            tr(i, restoreRangeFenwick(ft0, ft1));
            addFenwick(ft0, 0, i+1+1);
            addFenwick(ft0, i+1+1, -(i+1+1));
            addFenwick(ft1, 0, -1);
            addFenwick(ft1, i+1+1, 1);
//            tr(i, restoreRangeFenwick(ft0, ft1));
            
            while(qp < Q && qs[qp][1] <= i){
//                tr(qs[qp]);
                ans[qs[qp][2]] = sumRangeFenwick(ft0, ft1, qs[qp][0]);
                qp++;
            }
        }
        for(long an : ans){
            out.println(an);
        }
    }
    
    public static long sumFenwick(long[] ft, int i)
    {
        long sum = 0;
        for(i++;i > 0;i -= i&-i)sum += ft[i];
        return sum;
    }
    
    public static void addFenwick(long[] ft, int i, long v)
    {
        if(v == 0)return;
        int n = ft.length;
        for(i++;i < n;i += i&-i)ft[i] += v;
    }
    
    public static int firstGEFenwick(long[] ft, long v)
    {
        int i = 0, n = ft.length;
        for(int b = Integer.highestOneBit(n);b != 0;b >>= 1){
            if((i|b) < n && ft[i|b] < v){
                i |= b;
                v -= ft[i];
            }
        }
        return i;
    }
    
    public static long[] restoreFenwick(long[] ft)
    {
        int n = ft.length-1;
        long[] ret = new long[n];
        for(int i = 0;i < n;i++)ret[i] = sumFenwick(ft, i);
        for(int i = n-1;i >= 1;i--)ret[i] -= ret[i-1];
        return ret;
    }
    
    public static int findGFenwick(long[] ft, long v)
    {
        int i = 0;
        int n = ft.length;
        for(int b = Integer.highestOneBit(n);b != 0 && i < n;b >>= 1){
            if(i + b < n){
                int t = i + b;
                if(v >= ft[t]){
                    i = t;
                    v -= ft[t];
                }
            }
        }
        return v != 0 ? -(i+1) : i-1;
    }
    
    public static long[] buildFenwick(long[] a)
    {
        int n = a.length;
        long[] ft = new long[n+1];
        System.arraycopy(a, 0, ft, 1, n);
        for(int k = 2, h = 1;k <= n;k*=2, h*=2){
            for(int i = k;i <= n;i+=k){
                ft[i] += ft[i-h];
            }
        }
        return ft;
    }
    
    public static void addRangeFenwick(long[] ft0, long[] ft1, int i, long v)
    {
        addFenwick(ft1, i+1, -v);
        addFenwick(ft1, 0, v);
        addFenwick(ft0, i+1, v*(i+1));
    }
    
    public static void addRangeFenwick(long[] ft0, long[] ft1, int a, int b, long v)
    {
        if(a <= b){
            addFenwick(ft1, b+1, -v);
            addFenwick(ft0, b+1, v*(b+1));
            addFenwick(ft1, a, v);
            addFenwick(ft0, a, -v*a);
        }
    }
    
    public static long sumRangeFenwick(long[] ft0, long[] ft1, int i)
    {
        return sumFenwick(ft1, i) * (i+1) + sumFenwick(ft0, i);
    }
    
    public static long[] restoreRangeFenwick(long[] ft0, long[] ft1)
    {
        int n = ft0.length-1;
        long[] ret = new long[n];
        for(int i = 0;i < n;i++)ret[i] = sumRangeFenwick(ft0, ft1, i);
        for(int i = n-1;i >= 1;i--)ret[i] -= ret[i-1];
        return ret;
    }

    
    public static class SegmentTreeOverwrite {
        public int M, H, N;
        public int[] cover;
        public int I = Integer.MAX_VALUE;
        
        public SegmentTreeOverwrite(int len)
        {
            N = len;
            M = Integer.highestOneBit(Math.max(N-1, 1))<<2;
            H = M>>>1;
            cover = new int[M];
            Arrays.fill(cover, I);
            for(int i = 0;i < N;i++){
                cover[H+i] = -1;
            }
            for(int i = H-1;i >= 1;i--){
                propagate(i);
            }
        }
        
        private void propagate(int i){}
        
        public void update(int l, int r, int v){ update(l, r, v, 0, H, 1); }
        
        private void update(int l, int r, int v, int cl, int cr, int cur)
        {
            if(l <= cl && cr <= r){
                cover[cur] = v;
                propagate(cur);
            }else{
                int mid = cl+cr>>>1;
                if(cover[cur] != I){ // back-propagate
                    cover[2*cur] = cover[2*cur+1] = cover[cur];
                    cover[cur] = I;
                    propagate(2*cur);
                    propagate(2*cur+1);
                }
                if(cl < r && l < mid){
                    update(l, r, v, cl, mid, 2*cur);
                }
                if(mid < r && l < cr){
                    update(l, r, v, mid, cr, 2*cur+1);
                }
                propagate(cur);
            }
        }
        
        public int get(int x){ 
            int val = I;
            for(int i = H+x;i >= 1;i>>>=1){
                if(cover[i] != I)val = cover[i];
            }
            return val;
        }
    }

    
    public static class HeavyLightDecomposition {
        public int[] clus;
        public int[][] cluspath;
        public int[] clusiind;
        public int[] par, dep;
        
        public HeavyLightDecomposition(int[][] g, int[] par, int[] ord, int[] dep)
        {
            init(g, par, ord, dep);
        }
        
        public void init(int[][] g, int[] par, int[] ord, int[] dep)
        {
            clus = decomposeToHeavyLight(g, par, ord);
            cluspath = clusPaths(clus, ord);
            clusiind = clusIInd(cluspath, g.length);
            this.par = par;
            this.dep = dep;
        }
        
        public static int[] decomposeToHeavyLight(int[][] g, int[] par, int[] ord)
        {
            int n = g.length;
            int[] size = new int[n];
            Arrays.fill(size, 1);
            for(int i = n-1;i > 0;i--)size[par[ord[i]]] += size[ord[i]];
            
            int[] clus = new int[n];
            Arrays.fill(clus, -1);
            int p = 0;
            for(int i = 0;i < n;i++){
                int u = ord[i];
                if(clus[u] == -1)clus[u] = p++;
                // centroid path (not heavy path)
                int argmax = -1;
                for(int v : g[u]){
                    if(par[u] != v && (argmax == -1 || size[v] > size[argmax]))argmax = v;
                }
                if(argmax != -1)clus[argmax] = clus[u];
            }
            return clus;
        }
        
        public static int[][] clusPaths(int[] clus, int[] ord)
        {
            int n = clus.length;
            int[] rp = new int[n];
            int sup = 0;
            for(int i = 0;i < n;i++){
                rp[clus[i]]++;
                sup = Math.max(sup, clus[i]);
            }
            sup++;
            
            int[][] row = new int[sup][];
            for(int i = 0;i < sup;i++)row[i] = new int[rp[i]];
            
            for(int i = n-1;i >= 0;i--){
                row[clus[ord[i]]][--rp[clus[ord[i]]]] = ord[i];
            }
            return row;
        }
        
        public static int[] clusIInd(int[][] clusPath, int n)
        {
            int[] iind = new int[n];
            for(int[] path : clusPath){
                for(int i = 0;i < path.length;i++){
                    iind[path[i]] = i;
                }
            }
            return iind;
        }
        
        public int lca(int x, int y)
        {
            int rx = cluspath[clus[x]][0];
            int ry = cluspath[clus[y]][0];
            while(clus[x] != clus[y]){
                if(dep[rx] > dep[ry]){
                    x = par[rx];
                    rx = cluspath[clus[x]][0];
                }else{
                    y = par[ry];
                    ry = cluspath[clus[y]][0];
                }
            }
            return clusiind[x] > clusiind[y] ? y : x;
        }
        
        public int ancestor(int x, int v)
        {
            while(x != -1){
                if(v <= clusiind[x])return cluspath[clus[x]][clusiind[x]-v];
                v -= clusiind[x]+1;
                x = par[cluspath[clus[x]][0]];
            }
            return x;
        }
    }

    
    public static int lca2(int a, int b, int[][] spar, int[] depth) {
        if (depth[a] < depth[b]) {
            b = ancestor(b, depth[b] - depth[a], spar);
        } else if (depth[a] > depth[b]) {
            a = ancestor(a, depth[a] - depth[b], spar);
        }

        if (a == b)
            return a;
        int sa = a, sb = b;
        for (int low = 0, high = depth[a], t = Integer.highestOneBit(high), k = Integer
                .numberOfTrailingZeros(t); t > 0; t >>>= 1, k--) {
            if ((low ^ high) >= t) {
                if (spar[k][sa] != spar[k][sb]) {
                    low |= t;
                    sa = spar[k][sa];
                    sb = spar[k][sb];
                } else {
                    high = low | t - 1;
                }
            }
        }
        return spar[0][sa];
    }

    protected static int ancestor(int a, int m, int[][] spar) {
        for (int i = 0; m > 0 && a != -1; m >>>= 1, i++) {
            if ((m & 1) == 1)
                a = spar[i][a];
        }
        return a;
    }

    public static int[][] logstepParents(int[] par) {
        int n = par.length;
        int m = Integer.numberOfTrailingZeros(Integer.highestOneBit(n - 1)) + 1;
        int[][] pars = new int[m][n];
        pars[0] = par;
        for (int j = 1; j < m; j++) {
            for (int i = 0; i < n; i++) {
                pars[j][i] = pars[j - 1][i] == -1 ? -1 : pars[j - 1][pars[j - 1][i]];
            }
        }
        return pars;
    }

    
    public static int[][] parents3(int[][] g, int root) {
        int n = g.length;
        int[] par = new int[n];
        Arrays.fill(par, -1);

        int[] depth = new int[n];
        depth[0] = 0;

        int[] q = new int[n];
        q[0] = root;
        for (int p = 0, r = 1; p < r; p++) {
            int cur = q[p];
            for (int nex : g[cur]) {
                if (par[cur] != nex) {
                    q[r++] = nex;
                    par[nex] = cur;
                    depth[nex] = depth[cur] + 1;
                }
            }
        }
        return new int[][] { par, q, depth };
    }

    
    static int[][] packU(int n, int[] from, int[] to) {
        int[][] g = new int[n][];
        int[] p = new int[n];
        for (int f : from)
            p[f]++;
        for (int t : to)
            p[t]++;
        for (int i = 0; i < n; i++)
            g[i] = new int[p[i]];
        for (int i = 0; i < from.length; i++) {
            g[from[i]][--p[from[i]]] = to[i];
            g[to[i]][--p[to[i]]] = from[i];
        }
        return g;
    }
    
    public static class SuffixAutomatonOfBit {
        public Node t0;
        public int len;
        public Node[] nodes;
        public int gen;
        private boolean sortedTopologically = false;
        private boolean lexsorted = false;
        
        private SuffixAutomatonOfBit(int n)
        {
            gen = 0;
            nodes = new Node[2*n];
            this.t0 = makeNode(0, null);
        }
        
        private Node makeNode(int len, Node original)
        {
            Node node = new Node();
            node.id = gen;
            node.original = original;
            node.len = len;
            nodes[gen++] = node;
            return node;
        }
        
        public static class Node
        {
            public int id;
            public int len;
            public char key;
            public Node link;
            private Node[] next = new Node[3];
            public Node original;
            public int np = 0;
            public int hit = 0;
            
            public void putNext(char c, Node to)
            {
                to.key = c;
                if(hit<<~(c-'a')<0){
                    for(int i = 0;i < np;i++){
                        if(next[i].key == c){
                            next[i] = to;
                            return;
                        }
                    }
                }
                hit |= 1<<c-'a';
                if(np == next.length){
                    next = Arrays.copyOf(next, np*2);
                }
                next[np++] = to;
            }
            
            public boolean containsKeyNext(char c)
            {
                return hit<<~(c-'a')<0;
//                for(int i = 0;i < np;i++){
//                    if(next[i].key == c)return true;
//                }
//                return false;
            }
            
            public Node getNext(char c)
            {
                if(hit<<~(c-'a')<0){
                    for(int i = 0;i < np;i++){
                        if(next[i].key == c)return next[i];
                    }
                }
                return null;
            }
            
            public List<String> suffixes(char[] s)
            {
                List<String> list = new ArrayList<String>();
                if(id == 0)return list;
                int first = original != null ? original.len : len;
                for(int i = link.len + 1;i <= len;i++){
                    list.add(new String(s, first - i, i));
                }
                return list;
            }
        }

        public static SuffixAutomatonOfBit build(char[] str)
        {
            int n = str.length;
            SuffixAutomatonOfBit sa = new SuffixAutomatonOfBit(n);
            sa.len = str.length;
            
            Node last = sa.t0;
            for(char c : str){
                last = sa.extend(last, c);
            }
            
            return sa;
        }
        
        public Node extend(Node last, char c)
        {
            Node cur = makeNode(last.len+1, null);
            Node p;
            for(p = last; p != null && !p.containsKeyNext(c);p = p.link){
                p.putNext(c, cur);
            }
            if(p == null){
                cur.link = t0;
            }else{
                Node q = p.getNext(c); // not null
                if(p.len + 1 == q.len){
                    cur.link = q;
                }else{
                    Node clone = makeNode(p.len+1, q);
                    clone.next = Arrays.copyOf(q.next, q.next.length);
                    clone.hit = q.hit;
                    clone.np = q.np;
                    clone.link = q.link;
                    for(;p != null && q.equals(p.getNext(c)); p = p.link){
                        p.putNext(c, clone);
                    }
                    q.link = cur.link = clone;
                }
            }
            return cur;
        }
        
        public SuffixAutomatonOfBit lexSort()
        {
            for(int i = 0;i < gen;i++){
                Node node = nodes[i];
                Arrays.sort(node.next, 0, node.np, new Comparator<Node>() {
                    public int compare(Node a, Node b) {
                        return a.key - b.key;
                    }
                });
            }
            lexsorted = true;
            return this;
        }
        
        public SuffixAutomatonOfBit sortTopologically()
        {
            int[] indeg = new int[gen];
            for(int i = 0;i < gen;i++){
                for(int j = 0;j < nodes[i].np;j++){
                    indeg[nodes[i].next[j].id]++;
                }
            }
            Node[] sorted = new Node[gen];
            sorted[0] = t0;
            int p = 1;
            for(int i = 0;i < gen;i++){
                Node cur = sorted[i];
                for(int j = 0;j < cur.np;j++){
                    if(--indeg[cur.next[j].id] == 0){
                        sorted[p++] = cur.next[j];
                    }
                }
            }
            
            for(int i = 0;i < gen;i++)sorted[i].id = i;
            nodes = sorted;
            sortedTopologically = true;
            return this;
        }
        
        // visualizer
        
        public String toString()
        {
            StringBuilder sb = new StringBuilder();
            for(Node n : nodes){
                if(n != null){
                    sb.append(String.format("{id:%d, len:%d, link:%d, cloned:%b, ",
                            n.id,
                            n.len,
                            n.link != null ? n.link.id : null,
                            n.original.id));
                    sb.append("next:{");
                    for(int i = 0;i < n.np;i++){
                        sb.append(n.next[i].key + ":" + n.next[i].id + ",");
                    }
                    sb.append("}");
                    sb.append("}");
                    sb.append("\n");
                }
            }
            return sb.toString();
        }
        
        public String toGraphviz(boolean next, boolean suffixLink)
        {
            StringBuilder sb = new StringBuilder("http://chart.apis.google.com/chart?cht=gv:dot&chl=");
            sb.append("digraph{");
            for(Node n : nodes){
                if(n != null){
                    if(suffixLink && n.link != null){
                        sb.append(n.id)
                        .append("->")
                        .append(n.link.id)
                        .append("[style=dashed],");
                    }
                    
                    if(next && n.next != null){
                        for(int i = 0;i < n.np;i++){
                            sb.append(n.id)
                            .append("->")
                            .append(n.next[i].id)
                            .append("[label=")
                            .append(n.next[i].key)
                            .append("],");
                        }
                    }
                }
            }
            sb.append("}");
            return sb.toString();
        }
        
        public String label(Node n)
        {
            if(n.original != null){
                return n.id + "C";
            }else{
                return n.id + "";
            }
        }
        
        public String toDot(boolean next, boolean suffixLink)
        {
            StringBuilder sb = new StringBuilder("digraph{\n");
            sb.append("graph[rankdir=LR];\n");
            sb.append("node[shape=circle];\n");
            for(Node n : nodes){
                if(n != null){
                    if(suffixLink && n.link != null){
                        sb.append("\"" + label(n) + "\"")
                        .append("->")
                        .append("\"" + label(n.link) + "\"")
                        .append("[style=dashed];\n");
                    }
                    
                    if(next && n.next != null){
                        for(int i = 0;i < n.np;i++){
                            sb.append("\"" + label(n) + "\"")
                            .append("->")
                            .append("\"" + label(n.next[i]) + "\"")
                            .append("[label=\"")
                            .append(n.next[i].key)
                            .append("\"];\n");
                        }
                    }
                }
            }
            sb.append("}\n");
            return sb.toString();
        }
    }
    
    void run() throws Exception
    {
        is = INPUT.isEmpty() ? System.in : new ByteArrayInputStream(INPUT.getBytes());
        out = new PrintWriter(System.out);
        
        long s = System.currentTimeMillis();
        solve();
        out.flush();
        if(!INPUT.isEmpty())tr(System.currentTimeMillis()-s+"ms");
    }
    
    public static void main(String[] args) throws Exception { 
        new How_Many_Substrings().run(); 
    }
    
    private byte[] inbuf = new byte[1024];
    public int lenbuf = 0, ptrbuf = 0;
    
    private int readByte()
    {
        if(lenbuf == -1)throw new InputMismatchException();
        if(ptrbuf >= lenbuf){
            ptrbuf = 0;
            try { lenbuf = is.read(inbuf); } catch (IOException e) { throw new InputMismatchException(); }
            if(lenbuf <= 0)return -1;
        }
        return inbuf[ptrbuf++];
    }
    
    private boolean isSpaceChar(int c) { return !(c >= 33 && c <= 126); }
    private int skip() { int b; while((b = readByte()) != -1 && isSpaceChar(b)); return b; }
    
    private double nd() { return Double.parseDouble(ns()); }
    private char nc() { return (char)skip(); }
    
    private String ns()
    {
        int b = skip();
        StringBuilder sb = new StringBuilder();
        while(!(isSpaceChar(b))){ // when nextLine, (isSpaceChar(b) && b != ' ')
            sb.appendCodePoint(b);
            b = readByte();
        }
        return sb.toString();
    }
    
    private char[] ns(int n)
    {
        char[] buf = new char[n];
        int b = skip(), p = 0;
        while(p < n && !(isSpaceChar(b))){
            buf[p++] = (char)b;
            b = readByte();
        }
        return n == p ? buf : Arrays.copyOf(buf, p);
    }
    
    private char[][] nm(int n, int m)
    {
        char[][] map = new char[n][];
        for(int i = 0;i < n;i++)map[i] = ns(m);
        return map;
    }
    
    private int[] na(int n)
    {
        int[] a = new int[n];
        for(int i = 0;i < n;i++)a[i] = ni();
        return a;
    }
    
    private int ni()
    {
        int num = 0, b;
        boolean minus = false;
        while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
        if(b == '-'){
            minus = true;
            b = readByte();
        }
        
        while(true){
            if(b >= '0' && b <= '9'){
                num = num * 10 + (b - '0');
            }else{
                return minus ? -num : num;
            }
            b = readByte();
        }
    }
    
    private long nl()
    {
        long num = 0;
        int b;
        boolean minus = false;
        while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
        if(b == '-'){
            minus = true;
            b = readByte();
        }
        
        while(true){
            if(b >= '0' && b <= '9'){
                num = num * 10 + (b - '0');
            }else{
                return minus ? -num : num;
            }
            b = readByte();
        }
    }
    
    private static void tr(Object... o) { System.out.println(Arrays.deepToString(o)); }
}
C# C++ HackerRank Solutions java

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