HackerRank Cut the Tree Problem Solution
In this post, we will solve HackerRank Cut the Tree Problem Solution.
There is an undirected tree where each vertex is numbered from 1 ton, and each contains a data value. The sum of a tree is the sum of all its nodes’ data values. If an edge is cut, two smaller trees are formed. The difference between two trees is the absolute value of the difference in their sums.
Given a tree, determine which edge to cut so that the resulting trees have a minimal difference between them, then return that difference.
Example
data = [1, 2, 3, 4, 5, 6]
edges = [(1, 2), (1, 3), (2, 6), (3, 4), (3, 5)]
The values are calculated as follows:
Edge Tree 1 Tree 2 Absolute
Cut Sum Sum Difference
1 8 13 5
2 9 12 3
3 6 15 9
4 4 17 13
5 5 16 11
The minimum absolute difference is 3.
Note: The given tree is always rooted at vertex 1.
Function Description
Complete the cutTheTree function in the editor below.
cutTheTree has the following parameter(s):
- int data[n]: an array of integers that represent node values
- int edges[n-1][2]: an 2 dimensional array of integer pairs where each pair represents nodes connected by the edge
Returns
- int: the minimum achievable absolute difference of tree sums
Input Format
The first line contains an integer n, the number of vertices in the tree.
The second line contains n space-separated integers, where each integer u denotes the
node[u] data value, data[u].
Each of the n – 1 subsequent lines contains two space-separated integers u and that υ
describe edge u ↔ v in tree t.
Sample Input
STDIN Function
----- --------
6 data[] size n = 6
100 200 100 500 100 600 data = [100, 200, 100, 500, 100, 600]
1 2 edges = [[1, 2], [2, 3], [2, 5], [4, 5], [5, 6]]
2 3
2 5
4 5
5 6
Sample Output
400
Explanation
There are n − 1 = 5 edges we can cut:
- Edge 1 → 2 results in d1↔2 = 1500 – 100 = 1400
- Edge 2 → 3 results in d2↔3 = 1500 – 100 = 1400
- Edge 2 → 5 results in d2↔5 = 1200 – 400 = 800
- Edge 4 → 5 results in d4↔5=1100 – 500 = 600
- Edge 5 → 6 results in d5↔6 = 1000 – 600 = 400
The minimum difference is 400.
Cut the Tree C Solution
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
int main() {
//idea: accumulate costs bottom up from leaves to root
//the costs are c1 for one subtree, and totalcost-c1 for the other
long n;
long long vvalue[100000]={0}; //stores the value of each vertex
scanf("%ld",&n);
for(long i=0;i<n;i++){scanf("%lld",&vvalue[i]);}
//needs a good data structure to implement the tree
long treeedges[n-1][2];
for (long l=0;l<n-1;l++){
long i,j;
scanf("%ld %ld",&i,&j);
if(i>j) {treeedges[l][0]=j-1;treeedges[l][1]=i-1;}
else {treeedges[l][0]=i-1;treeedges[l][1]=j-1;}
}
//identify number of neighbors for each node
long ineighbors[100001]={0},in[100001]={0},nneighbors[100000]={0};
for (long l=0;l<n-1;l++){
nneighbors[treeedges[l][0]]++;nneighbors[treeedges[l][1]]++;
}
//for (long l=0;l<n;l++){printf("%ld ",nneighbors[l]); printf("\n");}
//use counting array to sort neighbors
for (long l=0;l<n+1;l++){
ineighbors[l+1]=ineighbors[l]+nneighbors[l];
in[l+1]=ineighbors[l+1];
}
//for (long l=0;l<n+1;l++){printf("%ld ",ineighbors[l]);}printf("\n");
long neighbortotal=ineighbors[n];
long allneighbors[neighbortotal];
for (long l=0;l<n-1;l++){
allneighbors[in[treeedges[l][0]]++]=treeedges[l][1];
allneighbors[in[treeedges[l][1]]++]=treeedges[l][0];
}
//for(long l=0;l<neighbortotal;l++){printf("%ld ",allneighbors[l]);}printf("\n");
//fill queue with leaves
long queue[100000], left=0,right=0;
for (long i=0;i<n;i++){
if(nneighbors[i]==1) queue[right++]=i;
}
//for (long i=left;i<right;i++){printf("%ld ",queue[i]);} printf("\n");
//process queue:
long root;
while(left<right){
//for (long i=0;i<n+1;i++){printf("%ld ",ineighbors[i]);} printf("\n");
//for (long i=left;i<right;i++){printf("%ld ",queue[i]);} printf("\n");
long thisv=queue[left++];
long neighbor=-1;
if(nneighbors[thisv]>0){
neighbor=allneighbors[ineighbors[thisv]++];
//remove duplicate edge in other direction
//printf("removing %ld %ld\n",neighbor, thisv);
long ptr=ineighbors[neighbor];
while(allneighbors[ptr]!=thisv) ptr++;
if(ptr>ineighbors[neighbor]){
long swap=allneighbors[ineighbors[neighbor]];
allneighbors[ineighbors[neighbor]]=allneighbors[ptr];
allneighbors[ptr]=swap;
}
ineighbors[neighbor]++;
nneighbors[thisv]--; //remove edge from thisv
nneighbors[neighbor]--;
vvalue[neighbor]+=vvalue[thisv]; //add value of thisv
if(nneighbors[neighbor]==1) queue[right++]=neighbor; //add to queue if now a leaf
}
else{
//found root as last node in this scheme
root=thisv;
}
}
//for (long i=0;i<n;i++){printf("%ld ",nneighbors[i]);} printf("\n");
//for(long i=0;i<n;i++){printf("%lld ",vvalue[i]);}printf("\n");
long long totalcost=vvalue[root];
//now, for each v have computed accumulated value of subtree rooted at v
long long mintreediff=totalcost;
for (long l=0;l<n-1;l++){
long i=treeedges[l][0], j=treeedges[l][1];
long long cost1=(vvalue[i]<vvalue[j])?vvalue[i]:vvalue[j];
long long cost2=totalcost-cost1;
long long treediff=abs(cost2-cost1);
//printf("edge %ld %ld cost %lld\n",i,j,treediff);
if(treediff<mintreediff) mintreediff=treediff;
}
printf("%lld",mintreediff);
return 0;
}
Cut the Tree C++ Solution
#include <cmath>
#include <cstdio>
#include <vector>
#include <map>
#include <iostream>
#include <algorithm>
using namespace std;
class node {
public:
int v;
int no;
multimap<int,node*> e;
bool visit;
int n;
node(int n,int no);
};
node::node (int n,int nn){
v=n;
no=nn;
visit=false;
}
long mind=999999999999;
int dfs(node *n,int tt){
int s=0,x=0,lmin;
multimap<int,node*> newe;
n->visit=true;
for(auto i:n->e){
if(!i.second->visit){
x = dfs(i.second,tt);
lmin=abs(2*x-tt);
if(lmin<mind){
mind=lmin;
//cout<<n->no<<" "<<mind<<endl;
}
s+=x;
newe.insert(make_pair(x,i.second));
}
}
n->e.clear();
n->e=newe;
return s+n->v;
}
int main() {
/* Enter your code here. Read input from STDIN. Print output to STDOUT */
int n,x,y,ss=0;
cin>>n;
node *a[n+1];
for(int i=1;i<=n;i++){
cin>>x;
a[i] = new node(x,i);
ss+=x;
}
for(int i=1;i<=n-1;i++){
cin>>x>>y;
a[x]->e.insert(make_pair(9999999999,a[y]));
a[y]->e.insert(make_pair(9999999999,a[x]));
}
int s = dfs(a[1],ss);
cout<<mind;
return 0;
}
Cut the Tree C Sharp Solution
using System;
using System.Collections.Generic;
using System.IO;
class Node {
public int Value;
public int SumValue;
public int ReachedTimes;
public List<Node> Sons = new List<Node>();
}
class Solution {
static int ReadInt() {
while (true) {
if (__line == null) __line = System.Console.ReadLine().Split(new[] {' '}, StringSplitOptions.RemoveEmptyEntries);
if (__line.Length > __index) {
__index++;
return Int32.Parse(__line[__index-1]);
} else { __line = null; __index = 0; continue; }
}
}
static string[] __line;
static int __index;
static int leastdiff = int.MaxValue;
static int everything;
static int[] values;
static List<Node> tree = new List<Node>();
static void dfs(int type, Node node, Node parent) {
int sum = 0;
if (type == 2 && parent != null) {
// What is my parent?
int parentis = everything - node.SumValue;
// What am I?
int iam = node.SumValue;
int diff = Math.Abs( parentis - iam );
if (diff < leastdiff) {
leastdiff = diff;
}
}
foreach(Node son in node.Sons) {
if (son != parent) {
dfs(type, son, node);
if (type == 1) {
sum += son.SumValue;
}
}
}
if (type == 1) {
node.SumValue = sum + node.Value;
}
}
static void Main(String[] args) {
int n = ReadInt();
values = new int[n];
for (int i =0;i<n;i++) {
values[i] = ReadInt();
Node node = new Node();
everything += values[i];
node.Value = values[i];
tree.Add(node);
}
// Read links
for (int i =0;i<n-1;i++) {
int from = ReadInt();
int to = ReadInt();
tree[from-1].Sons.Add(tree[to-1]);
tree[to-1].Sons.Add(tree[from-1]);
}
// DFS 1
dfs(1, tree[0], null);
// DFS 2
dfs(2, tree[0], null);
Console.WriteLine(leastdiff);
}
}
Cut the Tree Java Solution
import java.io.*;
import java.math.*;
import java.security.*;
import java.text.*;
import java.util.*;
import java.util.concurrent.*;
import java.util.function.*;
import java.util.regex.*;
import java.util.stream.*;
import static java.util.stream.Collectors.joining;
import static java.util.stream.Collectors.toList;
class Result {
/*
* Complete the 'cutTheTree' function below.
*
* The function is expected to return an INTEGER.
* The function accepts following parameters:
* 1. INTEGER_ARRAY data
* 2. 2D_INTEGER_ARRAY edges
*/
public static int cutTheTree(List<Integer> data, List<List<Integer>> edges) {
// Approach:
//
// (1) Build a graph with nodes containing value and sum subTree
// => Root will contain whole tree sum.
// 1:12
// / | \
// / | \
// 1:1 1:9 1:1
// / | \
// / | \
// 1:1 6:6 1:1
//
// (2) Iterate over the graph and figure out min dif using child sums.
final int n = data.size();
List<Integer>[] children = new List[n];
for (int i = 0; i < n; i++) children[i] = new ArrayList<>();
int i1, i2;
for (List<Integer> edge : edges) {
i1 = edge.get(0) - 1;
i2 = edge.get(1) - 1;
children[i1].add(i2);
children[i2].add(i1);
}
boolean[] visited = new boolean[n];
visited[0] = true;
Node root = new Node(0, data.get(0));
root.subTreeSum = addChildren(root, children, data, visited);
return minDif(root, root.subTreeSum, Integer.MAX_VALUE);
}
static int addChildren(
Node node,
List<Integer>[] children,
List<Integer> data,
boolean[] visited
) {
int subTreeSum = 0;
for (int i : children[node.id]) {
if (visited[i]) continue;
visited[i] = true;
Node child = new Node(i, data.get(i));
child.subTreeSum = addChildren(child, children, data, visited);
if (node.children == null) node.children = new ArrayList<>();
node.children.add(child);
subTreeSum += child.subTreeSum;
}
return node.val + subTreeSum;
}
static int minDif(Node node, int treeSum, int curMinDif) {
int dif = treeSum - 2*node.subTreeSum;
if (dif == 0) return 0; // edge case
if (dif < 0) dif *= -1;
if (curMinDif > dif) curMinDif = dif;
if (node.children != null)
for (Node child : node.children)
curMinDif = minDif(child, treeSum, curMinDif);
return curMinDif;
}
static final class Node {
int id;
int val;
int subTreeSum;
List<Node> children;
Node(int id, int val) {
this.id = id;
this.val = val;
this.subTreeSum = 0;
}
}
}
public class Solution {
public static void main(String[] args) throws IOException {
BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(System.in));
BufferedWriter bufferedWriter = new BufferedWriter(new FileWriter(System.getenv("OUTPUT_PATH")));
int n = Integer.parseInt(bufferedReader.readLine().trim());
List<Integer> data = Stream.of(bufferedReader.readLine().replaceAll("\\s+$", "").split(" "))
.map(Integer::parseInt)
.collect(toList());
List<List<Integer>> edges = new ArrayList<>();
IntStream.range(0, n - 1).forEach(i -> {
try {
edges.add(
Stream.of(bufferedReader.readLine().replaceAll("\\s+$", "").split(" "))
.map(Integer::parseInt)
.collect(toList())
);
} catch (IOException ex) {
throw new RuntimeException(ex);
}
});
int result = Result.cutTheTree(data, edges);
bufferedWriter.write(String.valueOf(result));
bufferedWriter.newLine();
bufferedReader.close();
bufferedWriter.close();
}
}
Cut the Tree JavaScript Solution
function processData(input) {
var lines = input.split('\n');
var n = lines[0];
var nodes = lines[1].split(' ').map(function(x){
return {
edges:[],
data:parseFloat(x),
sum:parseFloat(x),
children:0,
contributed:0
};
});
for (var i = 2; i < lines.length; ++i) {
var ab = lines[i].split(' ').map(function(x){return parseInt(x)-1;})
nodes[ab[0]].edges.push(ab[1]);
nodes[ab[1]].edges.push(ab[0]);
}
var queue = [0];
while (queue.length) {
var ptr = queue[queue.length-1];
var node = nodes[ptr];
if (!node.visited) {
for (var i = 0; i < node.edges.length; ++i) {
var childPtr = node.edges[i];
var child = nodes[childPtr];
if (child.visited)
continue;
node.children++;
child.parent = ptr;
queue.push(childPtr);
}
}
node.visited = true;
if (node.contributed == node.children) {
queue.pop();
if (node.hasOwnProperty('parent')) {
var parent = nodes[node.parent];
parent.sum += node.sum;
parent.contributed++;
}
}
}
var best = nodes[0].sum;
for (var i = 0; i < nodes.length; ++i) {
best = Math.min(best, Math.abs(nodes[0].sum-2*nodes[i].sum));
}
console.log(best);
}
process.stdin.resume();
process.stdin.setEncoding("ascii");
_input = "";
process.stdin.on("data", function (input) {
_input += input;
});
process.stdin.on("end", function () {
processData(_input);
});
Cut the Tree Python Solution
class Tree(object):
def __init__(self, value):
self.value = value
self.adj = []
self.parentTotal = 0
self.total = value
self.parent = -1
self.auxIndex = 0
self.sumMax = -1
def calculateTotalI(nodes, root):
stack = [int]*len(nodes)
stack[0] = root
stackIndex = 0
while stackIndex >= 0:
currentIndex = stack[stackIndex]
current = nodes[currentIndex]
if current.auxIndex >= len(current.adj):
stackIndex -= 1
nodes[current.parent].total += current.total
else:
while current.auxIndex < len(current.adj):
if current.adj[current.auxIndex] != current.parent:
stackIndex += 1
stack[stackIndex] = current.adj[current.auxIndex]
current.auxIndex += 1
nodes[stack[stackIndex]].parent = currentIndex
break
else:
current.auxIndex += 1
def calculateParentTotalI(nodes, root):
queue = [int]*len(nodes)
queue[0] = root
qIndex = 0
while qIndex >= 0:
current = nodes[queue[qIndex]]
qIndex -= 1
for n in current.adj:
if n != current.parent:
nodes[n].parentTotal = current.parentTotal + current.total - nodes[n].total
qIndex += 1
queue[qIndex] = n
def minCutI(nodes, root, min):
queue = [int]*len(nodes)
queue[0] = root
qIndex = 0
currentMin = min
while qIndex >= 0:
current = nodes[queue[qIndex]]
qIndex -= 1
for n in current.adj:
if n != current.parent:
qIndex += 1
queue[qIndex] = n
total1 = nodes[n].parentTotal
total2 = nodes[n].total
total = abs(total1 - total2)
if total < currentMin:
currentMin = total
return currentMin
if __name__=="__main__":
N = int(input())
values = input().split()
values = [ int(x) for x in values ]
nodes = [None]*N
root = -1
sumMax = sum(values)
for x in range(N):
nodes[x] = Tree(values[x])
nodes[x].min = sumMax
for z in range(N-1):
i, j = input().split()
i, j = int(i)-1, int(j)-1
if root == -1:
root = i
nodes[i].adj.append(j)
nodes[j].adj.append(i)
calculateTotalI(nodes,root)
calculateParentTotalI(nodes,root)
print(minCutI(nodes, root, sum(values)))