HackerRank Picking Numbers Problem Solution

In this post, We are going to solve HackerRank Picking Numbers Problem. Given an array of integers, find the longest subarray where the absolute difference between any two elements is less than or equal to 1.

Example

Example

a = [1,1,2,2,4,4,5,5,5]

There are two subarrays meeting the criterion: [1, 1, 2, 2] and [4, 4, 5, 5, 5]. The maximum length subarray has 5 elements.

Function Description

Complete the pickingNumbers function in the editor below.

pickingNumbers has the following parameter(s):

int a[n]: an array of integers

Returns

int: the length of the longest subarray that meets the criterion

Input Format

The first line contains a single integer n. the size of the array a. The second line contains n space-separated integers, each an a[i].

Constraints

2≤ n ≤ 100

Sample Input 0

6
4 6 5 3 3 1

Sample Output 0

Explanation 0

We choose the following multiset of integers from the array: (4, 3, 3). Each pair in the multiset has an absolute difference≤ 1 (e. [4-3] and [3-3]. so we print the number of chosen integers, 3, as our answer.

Sample Input 1

6
1 2 2 3 1 2

Sample Output 1

Explanation 1

We choose the following multiset of integers from the array (1, 2, 2, 1, 2). Each pair in the multiset has an absolute difference≤ 1 (ie.. [1-2]. and [2-2] =0). so we print the number of chosen integers. 5, as our answer.

Table of Contents

Picking Numbers C Solution

#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <limits.h>
#include <stdbool.h>
int a[101];
void bubble(int n)
    {
    int swap=0,i,j,k;
    for(i=1;i<n;i++){
        if(a[i]>a[i+1]){
            j=a[i+1];
            a[i+1]=a[i];
            a[i]=j;
            swap++;
        }
    }
    if(swap!=0) 
        bubble(n-1);  
}
int main(){
    int n,i,j,k,l,t,p[101];
    scanf("%d",&n);
    for(i = 1; i <= n; i++){
       scanf("%d",&a[i]);
        p[i]=0;}
    l=1;
    t=1;
    bubble(n);
    for(i=1;i<=n;i++){
        l=1;
        for(j=i+1;j<=n;j++){
            if(a[j]-a[i]==0)
                {
                i++;
                l++;
            }
            else if(a[j]-a[i]==1){
               l++; 
            }
        }
        if(t<l)
            t=l;
    }
    
    printf("%d\n",t);
    return 0;
}

Picking Numbers C++ Solution

#include <bits/stdc++.h>
using namespace std;
#define pii             pair<int , int >
#define inf             1111111111
#define in(a)           scanf("%d", &a)
#define ins(a)          scanf("%s", a)
#define in2(a, b)       scanf("%d%d", &a, &b)
#define in3(a, b, c)    scanf("%d%d%d", &a, &b, &c)
#define mp              make_pair
#define vi              vector<int >
#define _ceil(n, a)     ((n)%(a)==0?((n)/(a)):((n)/(a)+1))
#define cl              clear()
#define sz              size()
#define pn              printf("\n")
#define pr(a)           printf("%d\n", a)
#define prs(a)          printf("%d ", a)
#define pr2(a, b)       printf("%d %d\n", a, b)
#define pr3(a, b, c)    printf("%d %d %d\n", a, b, c)
#define pb              push_back
#define mem(a, b)       memset((a), (b), sizeof(a))
#define all(X)          (X).begin(), (X).end ()
#define iter(it, X)     for (__typeof((X).begin()) it = (X).begin(); it != (X).end(); it++)
#define ext(a)          {printf("%s\n", a); return 0;}
#define oka(x, y)       ((x)>=0&&(x)<row&&(y)>=0&&(y)<col)
#define x               first
#define y               second
#define lc              (2*i)
#define rc              (2*i+1)
#define sst             clock_t t = clock()
#define eed             printf ("It took me %d clicks.\n", (int)(clock() - t))

#define prr(args...) { vector<string> _v = split(#args, ','); err(_v.begin(), args); pn;}
vector<string> split(const string& s, char c) {
	vector<string> v;
	stringstream ss(s);
	string x;
	while (getline(ss, x, c))
		v.emplace_back(x);
	return move(v);
}
void err(vector<string>::iterator it) {}
template<typename T, typename... Args>
void err(vector<string>::iterator it, T a, Args... args) {
	cerr <<a<<" ";
	err(++it, args...);
}

typedef long long LL;
//int  dx[]={1,0,-1,0};int dy[]={0,1,0,-1}; //4 Direction
//int  dx[]={1,1,0,-1,-1,-1,0,1,0};int dy[]={0,1,1,1,0,-1,-1,-1,0};//8 direction
//int  dx[]={2,1,-1,-2,-2,-1,1,2};int dy[]={1,2,2,1,-1,-2,-2,-1};//Knight Direction
//bool check(int n, int pos) {return (n & (1<<pos))>>pos;} //typecast 1 in case of int
//int  on(int n, int pos) {return n | (1<<pos);} //typecast 1 in case of int
//int  off(int n, int pos) {return n & ~(1<<pos);} //typecast 1 in case of int
//bool operator < (const data &d) const{return cost<d.cost;} //reverse in priority queue

const int M = 300, mod = 1000000007;
int A[M];
int main()
{


    int n, i, j, k;

    in(n);
    for (i=0; i<n; i++) in(A[i]);

    int maxx = 0;

    sort(A, A+n);
    for (i=0; i<n; i++)
    {
        for (j=i+1; j<n; j++)
        {
            if (A[j] - A[i] <= 1) maxx = max(maxx, j-i+1);
        }
    }

    pr(maxx);

return 0;
}

Picking Numbers C Sharp Solution

using System;
using System.Collections.Generic;
using System.Linq;

class Solution {
    static void Main(string[] args) {
        int n = Convert.ToInt32(Console.ReadLine());
        int[] a = Array.ConvertAll(Console.ReadLine().Split(' '), int.Parse);
        int[] counts = new int[100];

        foreach (int x in a) {
            counts[x - 1]++;
        }

        int maxCount = 0;
        for (int i = 0; i < 99; i++) {
            maxCount = Math.Max(maxCount, counts[i] + counts[i + 1]);
        }

        Console.WriteLine(maxCount);
    }
}

Picking Numbers java Solution

import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int[] a = new int[n];
        for(int a_i=0; a_i < n; a_i++){
            a[a_i] = in.nextInt();
        }
        Arrays.sort(a);
        int max = 0, co;
        int i = 0;
        while(i<n-1)
            {
            int j = i+1;
            co = 1;
            while(j<n && a[j]-a[i] <= 1)
                {
                j++;
                co++;
            }
                if(co > max)
                    max = co;
            i++;
        }
        System.out.println(max);
    }
}

Picking Numbers JavaScript Solution

process.stdin.resume();
process.stdin.setEncoding('ascii');

var input_stdin = "";
var input_stdin_array = "";
var input_currentline = 0;

process.stdin.on('data', function (data) {
    input_stdin += data;
});

process.stdin.on('end', function () {
    input_stdin_array = input_stdin.split("\n");
    main();    
});

function readLine() {
    return input_stdin_array[input_currentline++];
}

/////////////// ignore above this line ////////////////////

function main() {
    var n = parseInt(readLine());
    a = readLine().split(' ');
    a = a.map(Number);
    
    
    var sorted = a.sort(function(a, b) {
        return a - b;
    });
    
    var i = 0;
    var j = 1;
    
    while(i < sorted.length && j < sorted.length) {
        if(Math.abs(sorted[i] - sorted[j]) > 1) {
            i++;
            j++;
        } else {
            j++;
        }
    }
    console.log(Math.abs(i - j));
    /*  
        console.log(selectedNumbers);
        console.log(selectedNumbers.length);
    */    
}

Picking Numbers Python Solution

#!/bin/python3

import sys


n = int(input().strip())
a = [int(a_temp) for a_temp in input().strip().split(' ')]

max = 0
a = sorted(a)
#print(a)
for i in range(n-1):
    nbr = 1
    for j in range(i+1, n):
        if a[j]-a[i]<=1:
            nbr += 1
            #print("i={0}, j={1}, a[{0}]={2}, a[{1}]={3}, diff = {4}, nbr={5}".format(i,j,a[i],a[j],a[j]-a[i],nbr))
            
            if nbr > max:
                max = nbr
        else:
            if nbr > max:
                max = nbr
            break

print(max)

Other Solution