HackerRank Array Manipulation problem solution

In this HackerRank Array Manipulation Interview preparation kit problem solution we have a Starting with a 1-indexed array of zeros and a list of operations, for each operation add a value to each array element between two given indices, inclusive. Once all operations have been performed, return the maximum value in the array.

Problem solution in Python programming.

#!/usr/bin/env python

import itertools, sys


if __name__ == '__main__':
    N, M = list(map(int, sys.stdin.readline().split()))
    x = [0] * N

    for _ in range(M):
        a, b, k = list(map(int, sys.stdin.readline().split()))

        x[a - 1] += k
        if b < N:
            x[b] -= k

    print(max(itertools.accumulate(x)))

Problem solution in Java Programming.

import java.io.*;
import java.util.*;

public class Solution {

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        long size = scanner.nextLong();

        Map<Long, Long> map = new HashMap<>();
        long operations = scanner.nextLong();

        for (long i = 0; i < operations; i++) {
            long start = scanner.nextLong();
            long end = scanner.nextLong();
            long value = scanner.nextLong();

            map.put(start, (map.containsKey(start) ? map.get(start) : 0) + value);
            map.put(end + 1, (map.containsKey(end + 1) ? map.get(end + 1) : 0) - value);
        }

        long max = 0;
        long value = 0;
        for (long i = 0; i < size; i++) {
            value += (map.containsKey(i + 1) ? map.get(i + 1) : 0);
            max = Math.max(max, value);
        }

        System.out.println(max);
    }
}

Problem solution in C++ programming.

#include <iostream>

using namespace std;
const int NMAX = 1e7+2;
long long a[NMAX];
int main()
{
    int n, m;
    cin >> n >> m;
    for(int i=1;i<=m;++i){
        int x, y, k;
        cin >> x >> y >> k;
        a[x] += k;
        a[y+1] -= k;
    }
    long long x = 0,sol=-(1LL<<60);
    for(int i=1;i<=n;++i){
        x += a[i];
        sol = max(sol,x);
    }
    cout<<sol<<"n";
    return 0;
}

Problem solution in C programming.

#include<stdio.h>
long A[10000009]={0},CF[10000009+1]={0};
int main()
{
 
  long N,Q;
  
    
    long val,left,right,i,count=0;
   long maxv=-1;
    scanf("%ld%ld",&N,&Q);
  for(i=0;i<Q;i++)
  {
   scanf("%ld%ld%ld",&left,&right,&val);
   CF[left-1]+=val;
   CF[right]-=val;
  }
    
  for(i=0;i<N;i++)
  {
   count+=CF[i];
   A[i]=count;
      if(count>maxv) maxv=count;
  }
       printf("%ldn",maxv);
 return 0;
}

Problem solution in JavaScript programming.

function processData(input) {
  var splitInput = input.split("n");
  var listSize = parseInt(splitInput[0].split(" ")[0]);
  var numInserts = parseInt(splitInput[0].split(" ")[1]);
  var max = 0;
  var amounts = Array(listSize);

  for (var i = 0; i < numInserts; i++) {
    // input is 1 based
    var start = parseInt(splitInput[i+1].split(" ")[0]) - 1;
    var end = parseInt(splitInput[i+1].split(" ")[1]);
    var amount = parseInt(splitInput[i+1].split(" ")[2]);

    amounts[start] = amounts[start] || 0;
    amounts[start] = amounts[start] + amount;

    amounts[end] = amounts[end] || 0;
    amounts[end] = amounts[end] - amount;
  }
  
  var current = 0;
  for (var i = 0; i < listSize; i++) {
    current += (amounts[i] || 0);
    if (current > max) {
      max = current;
    }
  }

  console.log(max);
}

process.stdin.resume();
process.stdin.setEncoding("ascii");
_input = "";
process.stdin.on("data", function (input) {
    _input += input;
});

process.stdin.on("end", function () {
   processData(_input);
});