HackerRank Array Manipulation solution in C/C++

 Array Manipulation


 

Starting with a 1-indexed array of zeros and a list of operations, for each operation add a value to each the array element between two given indices, inclusive. Once all operations have been performed, return the maximum value in the array.

Example

Queries are interpreted as follows:

    a b k
    1 5 3
    4 8 7
    6 9 1

Add the values of  between the indices  and  inclusive:

index->	 1 2 3  4  5 6 7 8 9 10
	[0,0,0, 0, 0,0,0,0,0, 0]
	[3,3,3, 3, 3,0,0,0,0, 0]
	[3,3,3,10,10,7,7,7,0, 0]
	[3,3,3,10,10,8,8,8,1, 0]

The largest value is  after all operations are performed.

Function Description

Complete the function arrayManipulation in the editor below.

arrayManipulation has the following parameters:

  • int n - the number of elements in the array
  • int queries[q][3] - a two dimensional array of queries where each queries[i] contains three integers, a, b, and k.

Returns

  • int - the maximum value in the resultant array

Input Format

The first line contains two space-separated integers  and , the size of the array and the number of operations.
Each of the next  lines contains three space-separated integers ,  and , the left index, right index and summand.

Constraints

Sample Input

5 3
1 2 100
2 5 100
3 4 100

Sample Output

200

Explanation

After the first update the list is 100 100 0 0 0.
After the second update list is 100 200 100 100 100.
After the third update list is 100 200 200 200 100.

The maximum value is .

Solution: C/C++ code

#include<stdio.h>

#include<stdlib.h>

int main()

{

    long long n,m,i,max=0,x=0;

    scanf("%lld %lld",&n,&m);

    int *arr=(int *)malloc(sizeof(int)* n+1);

    for(i=0;i<n;i++)

    {

        arr[i] =0;

    }

    for(i=1;i<=m;i++)

    {

        long long a,b,k;

        scanf("%lld%lld%lld",&a,&b,&k);

        arr[a]+=k;

        if(b+1<=n)

        {

            arr[b+1]-=k;

        }

    }

    for(i=1;i<=n;i++)

    {

        x += arr[i];

        if(max<x)

        {

            max=x;

        }

    }

    printf("%lld",max);

    return 0;

}


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