Count of elements such that difference between sum of left and right sub arrays is equal to a multiple of k

Given an array arr[] of length n and an integer k, the task is to find the number of indices from 2 to n-1 in an array having a difference of the sum of the left and right sub arrays equal to the multiple of the given number k.

Examples:

Input: arr[] = {1, 2, 3, 3, 1, 1}, k = 4 
Output: 2 
Explanation: The only possible indices are 4 and 5

Input: arr[] = {1, 2, 3, 4, 5}, k = 1 
Output: 3  

Approach:

• Create a prefix array that contains the sum of the elements on the left and the suffix array which contains the sum of elements on the right. 
• Check for every index the difference of the sum on the left and right and increase the counter if it is divisible by k.

Below is the implementation of the above approach: 

CPP

// C++ code to count of elements such that
// difference between the sum of left and right
// sub-arrays are equal to a multiple of k

#include <bits/stdc++.h>
using namespace std;

// Functions to find the no of elements
int noOfElement(int a[], int n, int k)
{
    // Creating a prefix array
    int prefix[n];

    // Starting element of prefix array
    // will be the first element
    // of given array
    prefix[0] = a[0];
    for (int i = 1; i < n; i++) {
        prefix[i] = prefix[i - 1] + a[i];
    }

    // Creating a suffix array;
    int suffix[n];
    // Last element of suffix array will
    // be the last element of given array
    suffix[n - 1] = a[n - 1];
    for (int i = n - 2; i >= 0; i--) {
        suffix[i] = suffix[i + 1] + a[i];
    }

    // Checking difference of left and right half
    // is divisible by k or not.
    int cnt = 0;
    for (int i = 1; i < n - 1; i++) {
        if ((prefix[i] - suffix[i]) % k == 0
            || (suffix[i] - prefix[i]) % k == 0) {
            cnt = cnt + 1;
        }
    }

    return cnt;
}

// Driver code
int main()
{
    int a[] = { 1, 2, 3, 3, 1, 1 };
    int k = 4;
    int n = sizeof(a) / sizeof(a[0]);
    cout << noOfElement(a, n, k);
    return 0;
}

Java

// Java code to count of elements such that
// difference between the sum of left and right
// sub-arrays are equal to a multiple of k
class GFG
{

// Functions to find the no of elements
static int noOfElement(int a[], int n, int k)
{
    // Creating a prefix array
    int []prefix = new int[n];

    // Starting element of prefix array
    // will be the first element
    // of given array
    prefix[0] = a[0];
    for (int i = 1; i < n; i++)
    {
        prefix[i] = prefix[i - 1] + a[i];
    }

    // Creating a suffix array;
    int []suffix = new int[n];

    // Last element of suffix array will
    // be the last element of given array
    suffix[n - 1] = a[n - 1];
    for (int i = n - 2; i >= 0; i--)
    {
        suffix[i] = suffix[i + 1] + a[i];
    }

    // Checking difference of left and right half
    // is divisible by k or not.
    int cnt = 0;
    for (int i = 1; i < n - 1; i++)
    {
        if ((prefix[i] - suffix[i]) % k == 0
            || (suffix[i] - prefix[i]) % k == 0)
        {
            cnt = cnt + 1;
        }
    }
    return cnt;
}

// Driver code
public static void main(String[] args)
{
    int a[] = { 1, 2, 3, 3, 1, 1 };
    int k = 4;
    int n = a.length;
    System.out.print(noOfElement(a, n, k));
}
}

// This code is contributed by Rajput-Ji

Python

# Python3 code to count of elements such that
# difference between the sum of left and right
# sub-arrays are equal to a multiple of k

# Functions to find the no of elements
def noOfElement(a, n, k):

    # Creating a prefix array
    prefix = [0] * n

    # Starting element of prefix array
    # will be the first element
    # of given array
    prefix[0] = a[0]
    for i in range(1, n):
        prefix[i] = prefix[i - 1] + a[i]

    # Creating a suffix array
    suffix = [0] * n

    # Last element of suffix array will
    # be the last element of given array
    suffix[n - 1] = a[n - 1]
    for i in range(n - 2, -1, -1):
        suffix[i] = suffix[i + 1] + a[i]

    # Checking difference of left and right half
    # is divisible by k or not.
    cnt = 0
    for i in range(1, n - 1):
        if ((prefix[i] - suffix[i]) % k == 0 or (suffix[i] - prefix[i]) % k == 0):
            cnt = cnt + 1

    return cnt

# Driver code

a = [ 1, 2, 3, 3, 1, 1 ]
k = 4
n = len(a)
print(noOfElement(a, n, k))

# This code is contributed by mohit kumar 29

C#

// C# code to count of elements such that
// difference between the sum of left and right
// sub-arrays are equal to a multiple of k
using System;

class GFG
{

    // Functions to find the no of elements
    static int noOfElement(int []a, int n, int k)
    {
        // Creating a prefix array
        int []prefix = new int[n];

        // Starting element of prefix array
        // will be the first element
        // of given array
        prefix[0] = a[0];
        for (int i = 1; i < n; i++)
        {
            prefix[i] = prefix[i - 1] + a[i];
        }

        // Creating a suffix array;
        int []suffix = new int[n];

        // Last element of suffix array will
        // be the last element of given array
        suffix[n - 1] = a[n - 1];
        for (int i = n - 2; i >= 0; i--)
        {
            suffix[i] = suffix[i + 1] + a[i];
        }

        // Checking difference of left and right half
        // is divisible by k or not.
        int cnt = 0;
        for (int i = 1; i < n - 1; i++)
        {
            if ((prefix[i] - suffix[i]) % k == 0
                || (suffix[i] - prefix[i]) % k == 0)
            {
                cnt = cnt + 1;
            }
        }
        return cnt;
    }

    // Driver code
    public static void Main()
    {
        int []a = { 1, 2, 3, 3, 1, 1 };
        int k = 4;
        int n = a.Length;
        Console.Write(noOfElement(a, n, k));
    }
}

// This code is contributed by AnkitRai01

Javascript

<script>
// Javascript code to count of elements such that
// difference between the sum of left and right
// sub-arrays are equal to a multiple of k

// Functions to find the no of elements
function noOfElement(a,n,k)
{
    // Creating a prefix array
    let prefix = new Array(n);

    // Starting element of prefix array
    // will be the first element
    // of given array
    prefix[0] = a[0];
    for (let i = 1; i < n; i++)
    {
        prefix[i] = prefix[i - 1] + a[i];
    }

    // Creating a suffix array;
    let suffix = new Array(n);

    // Last element of suffix array will
    // be the last element of given array
    suffix[n - 1] = a[n - 1];
    for (let i = n - 2; i >= 0; i--)
    {
        suffix[i] = suffix[i + 1] + a[i];
    }

    // Checking difference of left and right half
    // is divisible by k or not.
    let cnt = 0;
    for (let i = 1; i < n - 1; i++)
    {
        if ((prefix[i] - suffix[i]) % k == 0
            || (suffix[i] - prefix[i]) % k == 0)
        {
            cnt = cnt + 1;
        }
    }
    return cnt;
}

    // Driver code
    let a=[1, 2, 3, 3, 1, 1];
    let k = 4;
    let n = a.length;
    document.write(noOfElement(a, n, k));



// This code is contributed by patel2127
</script>

Output:

2

Time Complexity: O(n), where n is the size of the array
Auxiliary Space: O(n), as extra space of size n is used to create prefix and suffix array