mghamsar

def countPairsWithDiffK(arr, n, k):
    count = 0
    for i in range(0,n):
        for j in range(i,n):
            if (arr[i] - arr[j] == k or arr[j] - arr[i] == k ):
                count = count+1
    
    return count

mghamsar

int countPairsWithDiffK(int arr[], int n, int k)
{
    int count = 0;
     
    // Pick all elements one by one
    for (int i = 0; i < n; i++)
    {       
        // See if there is a pair of this picked element
        for (int j = i+1; j < n; j++)
            if (arr[i] - arr[j] == k || arr[j] - arr[i] == k )