Count distinct pairs with difference k Python

PROGRAM TO COUNT ALL DISTINCT PAIRS WITH DIFFERENCE EQUAL TO K

def countPairsWithDiffK(arr,n,k):

 

    count =0

     

    # Sort array elements

    arr.sort()

 

    l =0

    r=0

 

    while r<n:

        if arr[r]-arr[l]==k:

            count+=1

            l+=1

            r+=1

             

        # arr[r] - arr[l] < sum

        elif arr[r]-arr[l]>k:

            l+=1

        else:

            r+=1

    return count

 

# Driver code

if __name__=='__main__':

    arr = [1, 5, 3, 4, 2]

    n = len(arr)

    k = 3

    print("Count of pairs with given diff is ",

          countPairsWithDiffK(arr, n, k))







OUTPUT

Count of pairs with given diff is 2

CREDITS: https://www.geeksforgeeks.org/count-pairs-difference-equal-k/