Find sum of modulo K of first N natural number

Given two integer N ans K, the task is to find sum of modulo K of first N natural numbers i.e 1%K + 2%K + ..... + N%K.

Examples :

Input : N = 10 and K = 2.
Output : 5
Sum = 1%2 + 2%2 + 3%2 + 4%2 + 5%2 + 6%2 +
      7%2 + 8%2 + 9%2 + 10%2
   = 1 + 0 + 1 + 0 + 1 + 0 + 1 + 0 + 1 + 0
   = 5.

Method 1:

Iterate a variable i from 1 to N, evaluate and add i%K.

Below is the implementation of this approach:

C++

// C++ program to find sum of
// modulo K of first N natural numbers.
#include <bits/stdc++.h>
using namespace std;

// Return sum of modulo K of
// first N natural numbers.
int findSum(int N, int K)
{
    int ans = 0;

    // Iterate from 1 to N &&
    // evaluating and adding i % K.
    for (int i = 1; i <= N; i++)
        ans += (i % K);

    return ans;
}

// Driver Program
int main()
{
    int N = 10, K = 2;
    cout << findSum(N, K) << endl;
    return 0;
}

Java

// Java program to find sum of modulo
// K of first N natural numbers.
import java.io.*;

class GFG {

    // Return sum of modulo K of
    // first N natural numbers.
    static int findSum(int N, int K)
    {
        int ans = 0;

        // Iterate from 1 to N && evaluating
        // and adding i % K.
        for (int i = 1; i <= N; i++)
            ans += (i % K);

        return ans;
    }

    // Driver program
    static public void main(String[] args)
    {
        int N = 10, K = 2;
        System.out.println(findSum(N, K));
    }
}

// This code is contributed by vt_m.

Python3

# Python3 program to find sum 
# of modulo K of first N 
# natural numbers.

# Return sum of modulo K of 
# first N natural numbers.

def findSum(N, K):
    ans = 0;

    # Iterate from 1 to N &&
    # evaluating and adding i % K.
    for i in range(1, N + 1):
        ans += (i % K);

    return ans;

# Driver Code
N = 10; 
K = 2;
print(findSum(N, K));

# This code is contributed by mits

// C# program to find sum of modulo
// K of first N natural numbers.
using System;

class GFG {

    // Return sum of modulo K of
    // first N natural numbers.
    static int findSum(int N, int K)
    {
        int ans = 0;

        // Iterate from 1 to N && evaluating
        // and adding i % K.
        for (int i = 1; i <= N; i++)
            ans += (i % K);

        return ans;
    }

    // Driver program
    static public void Main()
    {
        int N = 10, K = 2;
        Console.WriteLine(findSum(N, K));
    }
}

// This code is contributed by vt_m.

PHP

<?php
// PHP program to find sum 
// of modulo K of first N 
// natural numbers.


// Return sum of modulo K of 
// first N natural numbers.

function findSum($N, $K)
{
    $ans = 0;

    // Iterate from 1 to N &&
    // evaluating and adding i % K.
    for ($i = 1; $i <= $N; $i++)
        $ans += ($i % $K);

    return $ans;
}

// Driver Code
$N = 10; $K = 2;
echo findSum($N, $K), "\n";

// This code is contributed by ajit
?>

JavaScript

<script>

// JavaScript program to find sum
// of modulo K of first N natural
// numbers. 

// Return sum of modulo K of
// first N natural numbers.
function findSum(N, K)
{
    let ans = 0;

    // Iterate from 1 to N && evaluating
    // and adding i % K.
    for(let i = 1; i <= N; i++)
        ans += (i % K);

    return ans;
}

// Driver Code
let N = 10, K = 2;

document.write(findSum(N, K));

// This code is contributed by code_hunt

</script>

Output :

Time Complexity : O(N).

Auxiliary Space: O(1)

Method 2 :
Two cases arise in this method.
Case 1: When N < K, for each number i, N >= i >= 1, will give i as result when operate with modulo K. So, the required sum will be the sum of the first N natural number, N*(N+1)/2.
Case 2: When N >= K, then integers from 1 to K in natural number sequence will produce, 1, 2, 3, ....., K - 1, 0 as result when operate with modulo K. Similarly, from K + 1 to 2K, it will produce same result. So, the idea is to count how many numbers of times this sequence appears and multiply it with the sum of first K - 1 natural numbers.