Find the number of pairs (a, b) such that a % b = K

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Given two integers N and K where N, K > 0, the task is to find the total number of pairs (a, b) where 1 ? a, b ? N such that a % b = K.

Examples:

Input: N = 4, K = 2
Output: 2
Only valid pairs are (2, 3) and (2, 4).

Input: N = 11, K = 5
Output: 7

Naive approach: Run two loops from 1 to n and count all the pairs (i, j) where i % j = K. The time complexity of this approach will be O(n²).
Efficient approach: Initially total count = N – K because all the numbers from the range which are > K will give K as the remainder after dividing it. After that, for all i > K there are exactly (N – K) / i numbers which will give the remainder as K after getting divided by i.

Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the count
// of required pairs
int CountAllPairs(int N, int K)
{
 
    int count = 0;
 
    if (N > K) {
 
        // Initial count
        count = N - K;
        for (int i = K + 1; i <= N; i++)
            count = count + ((N - K) / i);
    }
 
    return count;
}
 
// Driver code
int main()
{
    int N = 11, K = 5;
 
    cout << CountAllPairs(N, K);
 
    return 0;
}

Java

// Java implementation of the approach
import java.io.*;
class GFG {
 
    // Function to return the count
    // of required pairs
    static int CountAllPairs(int N, int K)
    {
 
        int count = 0;
 
        if (N > K) {
 
            // Initial count
            count = N - K;
            for (int i = K + 1; i <= N; i++)
                count = count + ((N - K) / i);
        }
 
        return count;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int N = 11, K = 5;
        System.out.println(CountAllPairs(N, K));
    }
}

Python3

# Python3 implementation of the approach
import math
 
# Function to return the count 
# of required pairs
def CountAllPairs(N, K):
    count = 0
    if( N > K):
         
        # Initial count
        count = N - K
        for i in range(K + 1, N + 1):
            count = count + ((N - K) // i)
             
    return count
     
# Driver code
N = 11
K = 5
print(CountAllPairs(N, K))

C#

// C# implementation of the approach
using System;
class GFG {
 
    // Function to return the count
    // of required pairs
    static int CountAllPairs(int N, int K)
    {
 
        int count = 0;
 
        if (N > K) {
 
            // Initial count
            count = N - K;
            for (int i = K + 1; i <= N; i++)
                count = count + ((N - K) / i);
        }
 
        return count;
    }
 
    // Driver code
    public static void Main()
    {
        int N = 11, K = 5;
        Console.WriteLine(CountAllPairs(N, K));
    }
}

PHP

<?php
// PHP implementation of the approach
 
// Function to return the count 
// of required pairs
function CountAllPairs($N, $K)
{
    $count = 0;
     
    if( $N > $K){
         
        // Initial count
        $count = $N - $K;
        for($i = $K+1; $i <= $N ; $i++)
        {
                $x = ((($N - $K) / $i));
                $count = $count + (int)($x);
        }
    }
 
    return $count;
}
 
    // Driver code
    $N = 11; 
    $K = 5;
    echo(CountAllPairs($N, $K));
 
?>

Javascript

<script>
 
// JavaScript implementation of the approach
 
 
    // Function to return the count
    // of required pairs
    function CountAllPairs( N,  K)
    {
 
        let count = 0;
 
        if (N > K) {
 
            // Initial count
            count = N - K;
            for (let i = K + 1; i <= N; i++)
                count = count + parseInt((N - K) / i);
        }
 
        return count;
    }
 
    // Driver code
        let N = 11, K = 5;
        document.write(CountAllPairs(N, K));
     
 
//contributed by sravan (171fa07058)
     
</script>

Output

Time Complexity: O(N – K), where N and K are the given inputs.
Auxiliary Space: O(1), no extra space is required, so it is a constant.

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