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

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 loop 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 remainder as K after getting divided by i.

Below is the implementataion of the above approach:

C++

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// 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;
}

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Java

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// 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));
    }
}

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Python3

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# 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))

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C#

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// 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));
    }
}

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PHP

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<?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));
  
?>

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Output:

7


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