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Find the number of pairs (a, b) such that a % b = K
• Difficulty Level : Easy
• Last Updated : 25 May, 2021

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:
Only valid pairs are (2, 3) and (2, 4).
Input: N = 11, K = 5
Output:

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 implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``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

 ` ``\$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

 ``
Output:
`7`

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