# 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

 `// 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``)); ` ` `  `?> `

Output:

```7
```

My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Article Tags :
Practice Tags :

2

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.