Given an array **A[] **and positive integer **K**, the task is to count total number of pairs in the array whose sum is divisible by **K**.

Note : This question is generalised version of this

**Examples:**

Input :A[] = {2, 2, 1, 7, 5, 3}, K = 4Output :5Explanation :There are five pairs possible whose sum is divisible by '4' i.e., (2, 2), (1, 7), (7, 5), (1, 3) and (5, 3)Input :A[] = {5, 9, 36, 74, 52, 31, 42}, K = 3Output :7

**Naive Approach**: The simplest approach is to iterate through every pair of the array but using two nested for loops and count those pairs whose sum is divisible by ‘K’. Time complexity of this approach is O(N^{2}).

**Efficient Approach**: An efficient approach is to use Hashing technique. We will separate elements into buckets depending on their (value mod K). When a number is divided by K then the remainder may be 0, 1, 2, upto (k-1). So take an array say **freq[]** of size K (initialized with Zero) and increase the value of freq[A[i]%K] so that we can calculate the number of values giving remainder j on division with K.

## C++

`// C++ Program to count pairs ` `// whose sum divisible by 'K' ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Program to count pairs whose sum divisible ` `// by 'K' ` `int` `countKdivPairs(` `int` `A[], ` `int` `n, ` `int` `K) ` `{ ` ` ` `// Create a frequency array to count ` ` ` `// occurrences of all remainders when ` ` ` `// divided by K ` ` ` `int` `freq[K] = { 0 }; ` ` ` ` ` `// Count occurrences of all remainders ` ` ` `for` `(` `int` `i = 0; i < n; i++) ` ` ` `++freq[A[i] % K]; ` ` ` ` ` `// If both pairs are divisible by 'K' ` ` ` `int` `sum = freq[0] * (freq[0] - 1) / 2; ` ` ` ` ` `// count for all i and (k-i) ` ` ` `// freq pairs ` ` ` `for` `(` `int` `i = 1; i <= K / 2 && i != (K - i); i++) ` ` ` `sum += freq[i] * freq[K - i]; ` ` ` `// If K is even ` ` ` `if` `(K % 2 == 0) ` ` ` `sum += (freq[K / 2] * (freq[K / 2] - 1) / 2); ` ` ` `return` `sum; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` ` ` `int` `A[] = { 2, 2, 1, 7, 5, 3 }; ` ` ` `int` `n = ` `sizeof` `(A) / ` `sizeof` `(A[0]); ` ` ` `int` `K = 4; ` ` ` `cout << countKdivPairs(A, n, K); ` ` ` ` ` `return` `0; ` `} ` |

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## Java

`// Java program to count pairs ` `// whose sum divisible by 'K' ` `import` `java.util.*; ` ` ` `class` `Count { ` ` ` `public` `static` `int` `countKdivPairs(` `int` `A[], ` `int` `n, ` `int` `K) ` ` ` `{ ` ` ` `// Create a frequency array to count ` ` ` `// occurrences of all remainders when ` ` ` `// divided by K ` ` ` `int` `freq[] = ` `new` `int` `[K]; ` ` ` ` ` `// Count occurrences of all remainders ` ` ` `for` `(` `int` `i = ` `0` `; i < n; i++) ` ` ` `++freq[A[i] % K]; ` ` ` ` ` `// If both pairs are divisible by 'K' ` ` ` `int` `sum = freq[` `0` `] * (freq[` `0` `] - ` `1` `) / ` `2` `; ` ` ` ` ` `// count for all i and (k-i) ` ` ` `// freq pairs ` ` ` `for` `(` `int` `i = ` `1` `; i <= K / ` `2` `&& i != (K - i); i++) ` ` ` `sum += freq[i] * freq[K - i]; ` ` ` `// If K is even ` ` ` `if` `(K % ` `2` `== ` `0` `) ` ` ` `sum += (freq[K / ` `2` `] * (freq[K / ` `2` `] - ` `1` `) / ` `2` `); ` ` ` `return` `sum; ` ` ` `} ` ` ` `public` `static` `void` `main(String[] args) ` ` ` `{ ` ` ` `int` `A[] = { ` `2` `, ` `2` `, ` `1` `, ` `7` `, ` `5` `, ` `3` `}; ` ` ` `int` `n = ` `6` `; ` ` ` `int` `K = ` `4` `; ` ` ` `System.out.print(countKdivPairs(A, n, K)); ` ` ` `} ` `} ` |

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## Python3

`# Python3 code to count pairs whose ` `# sum is divisible by 'K' ` ` ` `# Function to count pairs whose ` `# sum is divisible by 'K' ` `def` `countKdivPairs(A, n, K): ` ` ` ` ` `# Create a frequency array to count ` ` ` `# occurrences of all remainders when ` ` ` `# divided by K ` ` ` `freq ` `=` `[` `0` `] ` `*` `K ` ` ` ` ` `# Count occurrences of all remainders ` ` ` `for` `i ` `in` `range` `(n): ` ` ` `freq[A[i] ` `%` `K]` `+` `=` `1` ` ` ` ` `# If both pairs are divisible by 'K' ` ` ` `sum` `=` `freq[` `0` `] ` `*` `(freq[` `0` `] ` `-` `1` `) ` `/` `2` `; ` ` ` ` ` `# count for all i and (k-i) ` ` ` `# freq pairs ` ` ` `i ` `=` `1` ` ` `while` `(i <` `=` `K` `/` `/` `2` `and` `i !` `=` `(K ` `-` `i) ): ` ` ` `sum` `+` `=` `freq[i] ` `*` `freq[K` `-` `i] ` ` ` `i` `+` `=` `1` ` ` ` ` `# If K is even ` ` ` `if` `( K ` `%` `2` `=` `=` `0` `): ` ` ` `sum` `+` `=` `(freq[K` `/` `/` `2` `] ` `*` `(freq[K` `/` `/` `2` `]` `-` `1` `)` `/` `2` `); ` ` ` ` ` `return` `int` `(` `sum` `) ` ` ` `# Driver code ` `A ` `=` `[` `2` `, ` `2` `, ` `1` `, ` `7` `, ` `5` `, ` `3` `] ` `n ` `=` `len` `(A) ` `K ` `=` `4` `print` `(countKdivPairs(A, n, K)) ` |

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## C#

`// C# program to count pairs ` `// whose sum divisible by 'K' ` `using` `System; ` ` ` `class` `Count ` `{ ` ` ` `public` `static` `int` `countKdivPairs(` `int` `[]A, ` `int` `n, ` `int` `K) ` ` ` `{ ` ` ` `// Create a frequency array to count ` ` ` `// occurrences of all remainders when ` ` ` `// divided by K ` ` ` `int` `[]freq = ` `new` `int` `[K]; ` ` ` ` ` `// Count occurrences of all remainders ` ` ` `for` `(` `int` `i = 0; i < n; i++) ` ` ` `++freq[A[i] % K]; ` ` ` ` ` `// If both pairs are divisible by 'K' ` ` ` `int` `sum = freq[0] * (freq[0] - 1) / 2; ` ` ` ` ` `// count for all i and (k-i) ` ` ` `// freq pairs ` ` ` `for` `(` `int` `i = 1; i <= K / 2 && i != (K - i); i++) ` ` ` `sum += freq[i] * freq[K - i]; ` ` ` ` ` `// If K is even ` ` ` `if` `(K % 2 == 0) ` ` ` `sum += (freq[K / 2] * (freq[K / 2] - 1) / 2); ` ` ` `return` `sum; ` ` ` `} ` ` ` ` ` `// Driver code ` ` ` `static` `public` `void` `Main () ` ` ` `{ ` ` ` `int` `[]A = { 2, 2, 1, 7, 5, 3 }; ` ` ` `int` `n = 6; ` ` ` `int` `K = 4; ` ` ` `Console.WriteLine(countKdivPairs(A, n, K)); ` ` ` `} ` `} ` ` ` `// This code is contributed by akt_mit. ` |

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## PHP

`<?php ` `// PHP Program to count pairs ` `// whose sum divisible by 'K' ` ` ` `// Program to count pairs whose sum ` `// divisible by 'K' ` `function` `countKdivPairs(` `$A` `, ` `$n` `, ` `$K` `) ` `{ ` ` ` ` ` `// Create a frequency array to count ` ` ` `// occurrences of all remainders when ` ` ` `// divided by K ` ` ` `$freq` `= ` `array_fill` `(0, ` `$K` `, 0); ` ` ` ` ` `// Count occurrences of all remainders ` ` ` `for` `(` `$i` `= 0; ` `$i` `< ` `$n` `; ` `$i` `++) ` ` ` `++` `$freq` `[` `$A` `[` `$i` `] % ` `$K` `]; ` ` ` ` ` `// If both pairs are divisible by 'K' ` ` ` `$sum` `= (int)(` `$freq` `[0] * (` `$freq` `[0] - 1) / 2); ` ` ` ` ` `// count for all i and (k-i) ` ` ` `// freq pairs ` ` ` `for` `(` `$i` `= 1; ` `$i` `<= ` `$K` `/ 2 && ` ` ` `$i` `!= (` `$K` `- ` `$i` `); ` `$i` `++) ` ` ` `$sum` `+= ` `$freq` `[` `$i` `] * ` `$freq` `[` `$K` `- ` `$i` `]; ` ` ` ` ` `// If K is even ` ` ` `if` `(` `$K` `% 2 == 0) ` ` ` `$sum` `+= (int)(` `$freq` `[(int)(` `$K` `/ 2)] * ` ` ` `(` `$freq` `[(int)(` `$K` `/ 2)] - 1) / 2); ` ` ` `return` `$sum` `; ` `} ` ` ` `// Driver code ` `$A` `= ` `array` `( 2, 2, 1, 7, 5, 3 ); ` `$n` `= ` `count` `(` `$A` `); ` `$K` `= 4; ` `echo` `countKdivPairs(` `$A` `, ` `$n` `, ` `$K` `); ` ` ` `// This code is contributed by mits ` `?> ` |

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**Output :**

5

**Time complexity: **O(N)

**Auxiliary space: **O(1)

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