# Print k numbers where all pairs are divisible by m

Given an integer array and two numbers k and m. Print k numbers from the array, such that difference between any two pairs is divisible by m. If there are no k numbers, then print -1.

**Examples:**

Input: arr[] = {1, 8, 4} k = 2 m = 3 Output: 1 4 Explanation: Difference between 1 and 4 is divisible by 3. Input: arr[] = {1, 8, 4} k = 3 m = 3 Output: -1 Explanation: there are only two numbers whose difference is divisible by m, but k is three here which is not possible, hence we print -1.

A **naive approach ** is be to iterate for every element and check with all other elements, and if the count of numbers whose difference is divisible by m is greater then equal to k, then we print those k numbers by again iterating. But this won’t be efficient enough as it runs two nested loops.

Time complexity: O(n * n)

Auxiliary space: O(1)

An **efficient approach** is apply a mathematical approach, where we know if (x-y) % m is equal to x%m – y%m. So if we can store all the numbers who leave the same remainder when divided by m,

And if the count of numbers which leaves the same remainder is more than k or equal to k, then we have our answer as all those numbers who leave the same remainder.

Below is the implementation of the above approach

## C++

`// CPP program to find a list of k elements from ` `// an array such that difference between all of ` `// them is divisible by m. ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// function to generate k numbers whose difference ` `// is divisible by m ` `void` `print_result(` `int` `a[], ` `int` `n, ` `int` `k, ` `int` `m) ` `{ ` ` ` `// Using an adjacency list like representation ` ` ` `// to store numbers that lead to same ` ` ` `// remainder. ` ` ` `vector<` `int` `> v[m]; ` ` ` ` ` `for` `(` `int` `i = 0; i < n; i++) { ` ` ` ` ` `// stores the modulus when divided ` ` ` `// by m ` ` ` `int` `rem = a[i] % m; ` ` ` ` ` `v[rem].push_back(a[i]); ` ` ` ` ` `// If we found k elements which ` ` ` `// have same remainder. ` ` ` `if` `(v[rem].size() == k) ` ` ` `{ ` ` ` `for` `(` `int` `j = 0; j < k; j++) ` ` ` `cout << v[rem][j] << ` `" "` `; ` ` ` `return` `; ` ` ` `} ` ` ` `} ` ` ` ` ` `// If we could not find k elements ` ` ` `cout << ` `"-1"` `; ` `} ` ` ` `// driver program to test the above function ` `int` `main() ` `{ ` ` ` `int` `a[] = { 1, 8, 4 }; ` ` ` `int` `n = ` `sizeof` `(a) / ` `sizeof` `(a[0]); ` ` ` `print_result(a, n, 2, 3); ` ` ` `return` `0; ` `} ` |

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## Python3

`# Python3 program to find a list of k elements from ` `# an array such that difference between all of ` `# them is divisible by m. ` ` ` `# function to generate k numbers whose difference ` `# is divisible by m ` `def` `print_result(a, n, k, m): ` ` ` ` ` `# Using an adjacency list like representation ` ` ` `# to store numbers that lead to same ` ` ` `# remainder. ` ` ` `v ` `=` `[[] ` `for` `i ` `in` `range` `(m)] ` ` ` ` ` `for` `i ` `in` `range` `(` `0` `, n): ` ` ` ` ` `# stores the modulus when divided ` ` ` `# by m ` ` ` `rem ` `=` `a[i] ` `%` `m ` ` ` ` ` `v[rem].append(a[i]) ` ` ` ` ` `# If we found k elements which ` ` ` `# have same remainder. ` ` ` `if` `(` `len` `(v[rem]) ` `=` `=` `k): ` ` ` ` ` `for` `j ` `in` `range` `(` `0` `, k): ` ` ` `print` `(v[rem][j], end` `=` `" "` `) ` ` ` `return` ` ` ` ` `# If we could not find k elements ` ` ` `print` `(` `-` `1` `) ` ` ` `# driver program to test the above function ` `if` `__name__` `=` `=` `'__main__'` `: ` ` ` `a ` `=` `[` `1` `, ` `8` `, ` `4` `] ` ` ` `n ` `=` `len` `(a) ` ` ` `print_result(a, n, ` `2` `, ` `3` `) ` ` ` `# This code is contributed by ` `# Sanjit_Prasad ` |

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## PHP

**Output:**

1 4

**Time complexity:** O(n)

**Auxiliary Space:** O(m)

This article is contributed by **Raja Vikramaditya**. 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.

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