Find set of m-elements with difference of any two elements is divisible by k

Given an array of n-positive integers and a positive integer k, find a set of exactly m-elements such that difference of any two element is equal to k.

Examples :

Input : arr[] = {4, 7, 10, 6, 9},
        k = 3, m = 3
Output : Yes 
         4 7 10

Input : arr[] = {4, 7, 10, 6, 9}, 
        k = 12, m = 4
Output : No

Input : arr[] = {4, 7, 10, 6, 9}, 
        k = 3, m = 4
Output : No

Approach : To solve this question, just keep a record of remainders when an element is divided by k. Create a multi dimensional array remainder_set[][] of size k whose index shows remainder and elements of that array will element as per their corresponding remainder when divided by k. For example if arr[i] % k = 3 then arr[i] will be element of remainder_set[3] and so on for all elements. Now, traversing the remainder set, one can easily get a set whose size is greater than or equal to required size m if exist. And for sure difference of any elements of that set will be divisible by k.

C++

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// C++ program for finding reemainder set
#include <bits/stdc++.h>
using namespace std;
  
// function to find remainder set
void findSet(int arr[], int n, int k, int m) {
  
  vector<int> remainder_set[k];
  
  // calculate remainder set array
  // and push element as per their remainder
  for (int i = 0; i < n; i++) {
    int rem = arr[i] % k;
    remainder_set[rem].push_back(arr[i]);
  }
  
  // check whether sizeof any remainder set
  // is equal or greater than m
  for (int i = 0; i < k; i++) {
    if (remainder_set[i].size() >= m) {
      cout << "Yes \n";
      for (int j = 0; j < m; j++) 
        cout << remainder_set[i][j] << " ";      
      return;
    }
  }
  
  cout << "No";
}
  
// driver program
int main() {
  int arr[] = {5, 8, 9, 12, 13, 7, 11, 15};
  int k = 4;
  int m = 3;
  int n = sizeof(arr) / sizeof(arr[0]);
  findSet(arr, n, k, m);
}

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Java

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// Java program for finding reemainder set
import java.util.*;
class GFG 
{
  
// function to find remainder set
static void findSet(int arr[], int n,
                    int k, int m)
{
    Vector<Integer> []remainder_set = new Vector[k];
    for (int i = 0; i < k; i++)
    {
        remainder_set[i] = new Vector<Integer>();
    }
      
    // calculate remainder set array
    // and push element as per their remainder
    for (int i = 0; i < n; i++) 
    {
        int rem = arr[i] % k;
        remainder_set[rem].add(arr[i]);
    }
      
    // check whether sizeof any remainder set
    // is equal or greater than m
    for (int i = 0; i < k; i++)
    {
        if (remainder_set[i].size() >= m)
        {
            System.out.println("Yes");
            for (int j = 0; j < m; j++) 
                    System.out.print(remainder_set[i].get(j) + " ");     
            return;
        }
    }
    System.out.print("No");
}
  
// Driver Code
public static void main(String[] args)
{
    int arr[] = {5, 8, 9, 12, 13, 7, 11, 15};
    int k = 4;
    int m = 3;
    int n = arr.length;
    findSet(arr, n, k, m);
}
  
// This code is contributed by PrinciRaj1992

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Python3

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# Python3 program to find exacktly m-element set
# where differencec of any two is divisible by k
def findSet( arr, k, m):
  
    arr_size = len(arr);
    remainder_set=[0]*k;
  
    # initialize remainder set with blank array
    for i in range(k):
        remainder_set[i] = [];
  
    # calculate remainder set array
    # and push element as per their reainder
    for i in range(arr_size):
        rem = arr[i] % k;
        remainder_set[rem].append(arr[i]);
  
    # check whether sizeof any remainder set 
    # is equal or greater than m
    for i in range(k):
        # if size exist then print yes and all elements
        if(len(remainder_set[i]) >= m):
            print("Yes");
            for j in range(m):
                print(remainder_set[i][j],end="");
                print(" ",end="");
  
            # return if remainder set found
            return;
  
    # print no if no remiander set found
    print("No");
  
arr = [5, 8, 9, 12, 13, 7, 11, 15];
k = 4; # constant k
m = 3; # size of set required
findSet(arr, k, m);
  
# This code is contributed by mits.

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C#

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// C# program for finding 
// remainder set 
using System;
using System.Collections.Generic;
  
class GFG
{
// function to find 
// remainder set
static void findSet(int []arr, int n, 
                    int k, int m) 
{
    List<int>[] remainder_set = 
                       new List<int>[k];
    for(int i = 0; i < k; i++)
        remainder_set[i] = 
                       new List<int>();
      
    // calculate remainder set 
    // array and push element
    // as per their remainder
    for (int i = 0; i < n; i++) 
    {
        int rem = arr[i] % k;
        remainder_set[rem].Add(arr[i]);
    }
      
    // check whether sizeof 
    // any remainder set is
    // equal or greater than m
    for (int i = 0; i < k; i++) 
    {
        if (remainder_set[i].Count >= m)
        {
        Console.Write("Yes \n");
        for (int j = 0; j < m; j++) 
            Console.Write(remainder_set[i][j] + 
                                          " ");     
        return;
        }
    }
      
    Console.Write("No");
    }
      
// Driver Code
static void Main()
{
    int []arr = new int[]{5, 8, 9, 12, 
                          13, 7, 11, 15};
    int k = 4;
    int m = 3;
    int n = arr.Length;
    findSet(arr, n, k, m);
}
}
  
// This code is contributed by 
// Manish Shaw(manishshaw1)

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PHP

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<?php
// PHP program to find exacktly m-element set
// where differencec of any two is divisible by k
function findSet( $arr, $k, $m)
{
    $arr_size = sizeof($arr);
  
    // initialize remainder set with blank array
    for ($i = 0; $i < $k; $i++)
    {
        $remainder_set[$i] = array();
    }
  
    // calculate remainder set array
    // and push element as per their reainder
    for ($i = 0; $i < $arr_size; $i++)
    {
        $rem = $arr[$i] % $k;
        array_push($remainder_set[$rem], $arr[$i]);
    }
  
    // check whether sizeof any remainder set 
    // is equal or greater than m
    for($i = 0; $i < $k; $i++)
    {
        // if size exist then print yes and all elements
        if(sizeof($remainder_set[$i]) >= $m)
        {
            print("Yes\n");
            for($j = 0; $j < $m; $j++)
            {
                print($remainder_set[$i][$j]);
                print(" ");
            }
  
            // return if remainder set found
            return;
        }
    }
  
    // print no if no remiander set found
    print("No");
}
  
$arr = array(5, 8, 9, 12, 13, 7, 11, 15);
$k = 4; // constant k
$m = 3; // size of set required
findset($arr, $k, $m);
?>

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Output:

Yes 
5 9 13


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