Check if all array elements can be made divisible by K by replacing array elements with sum of pairs

Last Updated : 21 Apr, 2021

Given an array arr[] of size N and a positive integer K, the task is to make every element of the array divisible by K by repeatedly selecting a pair of indices (i, j) and perform arr[i] = arr[i] + arr[j].

Examples:

Input: arr[] = {3, 6, 3}, K = 6
Output: True
Explanation:
Update arr[0] = arr[0] + arr[0].
Update arr[2] = arr[2] + arr[2]

Input: arr[] = {6, 6, 8, 7, 3}, K = 6
Output: False

Approach: Follow the steps below to solve the problem:

Traverse the array arr[]:
- Check if arr[i] is divisible by K or not:
  - Initialize a variable, say temp, to store the value of arr[i] after each iteration.
  - Repeatedly Multiply arr[i] by 2 until arr[i] < K * temp.
  - Check if arr[i] % K != 0. If found to be true, print False and return.
After completing the above steps, print True as all the elements are now divisible by K.

Below is the implementation of the above approach:

C++

// C++ Program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Utility function to check if all array
// elements can be made divisible by K or not
bool makeArrayDivisibleByKUtil(
    int arr[], int N, int K)
{
    // Traverse the array
    for (int i = 0; i < N; i++) {
 
        // Check whether arr[i] is
        // divisible by K or not
        if (arr[i] % K != 0) {
 
            // Stores value of arr[i]
            int temp = arr[i];
 
            while (arr[i] < K * temp) {
 
                // Multiply by 2 until it
                // is less than K * temp
                arr[i] = arr[i] * 2;
            }
 
            if (arr[i] % K != 0) {
                return false;
            }
        }
    }
 
    // Return true as all array
    // elements are divisible by K
    return true;
}
 
// Function to check whether all array
// elements can be made divisible by K or not
void makeArrayDivisibleByK(
    int arr[], int N, int K)
{
    if (makeArrayDivisibleByKUtil(arr, N, K)) {
        cout << "True" << endl;
    }
    else {
        cout << "False" << endl;
    }
}
 
// Driver Code
int main()
{
    // Given array
    int arr[] = { 3, 6, 3 };
 
    // Size of the array
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Given value of K
    int K = 6;
 
    makeArrayDivisibleByK(arr, N, K);
 
    return 0;
}

C

#include<stdio.h>
 
// Utility function to check if all array
// elements can be made divisible by K or not
int makeArrayDivisibleByKUtil(
    int arr[], int N, int K)
{
   
    // Traverse the array
    for (int i = 0; i < N; i++) 
    {
 
        // Check whether arr[i] is
        // divisible by K or not
        if (arr[i] % K != 0)
        {
 
            // Stores value of arr[i]
            int temp = arr[i];
            while (arr[i] < K * temp) 
            {
 
                // Multiply by 2 until it
                // is less than K * temp
                arr[i] = arr[i] * 2;
            }
 
            if (arr[i] % K != 0)
            {
                return 0;
            }
        }
    }
 
    // Return true as all array
    // elements are divisible by K
    return 1;
}
 
// Function to check whether all array
// elements can be made divisible by K or not
void makeArrayDivisibleByK(
    int arr[], int N, int K)
{
    if (makeArrayDivisibleByKUtil(arr, N, K)) 
    {
        printf("True\n");
    }
    else
    {
        printf("False\n");
    }
}
 
// Driver Code
int main()
{
   
    // Given array
    int arr[] = { 3,6,3 };
 
    // Size of the array
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Given value of K
    int K = 6;
    makeArrayDivisibleByK(arr, N, K);
    return 0;
}
 
// This code is contributed by santoshcharan.

Java

import java.io.*;
class GFG
{
    public static boolean makeArrayDivisibleBykUtil(int arr[],
                                                    int n,int k)
    {
       
        // Traverse the array
        for(int i = 0; i < n; i++)
        {
           
            // Check whether arr[i] is
            // divisible by k or not
            if(arr[i] % k != 0)
            {
               
                // Stores value of arr[i]
                int temp = arr[i];
                while(arr[i] < k * arr[i])
                {
                   
                    // Multiply by 2 until it
                    // is less than k*temp
                    arr[i] = arr[i] * 2;
                }
                if(arr[i] % k != 0)
                {
                    return false;
                }
            }
        }
       
        // Return true as all array
        // elements are divisible by k
        return true;
    }
    public static void makeArrayDivisibleByk(int arr[],
                                             int n,int k)
    {
        if(makeArrayDivisibleBykUtil(arr, n, k))
        {
            System.out.println("True");
        }
        else
        {
            System.out.println("False");
        }
    }
   
    // Driver Code
    public static void main(String[] args) 
    {
       
        // Given array    
        int[] arr = { 3 , 6 , 3};
       
        // Size of the array
        int n = 3;
        int k = 6;
       
        // Calling the function
        makeArrayDivisibleByk(arr, n, k);
    }
}
 
// This code is contributed by santoshcharan.

Python3

# Python 3 Program for the above approach
 
# Utility function to check if all array
# elements can be made divisible by K or not
def makeArrayDivisibleByKUtil(arr, N, K):
   
  # Traverse the array
  for i in range(N):
     
    # Check whether arr[i] is
    # divisible by K or not
    if (arr[i] % K != 0):
       
      # Stores value of arr[i]
      temp = arr[i]
      while (arr[i] < K * temp):
         
        # Multiply by 2 until it
        # is less than K * temp
        arr[i] = arr[i] * 2
      if (arr[i] % K != 0):
        return False
 
  # Return true as all array
  # elements are divisible by K
  return True
 
# Function to check whether all array
# elements can be made divisible by K or not
def makeArrayDivisibleByK(arr, N, K):
  if (makeArrayDivisibleByKUtil(arr, N, K)):
    print("True")
  else:
    print("False")
 
# Driver Code
if __name__ == '__main__':
   
  # Given array
  arr =  [3, 6, 3]
 
  # Size of the array
  N =  len(arr)
 
  # Given value of K
  K = 6
  makeArrayDivisibleByK(arr, N, K)
   
  # This code is contributed by bgangwar59.

C#

using System;
 
class GFG
{
  public static bool makeArrayDivisibleBykUtil(int [] arr,
                                               int n,int k)
  {
 
    // Traverse the array
    for(int i = 0; i < n; i++)
    {
 
      // Check whether arr[i] is
      // divisible by k or not
      if(arr[i] % k != 0)
      {
 
        // Stores value of arr[i]
        //int temp = arr[i];
        while(arr[i] < k * arr[i])
        {
 
          // Multiply by 2 until it
          // is less than k*temp
          arr[i] = arr[i] * 2;
        }
        if(arr[i] % k != 0)
        {
          return false;
        }
      }
    }
 
    // Return true as all array
    // elements are divisible by k
    return true;
  }
  public static void makeArrayDivisibleByk(int []arr,
                                           int n,int k)
  {
    if(makeArrayDivisibleBykUtil(arr, n, k))
    {
      Console.WriteLine("True");
    }
    else
    {
      Console.WriteLine("False");
    }
  }
 
  // Driver Code
  public static void Main(string[] args) 
  {
 
    // Given array    
    int[] arr = { 3 , 6 , 3};
 
    // Size of the array
    int n = 3;
    int k = 6;
 
    // Calling the function
    makeArrayDivisibleByk(arr, n, k);
  }
}
 
// This code is contributed by chitranayal.

Javascript

<script>
 
// Javascript Program for the above approach
 
// Utility function to check if all array
// elements can be made divisible by K or not
function makeArrayDivisibleByKUtil(arr, N, K)
{
     
    // Traverse the array
    for(let i = 0; i < N; i++) 
    {
         
        // Check whether arr[i] is
        // divisible by K or not
        if (arr[i] % K != 0) 
        {
 
            // Stores value of arr[i]
            let temp = arr[i];
 
            while (arr[i] < K * temp)
            {
 
                // Multiply by 2 until it
                // is less than K * temp
                arr[i] = arr[i] * 2;
            }
 
            if (arr[i] % K != 0)
            {
                return false;
            }
        }
    }
 
    // Return true as all array
    // elements are divisible by K
    return true;
}
 
// Function to check whether all array
// elements can be made divisible by K or not
function makeArrayDivisibleByK(arr, N, K)
{
    if (makeArrayDivisibleByKUtil(arr, N, K))
    {
        document.write("True<br>");
    }
    else
    {
        document.write("False<br>");
    }
}
 
// Driver Code
 
// Given array
let arr = [ 3, 6, 3 ];
 
// Size of the array
let N = arr.length;
 
// Given value of K
let K = 6;
 
makeArrayDivisibleByK(arr, N, K);
 
// This code is contributed by rishavmahato348
 
</script>

Output:

True

Time Complexity: O(N * logN)
Auxiliary Space: O(1)

Check if sum of array can be made equal to X by removing either the first or last digits of every array element

santhoshcharan

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Check if all array elements can be made divisible by K by replacing array elements with sum of pairs

C++

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Python3

C#

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