Possible to make a divisible by 3 number using all digits in an array

Given an array of integers, the task is to find whether it’s possible to construct an integer using all the digits of these numbers such that it would be divisible by 3. If it is possible then print “Yes” and if not print “No”.

Examples:

Input : arr[] = {40, 50, 90}
Output : Yes
We can construct a number which is
divisible by 3, for example 945000. 
So the answer is Yes. 

Input : arr[] = {1, 4}
Output : No
The only possible numbers are 14 and 41,
but both of them are not divisible by 3, 
so the answer is No.



The idea is based on the fact that a number is divisible by 3 iff sum of its digits is divisible by 3. So we simply find sum of array elements. If the sum is divisible by 3, our answer is Yes, else No.

CPP

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// C++ program to find if it is possible
// to make a number divisible by 3 using
// all digits of given array
#include <bits/stdc++.h>
using namespace std;
  
bool isPossibleToMakeDivisible(int arr[], int n)
{
    // Find remainder of sum when divided by 3
    int remainder = 0;
    for (int i=0; i<n; i++)
       remainder = (remainder + arr[i]) % 3;
  
    // Return true if remainder is 0.
    return (remainder == 0);
}
  
// Driver code
int main()
{
    int arr[] = { 40, 50, 90 };
    int n = sizeof(arr) / sizeof(arr[0]);
    if (isPossibleToMakeDivisible(arr, n))
        printf("Yes\n");
    else
        printf("No\n");
  
    return 0;
}

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Java

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// Java program to find if it is possible
// to make a number divisible by 3 using
// all digits of given array
  
import java.io.*;
import java.util.*; 
  
class GFG
{
    public static boolean isPossibleToMakeDivisible(int arr[], int n)
    {
        // Find remainder of sum when divided by 3
        int remainder = 0;
        for (int i=0; i<n; i++)
            remainder = (remainder + arr[i]) % 3;
  
        // Return true if remainder is 0.
        return (remainder == 0);
    }
  
    public static void main (String[] args)
    {
        int arr[] = { 40, 50, 90 };
        int n = 3;
        if (isPossibleToMakeDivisible(arr, n))
            System.out.print("Yes\n");
        else
            System.out.print("No\n");
    }
}
  
// Code Contributed by Mohit Gupta_OMG <(0_o)>

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Python3

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# Python program to find if it is possible
# to make a number divisible by 3 using
# all digits of given array
  
def isPossibleToMakeDivisible(arr, n):
    # Find remainder of sum when divided by 3
    remainder = 0
    for i in range (0, n):
        remainder = (remainder + arr[i]) % 3
  
    # Return true if remainder is 0.
    return (remainder == 0)
  
# main()
  
arr = [40, 50, 90 ];
n = 3
if (isPossibleToMakeDivisible(arr, n)):
    print("Yes")
else:
    print("No")
  
# Code Contributed by Mohit Gupta_OMG <(0_o)>

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C#

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// C# program to find if it is possible
// to make a number divisible by 3 using
// all digits of given array
using System;
  
class GFG
{
    public static bool isPossibleToMakeDivisible(int []arr, int n)
    {
        // Find remainder of sum when divided by 3
        int remainder = 0;
          
        for (int i = 0; i < n; i++)
            remainder = (remainder + arr[i]) % 3;
  
        // Return true if remainder is 0.
        return (remainder == 0);
    }
  
    public static void Main ()
    {
        int []arr = { 40, 50, 90 };
        int n = 3;
          
        if (isPossibleToMakeDivisible(arr, n))
                        Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}
  
// This code is contributed by vt_m

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PHP

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<?php
// PHP program to find if it is possible
// to make a number divisible by 3 using
// all digits of given array
  
function isPossibleToMakeDivisible($arr, $n)
{
      
    // Find remainder of sum 
    // when divided by 3
    $remainder = 0;
    for ($i = 0; $i < $n; $i++)
    $remainder = ($remainder + $arr[$i]) % 3;
  
    // Return true if remainder is 0.
    return ($remainder == 0);
}
  
// Driver code
$arr = array( 40, 50, 90 );
$n = sizeof($arr);
if (isPossibleToMakeDivisible($arr, $n))
    echo("Yes\n");
else
    echo("No\n");
  
// This code is contributed by Ajit.
?>

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Output:

Yes

Time Complexity : O(n)



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Improved By : jit_t



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