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C/C++ Program to check whether it is possible to make a divisible by 3 number using all digits in an array
  • Last Updated : 05 Dec, 2018

Given an array of integers, the task is to find whether it is 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.




// 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;
}
Output:
Yes

Time Complexity: O(n)

Please refer complete article on Possible to make a divisible by 3 number using all digits in an array for more details!

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