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.

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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

 // C++ program to find if it is possible// to make a number divisible by 3 using// all digits of given array#include 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 codeint main(){    int arr[] = { 40, 50, 90 };    int n = sizeof(arr) / sizeof(arr);    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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