Es posible hacer un número divisible por 3 usando todos los dígitos en una array

Dada una array de números enteros, la tarea es encontrar si es posible construir un número entero usando todos los dígitos de estos números de modo que sea divisible por 3. Si es posible, escriba «Sí» y, si no, escriba «No». . 

Ejemplos: 

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.

La idea se basa en que un número es divisible por 3 si la suma de sus dígitos es divisible por 3 . Así que simplemente encontramos la suma de los elementos del arreglo. Si la suma es divisible por 3, nuestra respuesta es Sí, de lo contrario No

Implementación:

// 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;
}

// 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)>

# 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)>

// 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

<?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.
?>

<script>
 
// javascript 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
    var remainder = 0;
    for (i=0; i<n; i++)
        remainder = (remainder + arr[i]) % 3;
 
    // Return true if remainder is 0.
    return (remainder == 0);
}
 
var arr = [ 40, 50, 90 ];
var n = 3;
if (isPossibleToMakeDivisible(arr, n))
    document.write("Yes\n");
else
    document.write("No\n");
 
// This code contributed by Princi Singh
 
</script>

Yes

Complejidad de tiempo: O(n)
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