Check whether a large number represented as array is divisible by Y

Given a large integer X represented as an array arr[] where each arr[i] stores a digit in X. The task is to check whether the number represented by the array is divisible by given integer Y.

Examples:

Input: arr[] = {1, 2, 1, 5, 6}, Y = 4
Output: Yes
12156 / 4 = 3039



Input: arr[] = {1, 1, 1, 1, 1, 1, 1, 1, 1}, Y = 14
Output: No

Approach: Start traversing the digits of the given number from the left and take the largest number which is smaller than or equal to Y and divide it with Y. If the remainder is something other than 0 then it will be carried to the next possible number formed from the remaining digits just like in the long division. After the complete number is processed, if the remainder is still something other than 0 then the represented number is not divisible by Y else it is.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <iostream>
using namespace std;
  
// Function that returns true if the number represented
// by the given array is divisible by y
bool isDivisible(int* arr, int n, int y)
{
    int d = 0, i = 0;
  
    // While there are digits left
    while (i < n) {
  
        // Select the next part of the number
        // i.e. the maximum number which is <= y
        while (d < y && i < n)
            d = d * 10 + arr[i++];
  
        // Get the current remainder
        d = d % y;
    }
  
    // If the final remainder is 0
    if (d == 0)
        return true;
    return false;
}
  
// Driver code
int main()
{
    int arr[] = { 1, 2, 1, 5, 6 };
    int x = sizeof(arr) / sizeof(int);
    int y = 4;
  
    cout << (isDivisible(arr, x, y) ? "Yes" : "No");
  
    return 0;
}

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Java

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// Java implementation of the approach
class GFG
{
      
    // Function that returns true if the number represented
    // by the given array is divisible by y
    static boolean isDivisible(int [] arr, int n, int y)
    {
        int d = 0, i = 0;
      
        // While there are digits left
        while (i < n) 
        {
      
            // Select the next part of the number
            // i.e. the maximum number which is <= y
            while (d < y && i < n)
                d = d * 10 + arr[i++];
      
            // Get the current remainder
            d = d % y;
        }
      
        // If the final remainder is 0
        if (d == 0)
            return true;
        return false;
    }
      
    // Driver code
    public static void main (String[] args) 
    {
          
        int [] arr = { 1, 2, 1, 5, 6 };
        int x = arr.length;
        int y = 4;
      
        System.out.println(isDivisible(arr, x, y) ? "Yes" : "No");
    }
}
  
// This code is contributed by ihritik 

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Python3

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# Python3 implementation of the approach 
  
# Function that returns true if the number represented 
# by the given array is divisible by y 
def isDivisible(arr, n, y):
    d, i = 0, 0
      
    # While there are digits left 
    while i < n:
          
        # Select the next part of the number 
        # i.e. the maximum number which is <= y 
        while d < y and i < n:
            d = d * 10 + arr[i]
            i += 1
          
        # Get the current remainder 
        d = d % y
          
    # If the final remainder is 0 
    if d == 0:
        return True
    return False
  
# Driver code 
if __name__ == "__main__":
    arr = [ 1, 2, 1, 5, 6 ]
    x = len(arr)
    y = 4
    if (isDivisible(arr, x, y)):
        print("Yes"
    else
        print("No")
      
# This code is contributed by
# sanjeev2552

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C#

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// C# implementation of the approach
using System;
  
class GFG
{
      
    // Function that returns true if the number represented
    // by the given array is divisible by y
    static bool isDivisible(int [] arr, int n, int y)
    {
        int d = 0, i = 0;
      
        // While there are digits left
        while (i < n)
        {
      
            // Select the next part of the number
            // i.e. the maximum number which is <= y
            while (d < y && i < n)
                d = d * 10 + arr[i++];
      
            // Get the current remainder
            d = d % y;
        }
      
        // If the final remainder is 0
        if (d == 0)
            return true;
        return false;
    }
      
    // Driver code
    public static void Main () 
    {
          
        int [] arr = { 1, 2, 1, 5, 6 };
        int x = arr.Length;
        int y = 4;
      
        Console.WriteLine(isDivisible(arr, x, y) ? "Yes" : "No");
    }
}
  
// This code is contributed by ihritik 

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Output:

Yes


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Improved By : ihritik, sanjeev2552



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