Given an array arr[] of size N consisting of non-zero positive integers. The task is to determine whether the number that is formed by selecting the last digits of all the numbers is divisible by 10 or not. If the number is divisible by 10, then print Yes otherwise print No.
Examples:
Input: arr[] = {12, 65, 46, 37, 99}
Output: No
25679 is not divisible by 10.
Input: arr[] = {24, 37, 46, 50}
Output: Yes
4760 is divisible by 10.
Approach: In order for an integer to be divisible by 10, it must be ending with a 0. So, the last element of the array will decide whether the number formed will be divisible by 10 or not. If the last digit of the last element is 0 then print Yes otherwise print No.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <iostream>using namespace std;// Function that returns true if the// number formed by the last digits of// all the elements is divisible by 10bool isDivisible(int arr[], int n){ // Last digit of the last element int lastDigit = arr[n - 1] % 10; // Number formed will be divisible by 10 if (lastDigit == 0) return true; return false;}// Driver codeint main(){ int arr[] = { 12, 65, 46, 37, 99 }; int n = sizeof(arr) / sizeof(arr[0]); if (isDivisible(arr, n)) cout << "Yes"; else cout << "No"; return 0;} |
Java
// Java implementation of the approachimport java.util.*; class GFG{// Function that returns true if the// number formed by the last digits of// all the elements is divisible by 10static boolean isDivisible(int arr[], int n){ // Last digit of the last element int lastDigit = arr[n - 1] % 10; // Number formed will be divisible by 10 if (lastDigit == 0) return true; return false;}// Driver codestatic public void main ( String []arg){ int arr[] = { 12, 65, 46, 37, 99 }; int n = arr.length; if (isDivisible(arr, n)) System.out.println("Yes"); else System.out.println("No");}}// This code is contributed by Rajput-Ji |
Python3
# Python3 implementation of the approach # Function that returns true if the # number formed by the last digits of # all the elements is divisible by 10 def isDivisible(arr, n) : # Last digit of the last element lastDigit = arr[n - 1] % 10; # Number formed will be divisible by 10 if (lastDigit == 0) : return True; return False; # Driver code if __name__ == "__main__" : arr = [ 12, 65, 46, 37, 99 ]; n = len(arr); if (isDivisible(arr, n)) : print("Yes"); else : print("No"); # This code is contributed by AnkitRai01 |
C#
// C# implementation of the approachusing System; class GFG{// Function that returns true if the// number formed by the last digits of// all the elements is divisible by 10static bool isDivisible(int []arr, int n){ // Last digit of the last element int lastDigit = arr[n - 1] % 10; // Number formed will be divisible by 10 if (lastDigit == 0) return true; return false;}// Driver codestatic public void Main(String []arg){ int []arr = { 12, 65, 46, 37, 99 }; int n = arr.Length; if (isDivisible(arr, n)) Console.WriteLine("Yes"); else Console.WriteLine("No");}} // This code is contributed by Rajput-Ji |
Javascript
<script>// Javascript implementation of the approach// Function that returns true if the// number formed by the last digits of// all the elements is divisible by 10function isDivisible(arr, n){ // Last digit of the last element let lastDigit = arr[n - 1] % 10; // Number formed will be divisible by 10 if (lastDigit == 0) return true; return false;}// Driver code let arr = [ 12, 65, 46, 37, 99 ]; let n = arr.length; if (isDivisible(arr, n)) document.write("Yes"); else document.write("No");</script> |
Output:
No
Time Complexity: O(1)
Auxiliary Space: O(1)
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