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++ 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 10 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 code int 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 implementation of the approach import java.util.*;
class GFG
{ // Function that returns true if the // number formed by the last digits of // all the elements is divisible by 10 static 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 code static 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 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# implementation of the approach using System;
class GFG
{ // Function that returns true if the // number formed by the last digits of // all the elements is divisible by 10 static 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 code static 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 |
<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 10 function 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> |
No
Time Complexity: O(1)
Auxiliary Space: O(1)