Given a long integer, we need to find if the difference between sum of odd digits and sum of even digits is 0 or not. The indexes start from zero (0 index is for leftmost digit).
Examples:
Input : 1212112
Output : Yes
Explanation:-
the odd position element is 2+2+1=5
the even position element is 1+1+1+2=5
the difference is 5-5=0.so print yes.
Input :12345
Output : No
Explanation:-
the odd position element is 1+3+5=9
the even position element is 2+4=6
the difference is 9-6=3 not equal
to zero. So print no.
Method 1: One by one traverse digits and find the two sums. If difference between two sums is 0, print yes, else no.
Method 2 : This can be easily solved using divisibility of 11. This condition is only satisfied if the number is divisible by 11. So check the number is divisible by 11 or not.
CPP
#include <bits/stdc++.h>
using namespace std;
bool isDiff0( long long int n)
{
return (n % 11 == 0);
}
int main() {
long int n = 1243
if (isDiff0(n))
cout << "Yes" ;
else
cout << "No" ;
return 0;
}
|
Java
import java.io.*;
import java.util.*;
class GFG
{
public static boolean isDiff( int n)
{
return (n % 11 == 0 );
}
public static void main (String[] args)
{
int n = 1243 ;
if (isDiff(n))
System.out.print( "Yes" );
else
System.out.print( "No" );
}
}
|
Python
def isDiff(n):
return (n % 11 = = 0 )
n = 1243 ;
if (isDiff(n)):
print ( "Yes" )
else :
print ( "No" )
|
C#
using System;
class GFG {
public static bool isDiff( int n)
{
return (n % 11 == 0);
}
public static void Main ()
{
int n = 1243;
if (isDiff(n))
Console.WriteLine( "Yes" );
else
Console.WriteLine( "No" );
}
}
|
PHP
<?php
function isDiff0( $n )
{
return ( $n % 11 == 0);
}
$n = 1243;
if (isDiff0( $n ))
echo "Yes" ;
else
echo "No" ;
?>
|
Javascript
<script>
function isDiff0(n)
{
return (n % 11 == 0);
}
let n = 1243
if (isDiff0(n))
document.write( "Yes" );
else
document.write( "No" );
</script>
|
Output:
Yes
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