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# Difference between sums of odd and even digits

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.

## C++

 `// C++ program for above approach``#include``using` `namespace` `std;` `bool` `isDiff0(``int` `n){``    ``int` `first = 0;``    ``int` `second = 0;``    ``bool` `flag = ``true``;``    ``while``(n > 0){``        ``int` `digit = n % 10;``        ``if``(flag) first += digit;``        ``else` `second += digit;``        ``flag = !flag;``        ``n = n/10;``    ``}``    ``if``(first - second == 0) ``return` `true``;``    ``return` `false``;``}` `int` `main(){``    ``int` `n = 1243;``    ``if``(isDiff0(n)) cout<<``"Yes"``;``    ``else` `cout<<``"No"``;``    ``return` `0;``}``// This code is contributed by Kirti Agarwal(kirtiagarwal23121999)`

## Java

 `// Java equivalent of above code``public` `class` `Main {``    ``public` `static` `boolean` `isDiff0(``int` `n) {``        ``int` `first = ``0``;``        ``int` `second = ``0``;``        ``boolean` `flag = ``true``;``        ``while` `(n > ``0``) {``            ``int` `digit = n % ``10``;``            ``if` `(flag) first += digit;``            ``else` `second += digit;``            ``flag = !flag;``            ``n = n / ``10``;``        ``}``        ``if` `(first - second == ``0``) ``return` `true``;``        ``return` `false``;``    ``}` `    ``public` `static` `void` `main(String[] args) {``        ``int` `n = ``1243``;``        ``if` `(isDiff0(n)) System.out.println(``"Yes"``);``        ``else` `System.out.println(``"No"``);``    ``}``}`

## Python

 `# Python program for the above approach``def` `isDiff0(n):``    ``first ``=` `0``    ``second ``=` `0``    ``flag ``=` `True``    ``while``(n > ``0``):``        ``digit ``=` `n ``%` `10``        ``if``(flag):``            ``first ``+``=` `digit``        ``else``:``            ``second ``+``=` `digit``        ``if``(flag):``            ``flag ``=` `False``        ``else``:``            ``flag ``=` `True``        ``n ``=` `int``(n``/``10``)``    ``if``(first``-``second ``=``=` `0``):``        ``return` `True``    ``return` `False`  `# driver code``n ``=` `1243``if``(isDiff0(n)):``    ``print``(``"Yes"``)``else``:``    ``print``(``"No"``)`

## C#

 `// C# Program for the above approach``using` `System;` `public` `class` `BinaryTree{``    ``static` `bool` `isDiff0(``int` `n){``        ``int` `first = 0;``        ``int` `second = 0;``        ``bool` `flag = ``true``;``        ``while``(n > 0){``            ``int` `digit = n % 10;``            ``if``(flag) first += digit;``            ``else` `second += digit;``            ``flag = !flag;``            ``n = n/10;``        ``}``        ``if``(first - second == 0) ``return` `true``;``        ``return` `false``;``    ``}``    ` `    ``public` `static` `void` `Main(){``        ``int` `n = 1243;``        ``if``(isDiff0(n)) Console.Write(``"Yes"``);``        ``else` `Console.Write(``"No"``);``    ``}``}`

## Javascript

 `// JavaScript prgraom for the above approach``function` `isDiff0(n){``    ``let first = 0;``    ``let second = 0;``    ``let flag = ``true``;``    ``while``(n > 0){``        ``let digit = n % 10;``        ``if``(flag) first += digit;``        ``else` `second += digit;``        ``flag = !flag;``        ``n = parseInt(n/10);``        ` `    ``}``    ``if``(first - second == 0) ``return` `true``;``    ``return` `false``;``}` `let n = 1243;``if``(isDiff0(n))``    ``console.log(``"Yes"``);``else``    ``console.log(``"No"``);``    ` `    ``// THIS CODE IS CONTRIBUTED BY YASH AGARWAL(YASHAGAWRAL2852002)`

Output

`Yes`

Time Complexity:

The given approach traverses the digits of the input integer once. Hence the time complexity of this approach is O(log n), where n is the value of the input integer.

Space Complexity: O(1)

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

 `// C++ program to check if difference between sum of``// odd digits and sum of even digits is 0 or not``#include ``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

 `// Java program to check if difference between sum of``// odd digits and sum of even digits is 0 or not` `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

 `# Python program to check if difference between sum of``# odd digits and sum of even digits is 0 or not` `def` `isDiff(n):``    ``return` `(n ``%` `11` `=``=` `0``)` `# Driver code``n ``=` `1243``;``if` `(isDiff(n)):``    ``print``(``"Yes"``)``else``:``    ``print``(``"No"``)` `# Mohit Gupta_OMG <0_o>`

## C#

 `// C# program to check if difference``// between sum of odd digits and sum``// of even digits is 0 or not``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"``);``    ``}``}` `// This code is contributed by vt_m.`

## PHP

 ``

## Javascript

 ``

Output:

`Yes`

Time Complexity: O(1)

Auxiliary Space: O(1)

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