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Difference between sums of odd and even digits

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  • Difficulty Level : Basic
  • Last Updated : 05 Mar, 2023
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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<bits/stdc++.h>
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 <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




// 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




<?php
// PHP program to check if
// difference between sum of
// odd digits and sum of
// even digits is 0 or not
 
function isDiff0($n)
{
    return ($n % 11 == 0);
}
 
    // Driver Code
    $n = 1243;
    if (isDiff0($n))
        echo"Yes";
    else
        echo "No";
         
// This code is contributed by nitin mittal
?>


Javascript




<script>
 
// Javascript program to check if difference between sum of
// odd digits and sum of even digits is 0 or not
 
function isDiff0(n)
{
    return (n % 11 == 0);
}
 
     
    let n = 1243
    if (isDiff0(n))
        document.write("Yes");
    else
        document.write("No");
 
// This code is contributed by Mayank Tyagi
 
</script>


Output: 

Yes

Time Complexity: O(1)

Auxiliary Space: O(1)

This article is contributed by jaspal singh. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 


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