Find the sum of digits of a number at even and odd places

Given a number N, the task is to find the sum of digits of a number at even and odd places.

Examples:

Input: N = 54873
Output:
Sum odd = 16
Sum even = 11

Input: N = 457892
Output:
Sum odd = 20
Sum even = 15



Approach:

  • First, calculate the reverse of the given number.
  • To the reverse number we apply modulus operator and extract its last digit which is actually the first digit of a number so it is odd positioned digit.
  • The next digit will be even positioned digit, and we can take the sum in alternating turns.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the reverse of a number
int reverse(int n)
{
    int rev = 0;
    while (n != 0) {
        rev = (rev * 10) + (n % 10);
        n /= 10;
    }
    return rev;
}
  
// Function to find the sum of the odd
// and even positioned digits in a number
void getSum(int n)
{
    n = reverse(n);
    int sumOdd = 0, sumEven = 0, c = 1;
  
    while (n != 0) {
  
        // If c is even number then it means
        // digit extracted is at even place
        if (c % 2 == 0)
            sumEven += n % 10;
        else
            sumOdd += n % 10;
        n /= 10;
        c++;
    }
  
    cout << "Sum odd = " << sumOdd << "\n";
    cout << "Sum even = " << sumEven;
}
  
// Driver code
int main()
{
    int n = 457892;
    getSum(n);
  
    return 0;
}

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Java

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// Java implementation of the approach
import java.util.*;
  
class GFG
{
  
// Function to return the reverse of a number
static int reverse(int n)
{
    int rev = 0;
    while (n != 0
    {
        rev = (rev * 10) + (n % 10);
        n /= 10;
    }
    return rev;
}
  
// Function to find the sum of the odd
// and even positioned digits in a number
static void getSum(int n)
{
    n = reverse(n);
    int sumOdd = 0, sumEven = 0, c = 1;
  
    while (n != 0
    {
  
        // If c is even number then it means
        // digit extracted is at even place
        if (c % 2 == 0)
            sumEven += n % 10;
        else
            sumOdd += n % 10;
        n /= 10;
        c++;
    }
  
    System.out.println("Sum odd = "+sumOdd);
    System.out.println("Sum even = "+sumEven);
}
  
// Driver code
public static void main(String args[])
{
    int n = 457892;
    getSum(n);
}
}
  
// This code is contributed by
// Surendra_Gangwar

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Python3

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# Python3 implementation of the approach
  
# Function to return the 
# reverse of a number
def reverse(n):
    rev = 0
    while (n != 0):
        rev = (rev * 10) + (n % 10)
        n //= 10
    return rev
  
# Function to find the sum of the odd
# and even positioned digits in a number
def getSum(n):
  
    n = reverse(n)
    sumOdd = 0
    sumEven = 0
    c = 1
  
    while (n != 0):
  
        # If c is even number then it means
        # digit extracted is at even place
        if (c % 2 == 0):
            sumEven += n % 10
        else:
            sumOdd += n % 10
        n //= 10
        c += 1
  
    print("Sum odd =", sumOdd)
    print("Sum even =", sumEven)
  
# Driver code
n = 457892
getSum(n)
  
# This code is contributed 
# by mohit kumar

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C#

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// C# implementation of the approach
using System;
  
class GFG
{
  
// Function to return the reverse of a number
static int reverse(int n)
{
    int rev = 0;
    while (n != 0) 
    {
        rev = (rev * 10) + (n % 10);
        n /= 10;
    }
    return rev;
}
  
// Function to find the sum of the odd
// and even positioned digits in a number
static void getSum(int n)
{
    n = reverse(n);
    int sumOdd = 0, sumEven = 0, c = 1;
  
    while (n != 0) 
    {
  
        // If c is even number then it means
        // digit extracted is at even place
        if (c % 2 == 0)
            sumEven += n % 10;
        else
            sumOdd += n % 10;
        n /= 10;
        c++;
    }
  
    Console.WriteLine("Sum odd = "+sumOdd);
    Console.WriteLine("Sum even = "+sumEven);
}
  
// Driver code
public static void Main()
{
    int n = 457892;
    getSum(n);
}
}
  
// This code is contributed by
// Akanksha Rai

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PHP

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<?php
// PHP implementation of the above approach 
  
// Function to return the reverse of a number 
function reverse($n
    $rev = 0; 
    while ($n != 0)
    
        $rev = ($rev * 10) + ($n % 10); 
        $n = floor($n / 10); 
    
    return $rev
  
// Function to find the sum of the odd 
// and even positioned digits in a number 
function getSum($n
    $n = reverse($n); 
    $sumOdd = 0; $sumEven = 0; $c = 1; 
  
    while ($n != 0) 
    
  
        // If c is even number then it means 
        // digit extracted is at even place 
        if ($c % 2 == 0) 
            $sumEven += $n % 10; 
        else
            $sumOdd += $n % 10; 
              
        $n = floor($n / 10); 
        $c++; 
    
  
    echo "Sum odd = ", $sumOdd, "\n"
    echo "Sum even = ", $sumEven
  
// Driver code 
$n = 457892; 
getSum($n); 
  
// This code is contributed by Ryuga
?>

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Output:

Sum odd = 20
Sum even = 15


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