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

Another approach: The problem can be solved without reversing the number. We can extract all the digits from the number one by one from the end. If the original number was odd then the last digit must be odd positioned else it will be even positioned. After processing a digit, we can invert the state from odd to even and vice versa.

Below is the implementation of the above approach:

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to find the sum of the odd
// and even positioned digits in a number
void getSum(int n)
{
  
    // If n is odd then the last digit
    // will be odd positioned
    bool isOdd = (n % 2 == 1) ? true : false;
  
    // To store the respective sums
    int sumOdd = 0, sumEven = 0;
  
    // While there are digits left process
    while (n != 0) {
  
        // If current digit is odd positioned
        if (isOdd)
            sumOdd += n % 10;
  
        // Even positioned digit
        else
            sumEven += n % 10;
  
        // Invert state
        isOdd = !isOdd;
  
        // Remove last digit
        n /= 10;
    }
  
    cout << "Sum odd = " << sumOdd << "\n";
    cout << "Sum even = " << sumEven;
}
  
// Driver code
int main()
{
    int n = 457892;
    getSum(n);
  
    return 0;
}

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Output:

Sum odd = 20
Sum even = 15


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