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++

// 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;
}

Java

// 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

Python3

# 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

C#

// 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

PHP

<?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
?>

Javascript

<script>

//JavaScript implementation of the approach

    // Function to return the
    // reverse of a number
    function reverse(n) {
    let rev = 0;
    while (n != 0) {
        rev = (rev * 10) + (n % 10);
        n = Math.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);
        let 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 = Math.floor(n / 10);
        c++;
    }

    document.write("Sum odd = " + sumOdd);
    document.write("<br>");
    document.write("Sum even = " + sumEven);
}
      let n = 457892;
      // function call
      getSum(n);

// This code is contributed by Surbhi Tyagi

</script>

Output:

Sum odd = 20
Sum even = 15

Time Complexity: O(log₁₀n)
Auxiliary Space: O(1)

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: 

C++

// 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;
}

Java

// Java implementation of the above approach
class GFG{

// Function to find the sum of the odd
// and even positioned digits in a number
static void getSum(int n)
{

    // If n is odd then the last digit
    // will be odd positioned
    boolean 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;
    }
    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 jrishabh99

Python3

# Python3 implementation of the approach

# Function to find the sum of the odd
# and even positioned digits in a number
def getSum(n):

    # If n is odd then the last digit
    # will be odd positioned
    if (n % 2 == 1) :
        isOdd = True
    else:
        isOdd = False

    # To store the respective sums
    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 = not isOdd

        # Remove last digit
        n //= 10

    print( "Sum odd = " , sumOdd )
    print("Sum even = " ,sumEven)

# Driver code
if __name__ =="__main__":
    n = 457892
    getSum(n)

# This code is contributed by chitranayal

C#

// C# implementation of the above approach
using System;

class GFG{

// Function to find the sum of the odd
// and even positioned digits in a number
static 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;
    }
    Console.WriteLine("Sum odd = " + sumOdd);
    Console.Write("Sum even = " + sumEven);
}

// Driver code
static public void Main ()
{
    int n = 457892;

    getSum(n);
}
}

// This code is contributed by offbeat

Javascript

<script>

// Javascript implementation of the approach

// Function to find the sum of the odd
// and even positioned digits in a number
function getSum(n)
{

    // If n is odd then the last digit
    // will be odd positioned
    let isOdd = (n % 2 == 1) ? true : false;

    // To store the respective sums
    let 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 = Math.floor(n/10);
    }

    document.write("Sum odd = " + sumOdd + "<br>");
    document.write("Sum even = " + sumEven);
}

// Driver code

    let n = 457892;
    getSum(n);


// This code is contributed by Mayank Tyagi

</script>

Output:

Sum odd = 20
Sum even = 15

Time Complexity: O(log₁₀n) as while loop would run for log₁₀n times
Auxiliary Space: O(1)

Method #3:Using string() method:

1. Convert the integer to string. Traverse the string and store all even indices sum in one variable and all odd indices sum in another variable.

Below is the implementation:

C++

// 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)
{

    // To store the respective sums
    int sumOdd = 0, sumEven = 0;

    // Converting integer to string
    string num = to_string(n);

    // Traversing the string
    for(int i = 0; i < num.size(); i++)
    {
        if (i % 2 == 0)
            sumOdd = sumOdd + (int(num[i]) - 48);
        else
            sumEven = sumEven + (int(num[i]) - 48);
    }
    cout << "Sum odd = " << sumOdd << "\n";
    cout << "Sum even = " << sumEven << "\n";
}

// Driver code
int main()
{
    int n = 457892;
    getSum(n);

    return 0;
}

// This code is contributed by souravmahato348

Java

// Java implementation of the approach

import java.util.*;

class GFG{

static void getSum(int n)
{
    // To store the respective sum
    int sumOdd = 0;
    int sumEven = 0;

    // Converting integer to String
    String num = String.valueOf(n);

    // Traversing the String
    for(int i = 0; i < num.length(); i++)
        if (i % 2 == 0)
            sumOdd = sumOdd + (num.charAt(i) - '0');
        else
            sumEven = sumEven + (num.charAt(i) - '0');

    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);
}
}

// Code contributed by swarnalii

Python3

# Python3 implementation of the approach

# Function to find the sum of the odd
# and even positioned digits in a number
def getSum(n):

    # To store the respective sums
    sumOdd = 0
    sumEven = 0

    # Converting integer to string
    num = str(n)

    # Traversing the string
    for i in range(len(num)):
        if(i % 2 == 0):
            sumOdd = sumOdd+int(num[i])
        else:
            sumEven = sumEven+int(num[i])

    print("Sum odd = ", sumOdd)
    print("Sum even = ", sumEven)


# Driver code
if __name__ == "__main__":
    n = 457892
    getSum(n)

# This code is contributed by vikkycirus

C#

// C# implementation of the approach
using System;

class GFG{

static void getSum(int n)
{

    // To store the respective sum
    int sumOdd = 0;
    int sumEven = 0;

    // Converting integer to String
    String num = n.ToString();

    // Traversing the String
    for(int i = 0; i < num.Length; i++)
        if (i % 2 == 0)
            sumOdd = sumOdd + (num[i] - '0');
        else
            sumEven = sumEven + (num[i] - '0');

    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 subhammahato348

Javascript

<script>
// Javascript implementation of the approach

function getSum(n)
{
    // To store the respective sum
    let sumOdd = 0;
    let sumEven = 0;

    // Converting integer to String
    let num = (n).toString();

    // Traversing the String
    for(let i = 0; i < num.length; i++)
        if (i % 2 == 0)
            sumOdd = sumOdd + (num[i] - '0');
        else
            sumEven = sumEven + (num[i] - '0');

    document.write("Sum odd = " + sumOdd+"<br>");
    document.write("Sum even = " + sumEven+"<br>");
}

// Driver code
let n = 457892;
getSum(n);

// This code is contributed by unknown2108
</script>

Output

Sum odd = 20
Sum even = 15

Time Complexity: O(log₁₀n), as the length of the string will be log₁₀n.
Auxiliary Space: O(log₁₀n)