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

• Difficulty Level : Medium
• Last Updated : 10 Oct, 2022

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 ` `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

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## Javascript

 ``

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 ` `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

 ``

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 ` `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

 ``

Output

```Sum odd = 20
Sum even = 15```

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

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