# Minimize the sum of digits of A and B such that A + B = N

• Last Updated : 01 Mar, 2022

Given an integer N, the task is to find two positive integers A and B such that A + B = N and the sum of digits of A and B is minimum. Print the sum of digits of A and B.
Examples:

Input: N = 16
Output:
(10 + 6) = 16 and (1 + 0 + 6) = 7
is minimum possible.
Input: N = 1000
Output: 10
(900 + 100) = 1000

Approach: If N is a power of 10 then the answer will be 10 otherwise the answer will be the sum of digits of N. It is clear that the answer can not be smaller than the sum of digits of N because the sum of digits decreases whenever a carry is generated. Moreover, when N is a power of 10, obviously the answer can not be 1, so the answer will be 10. Because A or B can not be 0 as both of them must be positive numbers.
Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `// Function to return the minimum``// possible sum of digits of A``// and B such that A + B = n``int` `minSum(``int` `n)``{``    ``// Find the sum of digits of n``    ``int` `sum = 0;``    ``while` `(n > 0) {``        ``sum += (n % 10);``        ``n /= 10;``    ``}` `    ``// If num is a power of 10``    ``if` `(sum == 1)``        ``return` `10;` `    ``return` `sum;``}` `// Driver code``int` `main()``{``    ``int` `n = 1884;` `    ``cout << minSum(n);` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach` `class` `GFG``{` `// Function to return the minimum``// possible sum of digits of A``// and B such that A + B = n``static` `int` `minSum(``int` `n)``{``    ``// Find the sum of digits of n``    ``int` `sum = ``0``;``    ``while` `(n > ``0``)``    ``{``        ``sum += (n % ``10``);``        ``n /= ``10``;``    ``}` `    ``// If num is a power of 10``    ``if` `(sum == ``1``)``        ``return` `10``;` `    ``return` `sum;``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``int` `n = ``1884``;` `    ``System.out.print(minSum(n));` `}``}` `// This code is contributed by 29AjayKumar`

## Python3

 `# Python implementation of the approach` `# Function to return the minimum``# possible sum of digits of A``# and B such that A + B = n``def` `minSum(n) :` `    ``# Find the sum of digits of n``    ``sum` `=` `0``;``    ``while` `(n > ``0``) :``        ``sum` `+``=` `(n ``%` `10``);``        ``n ``/``/``=` `10``;` `    ``# If num is a power of 10``    ``if` `(``sum` `=``=` `1``) :``        ``return` `10``;` `    ``return` `sum``;` `# Driver code``if` `__name__ ``=``=` `"__main__"` `:``    ``n ``=` `1884``;` `    ``print``(minSum(n));` `# This code is contributed by AnkitRai01`

## C#

 `// C# implementation of the approach``using` `System;` `class` `GFG``{` `// Function to return the minimum``// possible sum of digits of A``// and B such that A + B = n``static` `int` `minSum(``int` `n)``{``    ``// Find the sum of digits of n``    ``int` `sum = 0;``    ``while` `(n > 0)``    ``{``        ``sum += (n % 10);``        ``n /= 10;``    ``}` `    ``// If num is a power of 10``    ``if` `(sum == 1)``        ``return` `10;` `    ``return` `sum;``}` `// Driver code``public` `static` `void` `Main(String[] args)``{``    ``int` `n = 1884;` `    ``Console.Write(minSum(n));``}``}` `// This code is contributed by PrinciRaj1992`

## Javascript

 ``

Output:

`21`

Time Complexity: O(log n)

Auxiliary Space: O(1)

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