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

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: 7 
(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 <bits/stdc++.h>
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

<script>

// Javascript implementation of the approach

// Function to return the minimum
// possible sum of digits of A
// and B such that A + B = n
function minSum(n)
{

    // Find the sum of digits of n
    var sum = 0;
    while (n > 0) {
        sum += (n % 10);
        n = parseInt(n/10);
    }

    // If num is a power of 10
    if (sum == 1)
        return 10;

    return sum;
}

// Driver code
var n = 1884;
document.write( minSum(n));

// This code is contributed by famously.
</script>

Output:

21

Time Complexity: O(log n)

Auxiliary Space: O(1)