Divide a number into two parts such that sum of digits is maximum

Given a number N. The task is to find the maximum possible value of SumOfDigits(A) + SumOfDigits(B) such that A + B = n (0<=A, B<=n).

Examples:

Input: N = 35
Output: 17
35 = 9 + 26
SumOfDigits(26) = 8, SumOfDigits(9) = 9
So, 17 is the answer.

Input: N = 7
Output: 7

Approach: The idea is to divide the number into two parts A and B such that A is in terms of 9 i.e. closest number to N and B = N-A. For example, N = 35, Smallest Number that is closest to 35 is 29. 

C++

// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;

// Returns sum of digits of x
int sumOfDigitsSingle(int x)
{
    int ans = 0;
    while (x) {
        ans += x % 10;
        x /= 10;
    }
    return ans;
}

// Returns closest number to x in terms of 9's.
int closest(int x)
{
    int ans = 0;
    while (ans * 10 + 9 <= x)
        ans = ans * 10 + 9;

    return ans;
}

int sumOfDigitsTwoParts(int N)
{
    int A = closest(N);
    return sumOfDigitsSingle(A) + sumOfDigitsSingle(N - A);
}

// Driver code
int main()
{
    int N = 35;
    cout << sumOfDigitsTwoParts(N);
    return 0;
}

C

// C implementation of above approach
#include <stdio.h>

// Returns sum of digits of x
int sumOfDigitsSingle(int x)
{
    int ans = 0;
    while (x) {
        ans += x % 10;
        x /= 10;
    }
    return ans;
}

// Returns closest number to x in terms of 9's.
int closest(int x)
{
    int ans = 0;
    while (ans * 10 + 9 <= x)
        ans = ans * 10 + 9;

    return ans;
}

int sumOfDigitsTwoParts(int N)
{
    int A = closest(N);
    return sumOfDigitsSingle(A) + sumOfDigitsSingle(N - A);
}

// Driver code
int main()
{
    int N = 35;
    printf("%d",sumOfDigitsTwoParts(N));
    return 0;
}

// This code is contributed by kothavvsaakash.

Java

// Java implementation of above approach
// Returns sum of digits of x
import java.util.*;
import java.lang.*;
import java.io.*;

class GFG
{
static int sumOfDigitsSingle(int x)
{
    int ans = 0;
    while (x != 0)
    {
        ans += x % 10;
        x /= 10;
    }
    return ans;
}

// Returns closest number to x
// in terms of 9's.
static int closest(int x)
{
    int ans = 0;
    while (ans * 10 + 9 <= x)
        ans = ans * 10 + 9;

    return ans;
}

static int sumOfDigitsTwoParts(int N)
{
    int A = closest(N);
    return sumOfDigitsSingle(A) +
           sumOfDigitsSingle(N - A);
}

// Driver code
public static void main(String args[])
{
    int N = 35;
    System.out.print(sumOfDigitsTwoParts(N));
}
}

// This code is contributed by
// Subhadeep Gupta

Python3

# Python 3 implementation of above approach

#  Returns sum of digits of x
def sumOfDigitsSingle(x) :
    ans = 0
    while x :
        ans += x % 10
        x //= 10

    return ans

# Returns closest number to x in terms of 9's
def closest(x) :
    ans = 0
    while (ans * 10 + 9 <= x) :
        ans = ans * 10 + 9

    return ans

def sumOfDigitsTwoParts(N) :
    A = closest(N)

    return sumOfDigitsSingle(A) + sumOfDigitsSingle(N - A)


# Driver Code
if __name__ == "__main__" :

    N = 35
    print(sumOfDigitsTwoParts(N))

# This code is contributed by ANKITRAI1

C#

// C# implementation of above approach
// Returns sum of digits of x

using System;
class GFG
{
static int sumOfDigitsSingle(int x)
{
    int ans = 0;
    while (x != 0)
    {
        ans += x % 10;
        x /= 10;
    }
    return ans;
}

// Returns closest number to x
// in terms of 9's.
static int closest(int x)
{
    int ans = 0;
    while (ans * 10 + 9 <= x)
        ans = ans * 10 + 9;

    return ans;
}

static int sumOfDigitsTwoParts(int N)
{
    int A = closest(N);
    return sumOfDigitsSingle(A) +
           sumOfDigitsSingle(N - A);
}

// Driver code
public static void Main()
{
    int N = 35;
    Console.Write(sumOfDigitsTwoParts(N));
}
}

PHP

<?php
// PHP implementation of above approach

// Returns sum of digits of x
function sumOfDigitsSingle($x)
{
    $ans = 0;
    while ($x)
    {
        $ans += $x % 10;
        $x /= 10;
    }
    return $ans;
}

// Returns closest number
// to x in terms of 9's.
function closest($x)
{
    $ans = 0;
    while ($ans * 10 + 9 <= $x)
        $ans = $ans * 10 + 9;

    return $ans;
}

function sumOfDigitsTwoParts($N)
{
    $A = closest($N);
    return sumOfDigitsSingle($A) +
           sumOfDigitsSingle($N - $A);
}

// Driver code
$N = 35;
echo sumOfDigitsTwoParts($N);

// This code is contributed
// by inder_verma
?>

Javascript

<script>

// JavaScript implementation of above approach

// Returns sum of digits of x
function sumOfDigitsSingle(x)
{
    let ans = 0;
    while (x)
    {
        ans += x % 10;
        x = Math.floor(x / 10);
    }
    return ans;
}

// Returns closest number to x
// in terms of 9's.
function closest(x)
{
    let ans = 0;

    while(ans * 10 + 9 <= x)
        ans = ans * 10 + 9;

    return ans;
}

function sumOfDigitsTwoParts(N)
{
    let A = closest(N);
    return sumOfDigitsSingle(A) +
           sumOfDigitsSingle(N - A);
}

// Driver code
let N = 35;

document.write(sumOfDigitsTwoParts(N));

// This code is contributed by Surbhi Tyagi.

</script>

Output:

17

Time Complexity: O(log₁₀N)
Auxiliary Space: O(1)