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++

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// 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;
}

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Java














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// 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 



















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Python 3














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# 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 



















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C#














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// 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)); 


} 


} 


  



















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PHP














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<?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 


?> 



















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Output:


17







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