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Maximize ascii sum removing K characters in String

Given a string s and an integer K, the task is to remove K characters from the string such that the sum of ASCII (American Standard Code for Information Interchange) values of the remaining characters is maximized.

Examples:



Input: s = “word”, K = 2
Output: 233
Explanation: We need to remove exactly 2 characters from the string “word” to maximize the ASCII sum of the remaining characters. By removing ‘o’ and ‘d’, we obtain the maximum sum of 233, which corresponds to the ASCII sum of the remaining characters (‘r’ and ‘o’).

Input: s = “abcABd”, K = 3
Output: 297
Explanation: We need to remove exactly 3 characters from the string “abcABd” to maximize the ASCII sum of the remaining characters. By removing ‘a’, ‘A’, and ‘B’, we obtain the maximum sum of 297, which corresponds to the ASCII sum of the remaining characters (‘b’, ‘c’, and ‘d’).



Approach: Bruteforce Approach 

This can be solved with the following idea:

This can be solved by keeping some observations in the notice.

Below are the steps involved in the implementation:

Below is the implementation of the above approach:




// C++ Implementation
#include <bits/stdc++.h>
using namespace std;
 
// Function to find maximum value by
// removing characters
int maxAsciiSum(string s, int k)
{
 
    int n = s.length();
    int maxSum = 0;
 
    for (int mask = 0; mask < (1 << n); mask++) {
 
        // Check if exactly k characters
        // are removed
        if (__builtin_popcount(mask) == k) {
            int sum = 0;
            for (int i = 0; i < n; i++) {
 
                // Check if i-th character
                // is not removed
                if ((mask & (1 << i)) == 0) {
                    sum += int(s[i]);
                }
            }
            maxSum = max(maxSum, sum);
        }
    }
    return maxSum;
}
 
// Driver code
int main()
{
    string s = "abcABd";
    int k = 3;
 
    // Function call
    cout << maxAsciiSum(s, k) << endl;
    return 0;
}




// java code implementation
import java.util.*;
 
public class Main {
    // Function to find maximum value by
    // removing characters
    public static int maxAsciiSum(String s, int k) {
        int n = s.length();
        int maxSum = 0;
 
        for (int mask = 0; mask < (1 << n); mask++) {
            // Check if exactly k characters
            // are removed
            if (Integer.bitCount(mask) == k) {
                int sum = 0;
                for (int i = 0; i < n; i++) {
                    // Check if i-th character
                    // is not removed
                    if ((mask & (1 << i)) == 0) {
                        sum += (int) s.charAt(i);
                    }
                }
                maxSum = Math.max(maxSum, sum);
            }
        }
        return maxSum;
    }
 
    // Driver code
    public static void main(String[] args) {
        String s = "abcABd";
        int k = 3;
 
        // Function call
        System.out.println(maxAsciiSum(s, k));
    }
}




def max_ascii_sum(s, k):
    n = len(s)
    max_sum = 0
 
    for mask in range(1 << n):
        if bin(mask).count('1') == k:
            current_sum = 0
            for i in range(n):
                if (mask & (1 << i)) == 0:
                    current_sum += ord(s[i])
            max_sum = max(max_sum, current_sum)
 
    return max_sum
 
# Driver code
if __name__ == "__main__":
    s = "abcABd"
    k = 3
 
    # Function call
    print(max_ascii_sum(s, k))
    




using System;
 
class Program
{
    // Function to find maximum value by
    // removing characters
    static int MaxAsciiSum(string s, int k)
    {
        int n = s.Length;
        int maxSum = 0;
 
        for (int mask = 0; mask < (1 << n); mask++)
        {
            // Check if exactly k characters
            // are removed
            if (CountSetBits(mask) == k)
            {
                int sum = 0;
                for (int i = 0; i < n; i++)
                {
                    // Check if i-th character
                    // is not removed
                    if ((mask & (1 << i)) == 0)
                    {
                        sum += (int)s[i];
                    }
                }
                maxSum = Math.Max(maxSum, sum);
            }
        }
        return maxSum;
    }
 
    // Function to count the number of set bits in an integer
    static int CountSetBits(int n)
    {
        int count = 0;
        while (n > 0)
        {
            count += n & 1;
            n >>= 1;
        }
        return count;
    }
 
    static void Main()
    {
        string s = "abcABd";
        int k = 3;
 
        // Function call
        Console.WriteLine(MaxAsciiSum(s, k));
    }
}




function maxAsciiSum(s, k) {
    const n = s.length;
    let maxSum = 0;
 
    // Iterate through all possible combinations of characters to remove
    for (let mask = 0; mask < (1 << n); mask++) {
 
        // Check if exactly k characters are removed
        if (countBits(mask) === k) {
            let sum = 0;
 
            // Calculate the sum of ASCII values for characters that are not removed
            for (let i = 0; i < n; i++) {
 
                // Check if i-th character is not removed
                if ((mask & (1 << i)) === 0) {
                    sum += s.charCodeAt(i);
                }
            }
 
            maxSum = Math.max(maxSum, sum);
        }
    }
 
    return maxSum;
}
 
// Helper function to count the number of set bits (1s) in a binary representation
function countBits(mask) {
    let count = 0;
    while (mask > 0) {
        count += mask & 1;
        mask >>= 1;
    }
    return count;
}
 
// Driver code
const s = "abcABd";
const k = 3;
 
// Function call
console.log(maxAsciiSum(s, k));

Output
297

Time Complexity: O(2^n * n)
Auxiliary Space: O(1)

Approach: Greedy Approach 

This can be solved with the following idea:

If we remove minimum ascii values characters then we will get maximized sum of remaining character.

Below is the implementation of the above approach:




// C++ Implementation
#include <bits/stdc++.h>
using namespace std;
 
// Function to find maximum value by
// removing characters
int maxAsciiSum(string s, int k)
{
 
    int n = s.length();
    int maxSum = 0;
 
    // sorting the string s
    sort(s.begin(), s.end());
 
    // we start loop from k because we removed starting k
    // characters so we will not add their ascii values
    for (int i = k; i < n; i++) {
        maxSum += s[i];
    }
 
    // return maxSum of remaining characters
    return maxSum;
}
 
// Driver code
int main()
{
    string s = "abcABd";
    int k = 3;
 
    // Function call
    cout << maxAsciiSum(s, k) << endl;
    return 0;
}




// java code implementation
import java.util.*;
 
public class Main {
    // Function to find maximum value by
    // removing characters
    public static int maxAsciiSum(String s, int k)
    {
        int n = s.length();
        int maxSum = 0;
 
        // sorting the string s
        char array[] = s.toCharArray();
        Arrays.sort(array);
        s = new String(array);
 
        // we start loop from k because we removed starting
        // k characters so we will not add their ascii
        // values
        for (int i = k; i < n; i++) {
            maxSum += s.charAt(i);
        }
 
        // return maxSum of remaining characters
        return maxSum;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        String s = "abcABd";
        int k = 3;
 
        // Function call
        System.out.println(maxAsciiSum(s, k));
    }
}




def max_ascii_sum(s, k):
    n = len(s)
    max_sum = 0
 
    # Sorting the string s
    s = sorted(s)
 
    # We start the loop from k because we removed the starting k characters
    # so we will not add their ASCII values
    for i in range(k, n):
        max_sum += ord(s[i])
 
    # Return max_sum of remaining characters
    return max_sum
 
# Driver code
if __name__ == "__main__":
    s = "abcABd"
    k = 3
 
    # Function call
    print(max_ascii_sum(s, k))




using System;
using System.Linq;
 
class Program
{
    static int MaxAsciiSum(string s, int k)
    {
        int n = s.Length;
        int maxSum = 0;
 
        // Sorting the string s
        char[] charArray = s.ToCharArray();
        Array.Sort(charArray);
        s = new string(charArray);
 
        // We start the loop from k because we removed the starting k characters,
        // so we will not add their ASCII values.
        for (int i = k; i < n; i++)
        {
            maxSum += (int)s[i];
        }
 
        // Return the maxSum of the remaining characters
        return maxSum;
    }
 
    static void Main()
    {
        string s = "abcABd";
        int k = 3;
 
        // Function call
        Console.WriteLine(MaxAsciiSum(s, k));
    }
}




// JavaScript code for the above approach
 
function maxAsciiSum(s, k) {
    let n = s.length;
    let maxSum = 0;
 
    // Sorting the string s
    let sortedString = s.split('').sort().join('');
 
    // We start the loop from k because we removed the first k characters
    // so we will not add their ASCII values
    for (let i = k; i < n; i++) {
        maxSum += sortedString.charCodeAt(i);
    }
 
    // Return the maxSum of the remaining characters
    return maxSum;
}
 
// Driver code
let s = "abcABd";
let k = 3;
 
// Function call
console.log(maxAsciiSum(s, k));
 
// This code is contributed by Abhinav Mahajan (abhinav_m22)

Output
297

Time Complexity: O( n*log(n) )
Auxiliary Space: O(n)


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