Maximize ascii sum removing K characters in String
Last Updated :
08 Feb, 2024
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:
- We start by initializing a variable maxSum to the sum of the ASCII values of all the characters in the string. This is because the maximum ASCII sum we can get by removing K characters from the string cannot be greater than the sum of all the characters in the string.
- We then iterate over all possible combinations of K characters to remove from the string. Since we want to remove exactly K characters, we use a bitmask approach to generate all possible combinations of k characters from the string. We iterate over all numbers from 0 to 2^n – 1, where n is the length of the string, and use the binary representation of each number as a mask to indicate which characters to remove.
- For each combination of K characters, we calculate the ASCII sum of the remaining characters in the string. To do this, we iterate over all the characters in the string and check if the corresponding bit in the mask is set or not. If the bit is not set, we add the ASCII value of the character to the sum.
- If the calculated ASCII sum is greater than the current value of maxSum, we update maxSum to the new value.
- Finally, we return maxSum as the maximum ASCII sum that can be obtained by removing k characters from the string.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int maxAsciiSum(string s, int k)
{
int n = s.length();
int maxSum = 0;
for (int mask = 0; mask < (1 << n); mask++) {
if (__builtin_popcount(mask) == k) {
int sum = 0;
for (int i = 0; i < n; i++) {
if ((mask & (1 << i)) == 0) {
sum += int(s[i]);
}
}
maxSum = max(maxSum, sum);
}
}
return maxSum;
}
int main()
{
string s = "abcABd";
int k = 3;
cout << maxAsciiSum(s, k) << endl;
return 0;
}
|
Java
import java.util.*;
public class Main {
public static int maxAsciiSum(String s, int k) {
int n = s.length();
int maxSum = 0;
for (int mask = 0; mask < (1 << n); mask++) {
if (Integer.bitCount(mask) == k) {
int sum = 0;
for (int i = 0; i < n; i++) {
if ((mask & (1 << i)) == 0) {
sum += (int) s.charAt(i);
}
}
maxSum = Math.max(maxSum, sum);
}
}
return maxSum;
}
public static void main(String[] args) {
String s = "abcABd";
int k = 3;
System.out.println(maxAsciiSum(s, k));
}
}
|
Python
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
if __name__ == "__main__":
s = "abcABd"
k = 3
print(max_ascii_sum(s, k))
|
C#
using System;
class Program
{
static int MaxAsciiSum(string s, int k)
{
int n = s.Length;
int maxSum = 0;
for (int mask = 0; mask < (1 << n); mask++)
{
if (CountSetBits(mask) == k)
{
int sum = 0;
for (int i = 0; i < n; i++)
{
if ((mask & (1 << i)) == 0)
{
sum += (int)s[i];
}
}
maxSum = Math.Max(maxSum, sum);
}
}
return maxSum;
}
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;
Console.WriteLine(MaxAsciiSum(s, k));
}
}
|
Javascript
function maxAsciiSum(s, k) {
const n = s.length;
let maxSum = 0;
for (let mask = 0; mask < (1 << n); mask++) {
if (countBits(mask) === k) {
let sum = 0;
for (let i = 0; i < n; i++) {
if ((mask & (1 << i)) === 0) {
sum += s.charCodeAt(i);
}
}
maxSum = Math.max(maxSum, sum);
}
}
return maxSum;
}
function countBits(mask) {
let count = 0;
while (mask > 0) {
count += mask & 1;
mask >>= 1;
}
return count;
}
const s = "abcABd";
const k = 3;
console.log(maxAsciiSum(s, k));
|
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.
- Initialize maxSum variable to 0;
- Sort the given string.
- Add the ASCII value of characters from index k to the last index in the maxSum.
- Return maxSum variable.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int maxAsciiSum(string s, int k)
{
int n = s.length();
int maxSum = 0;
sort(s.begin(), s.end());
for (int i = k; i < n; i++) {
maxSum += s[i];
}
return maxSum;
}
int main()
{
string s = "abcABd";
int k = 3;
cout << maxAsciiSum(s, k) << endl;
return 0;
}
|
Java
import java.util.*;
public class Main {
public static int maxAsciiSum(String s, int k)
{
int n = s.length();
int maxSum = 0;
char array[] = s.toCharArray();
Arrays.sort(array);
s = new String(array);
for (int i = k; i < n; i++) {
maxSum += s.charAt(i);
}
return maxSum;
}
public static void main(String[] args)
{
String s = "abcABd";
int k = 3;
System.out.println(maxAsciiSum(s, k));
}
}
|
Python
def max_ascii_sum(s, k):
n = len(s)
max_sum = 0
s = sorted(s)
for i in range(k, n):
max_sum += ord(s[i])
return max_sum
if __name__ == "__main__":
s = "abcABd"
k = 3
print(max_ascii_sum(s, k))
|
C#
using System;
using System.Linq;
class Program
{
static int MaxAsciiSum(string s, int k)
{
int n = s.Length;
int maxSum = 0;
char[] charArray = s.ToCharArray();
Array.Sort(charArray);
s = new string(charArray);
for (int i = k; i < n; i++)
{
maxSum += (int)s[i];
}
return maxSum;
}
static void Main()
{
string s = "abcABd";
int k = 3;
Console.WriteLine(MaxAsciiSum(s, k));
}
}
|
Javascript
function maxAsciiSum(s, k) {
let n = s.length;
let maxSum = 0;
let sortedString = s.split('').sort().join('');
for (let i = k; i < n; i++) {
maxSum += sortedString.charCodeAt(i);
}
return maxSum;
}
let s = "abcABd";
let k = 3;
console.log(maxAsciiSum(s, k));
|
Time Complexity: O( n*log(n) )
Auxiliary Space: O(n)
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