Minimize ASCII values sum after removing all occurrences of one character

Given string str, the task is to minimize the sum of ASCII values of each character of str after removing every occurrence of a particular character.

Examples:

Input: str = “geeksforgeeks” 
Output: 977 
‘g’ occurs twice -> 2 * 103 = 206 
‘e’ occurs 4 times -> 4 * 101 = 404 
‘k’ occurs twice -> 2 * 107 = 214 
‘s’ occurs twice -> 2 * 115 = 230 
‘f’ occurs once -> 1 * 102 = 102 
‘o’ occurs once -> 1 * 111 = 111 
‘r’ occurs once -> 1 * 114 = 114 
Total sum = 1381 
In order to minimize the sum, remove all the occurrences of ‘e’ from the string 
And the new sum becomes 1381 – 404 = 977
Input: str = “abcd” 
Output: 294 

Brute Force Approach:

The brute force approach to solve this problem would be to iterate over all the characters of the string and remove every occurrence of each character one by one, calculate the sum of ASCII values of the remaining string, and keep track of the minimum sum obtained. Finally, return the minimum sum obtained.

Below is the implementation of the above approach: 

C++

#include <bits/stdc++.h>
using namespace std;

// Function to return the minimized sum
int getMinimizedSum(string str, int len)
{
    int minSum = INT_MAX;

    for(int i = 0; i < len; i++)
    {
        string temp = "";
        for(int j = 0; j < len; j++)
        {
            if(str[j] != str[i])
            {
                temp += str[j];
            }
        }
        int sum = 0;
        for(int j = 0; j < temp.length(); j++)
        {
            sum += temp[j];
        }
        minSum = min(minSum, sum);
    }

    return minSum;
}

// Driver code
int main()
{
    string str = "geeksforgeeks";
    int len = str.length();
    cout << getMinimizedSum(str, len);

    return 0;
}

Java

import java.util.*;

public class Main {

    // Function to return the minimized sum
    static int getMinimizedSum(String str, int len)
    {
        int minSum = Integer.MAX_VALUE;

        for (int i = 0; i < len; i++) {
            String temp = "";
            for (int j = 0; j < len; j++) {
                if (str.charAt(j) != str.charAt(i)) {
                    temp += str.charAt(j);
                }
            }
            int sum = 0;
            for (int j = 0; j < temp.length(); j++) {
                sum += temp.charAt(j);
            }
            minSum = Math.min(minSum, sum);
        }

        return minSum;
    }

    // Driver code
    public static void main(String[] args)
    {
        String str = "geeksforgeeks";
        int len = str.length();
        System.out.println(getMinimizedSum(str, len));
    }
}

Python3

# code
def get_minimized_sum(s):
    min_sum = float('inf')
    n = len(s)

    for i in range(n):
        temp = ""
        for j in range(n):
            if s[j] != s[i]:
                temp += s[j]
        curr_sum = sum(ord(c) for c in temp)
        min_sum = min(min_sum, curr_sum)

    return min_sum


# Driver code
s = "geeksforgeeks"
print(get_minimized_sum(s))

C#

using System;

class Program {

    // Function to return the minimized sum
    static int GetMinimizedSum(string str, int len)
    {
        int minSum = int.MaxValue;

        for (int i = 0; i < len; i++) {
            string temp = "";
            for (int j = 0; j < len; j++) {
                if (str[j] != str[i]) {
                    temp += str[j];
                }
            }
            int sum = 0;
            for (int j = 0; j < temp.Length; j++) {
                sum += temp[j];
            }
            minSum = Math.Min(minSum, sum);
        }

        return minSum;
    }

    // Driver code
    static void Main(string[] args)
    {
        string str = "geeksforgeeks";
        int len = str.Length;
        Console.WriteLine(GetMinimizedSum(str, len));
    }
}

Javascript

// Function to return the minimized sum

function getMinimizedSum(str, len) {
  let minSum = Number.MAX_SAFE_INTEGER;
  for(let i = 0; i < len; i++) {
    let temp = "";
    for(let j = 0; j < len; j++) {
      if(str[j] != str[i]) {
        temp += str[j];
      }
    }
    let sum = 0;
    for(let j = 0; j < temp.length; j++) {
      sum += temp.charCodeAt(j);
    }
    minSum = Math.min(minSum, sum);
  }
  return minSum;
}

// Driver code

let str = "geeksforgeeks";
let len = str.length;
console.log(getMinimizedSum(str, len));

Output

977

Time Complexity: O(N^2)
Space Complexity: O(1)

Approach:

1. Take the sum of all ASCII values in the given string.
2. Also, store the occurrences of each of the characters of the string.
3. Remove every occurrence of the character which is contributing the maximum value to the sum i.e. whose occurrence * ASCII is maximum.

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 minimized sum
int getMinimizedSum(string str, int len)
{
    int i, maxVal = INT_MIN, sum = 0;

    // To store the occurrences of
    // each character of the string
    int occurrences[26] = { 0 };
    for (i = 0; i < len; i++) {

        // Update the occurrence
        occurrences[str[i] - 'a']++;

        // Calculate the sum
        sum += (int)str[i];
    }

    // Get the character which is contributing
    // the maximum value to the sum
    for (i = 0; i < 26; i++)

        // Count of occurrence of the character
        // multiplied by its ASCII value
        maxVal = max(maxVal, occurrences[i] * (i + 'a'));

    // Return the minimized sum
    return (sum - maxVal);
}

// Driver code
int main()
{
    string str = "geeksforgeeks";
    int len = str.length();
    cout << getMinimizedSum(str, len);

    return 0;
}

Java

// Java implementation of the approach
import java.util.Arrays;
import java.lang.Math;

class GfG
{

    // Function to return the minimized sum
    static int getMinimizedSum(String str, int len)
    {
        int i, maxVal = Integer.MIN_VALUE, sum = 0;

        // To store the occurrences of
        // each character of the string
        int occurrences[] = new int[26];
        Arrays.fill(occurrences, 0);

        for (i = 0; i < len; i++)
        {

            // Update the occurrence
            occurrences[str.charAt(i) - 'a']++;

            // Calculate the sum
            sum += (int)str.charAt(i);
        }

        // Get the character which is contributing
        // the maximum value to the sum
        for (i = 0; i < 26; i++)

            // Count of occurrence of the character
            // multiplied by its ASCII value
            maxVal = Math.max(maxVal, occurrences[i] * (i + 'a'));

        // Return the minimized sum
        return (sum - maxVal);
    }

    // Driver code
    public static void main(String []args){

        String str = "geeksforgeeks";
        int len = str.length();
        System.out.println(getMinimizedSum(str, len));
    }
}

// This code is contributed by Rituraj Jain

Python3

# Python3 implementation of the approach
import sys

# Function to return the minimized sum
def getMinimizedSum(string, length) :

    maxVal = -(sys.maxsize - 1)
    sum = 0;

    # To store the occurrences of
    # each character of the string
    occurrences = [0] * 26;

    for i in range(length) :

        # Update the occurrence
        occurrences[ord(string[i]) -
                    ord('a')] += 1;

        # Calculate the sum
        sum += ord(string[i]);

    # Get the character which is contributing
    # the maximum value to the sum
    for i in range(26) :

        # Count of occurrence of the character
        # multiplied by its ASCII value
        count = occurrences[i] * (i + ord('a'))
        maxVal = max(maxVal, count);

    # Return the minimized sum
    return (sum - maxVal);

# Driver code
if __name__ == "__main__" :

    string = "geeksforgeeks";
    length = len(string);

    print(getMinimizedSum(string, length));

# This code is contributed by Ryuga

C#

// C# implementation of the approach
using System;

class GfG
{

    // Function to return the minimized sum
    static int getMinimizedSum(string str, int len)
    {
        int i, maxVal = Int32.MinValue, sum = 0;

        // To store the occurrences of
        // each character of the string
        int [] occurrences = new int[26];

        for (i = 0; i < len; i++)
        {

            // Update the occurrence
            occurrences[str[i] - 'a']++;

            // Calculate the sum
            sum += (int)str[i];
        }

        // Get the character which is contributing
        // the maximum value to the sum
        for (i = 0; i < 26; i++)

            // Count of occurrence of the character
            // multiplied by its ASCII value
            maxVal = Math.Max(maxVal, occurrences[i] * (i + 'a'));

        // Return the minimized sum
        return (sum - maxVal);
    }

    // Driver code
    public static void Main()
    {
        string str = "geeksforgeeks";
        int len = str.Length;
        Console.WriteLine(getMinimizedSum(str, len));
    }
}

// This code is contributed by ihritik

Javascript

<script>
      // JavaScript implementation of the approach
      // Function to return the minimized sum
      function getMinimizedSum(str, len) {
        var i,
          maxVal = -2147483648,
          sum = 0;

        // To store the occurrences of
        // each character of the string
        var occurrences = new Array(26).fill(0);

        for (i = 0; i < len; i++) {
          // Update the occurrence
          occurrences[str[i].charCodeAt(0) - "a".charCodeAt(0)]++;

          // Calculate the sum
          sum += str[i].charCodeAt(0);
        }

        // Get the character which is contributing
        // the maximum value to the sum
        for (i = 0; i < 26; i++) {
          // Count of occurrence of the character
          // multiplied by its ASCII value
          maxVal = Math.max(maxVal, occurrences[i] * (i + "a".charCodeAt(0)));
        }
        // Return the minimized sum
        return sum - maxVal;
      }

      // Driver code
      var str = "geeksforgeeks";
      var len = str.length;
      document.write(getMinimizedSum(str, len));
    </script>

Output:

977

Time Complexity: O(n), length of the string

Auxiliary Space: O(26)