Generate longest String with character sum at most K by deleting letters

Last Updated : 21 Feb, 2023

Given a string str and an integer K, the task is to find the longest string that can be made by deleting letters from the given string such that the sum of the characters of the remaining string is at most K.

Note: The value of the string is calculated as the sum of the value of characters (a value is 1, b value is 2...., z value is 26).

Examples:

Input: str = "geeksforgeeks", K = 15
Output: eee
Explanation: After deleting the characters string gets reduced to eee whose value is 5 + 5 + 5 = 15 which is less than or equal to K i.e. 15

Input: str = "abca", K = 6
Output: aba
Explanation: Initial value of str 1 + 2 + 3 + 1 = 7, after deleting the letter c, the string gets reduced to aba whose value is 1+2+1 = 4 which is less than K i.e. 6.

Approach: The problem can be solved using Greedy approach based on the following idea:

We must delete letters with the highest value first. So, we sort the string str in decreasing order and will be deleting letters from the starting of string str as long as the initial value of string str is greater than the given value K.

Follow the steps mentioned below to implement the idea:

Calculate the initial value of the given string.
Sort the string str in decreasing order
Start removing the value of the current letter from the initial value as long as the initial value is greater than K.
- Store the removed characters in the map
Iterate through the given string once again and
- If the current letter does not exist in the map take it into the resultant string
- Else decrease the frequency of the letter
Return the resultant string

Below is the implementation of the above approach:

C++

// C++ code to implement the approach  #include <bits/stdc++.h> using namespace std;  // Function for finding out string with value // less than or equal to K string minimise_str(string str, int K) {     int initial_value = 0;      // Calculate initial value of string     for (int i = 0; i < str.length(); i++) {         initial_value += (str[i] - 'a') + 1;     }      string temp = str;      // Sort the string in decreasing order     sort(str.begin(), str.end());     reverse(str.begin(), str.end());      // Store the deleted letters     unordered_map<char, int> mpp;     int i = 0;      // Remove letters as long as the initial     // value is greater than K     while (initial_value > K) {         initial_value -= (str[i] - 'a') + 1;         mpp[str[i]]++;         i++;     }      // Store resultant string     string ans = "";     for (int i = 0; i < temp.size(); i++) {          // If letter do exist in map, decrease         // frequency         if (mpp[temp[i]] > 0) {             mpp[temp[i]]--;         }         // Else store the letter in resultant         // string         else {             ans += temp[i];         }     }      // Return resultant string     return ans; }  // Driver code int main() {     string str = "geeksforgeeks";     int K = 15;      // Function call     cout << minimise_str(str, K);      return 0; }

Java

// Java code to implement the approach  import java.io.*; import java.util.*;  class GFG {      // Function for finding out string with value less than     // or equal to K     static String minimiseStr(String str, int K)     {         int initialValue = 0;          // Calculate initial value of string         for (int i = 0; i < str.length(); i++) {             initialValue += (str.charAt(i) - 'a') + 1;         }          String temp = str;          // Sort the string in decreasing order         char[] chars = str.toCharArray();         Arrays.sort(chars);         str = new StringBuilder(new String(chars))                   .reverse()                   .toString();          // Store the deleted letters         Map<Character, Integer> mpp = new HashMap<>();         int i = 0;          // Remove letters as long as the initial value is         // greater than K         while (initialValue > K) {             initialValue -= (str.charAt(i) - 'a') + 1;             mpp.put(str.charAt(i),                     mpp.getOrDefault(str.charAt(i), 0) + 1);             i++;         }          // Store resultant string         StringBuilder ans = new StringBuilder();         for (int j = 0; j < temp.length(); j++) {              // If letter do exist in map, decrease frequency             if (mpp.containsKey(temp.charAt(j))                 && mpp.get(temp.charAt(j)) > 0) {                 int freq = mpp.get(temp.charAt(j));                 mpp.put(temp.charAt(j), freq - 1);             }             // Else store the letter in resultant string             else {                 ans.append(temp.charAt(j));             }         }          // Return resultant string         return ans.toString();     }      public static void main(String[] args)     {         String str = "geeksforgeeks";         int K = 15;          // Function call         System.out.println(minimiseStr(str, K));     } }  // This code is contributed by karthik.

Python3

# python code imple.  import collections  def minimise_str(str, K):     initial_value = 0      # Calculate initial value of string     for i in range(len(str)):         initial_value += (ord(str[i]) - ord('a') + 1)      temp = str      # Sort the string in decreasing order     str = ''.join(sorted(str, reverse=True))      # Store the deleted letters     mpp = collections.defaultdict(int)     i = 0      # Remove letters as long as the initial     # value is greater than K     while initial_value > K:         initial_value -= (ord(str[i]) - ord('a') + 1)         mpp[str[i]] += 1         i += 1      # Store resultant string     ans = ""     for i in range(len(temp)):          # If letter do exist in map, decrease         # frequency         if mpp[temp[i]] > 0:             mpp[temp[i]] -= 1         # Else store the letter in resultant         # string         else:             ans += temp[i]      # Return resultant string     return ans  # Driver code str = "geeksforgeeks" K = 15  # Function call print(minimise_str(str, K))  #code by ksam24000

using System; using System.Collections.Generic; using System.Linq;  class GFG { // Function for finding out string with value // less than or equal to K static string MinimiseString(string str, int K) { int initial_value = 0;       // Calculate initial value of string     foreach (char c in str)     {         initial_value += (c - 'a') + 1;     }      string temp = str;      // Sort the string in decreasing order     char[] arr = str.ToCharArray();     Array.Sort(arr);     Array.Reverse(arr);     str = new string(arr);      // Store the deleted letters     Dictionary<char, int> mpp = new Dictionary<char, int>();     int i = 0;      // Remove letters as long as the initial     // value is greater than K     while (initial_value > K)     {         initial_value -= (str[i] - 'a') + 1;         if (mpp.ContainsKey(str[i]))         {             mpp[str[i]]++;         }         else         {             mpp[str[i]] = 1;         }         i++;     }      // Store resultant string     string ans = "";     foreach (char c in temp)     {         // If letter do exist in map, decrease         // frequency         if (mpp.ContainsKey(c) && mpp[c] > 0)         {             mpp[c]--;         }         // Else store the letter in resultant         // string         else         {             ans += c;         }     }      // Return resultant string     return ans; }  // Driver code static void Main(string[] args) {     string str = "geeksforgeeks";     int K = 15;      // Function call     Console.WriteLine(MinimiseString(str, K)); } }

JavaScript

function minimiseStr(str, K) {   let initialValue = 0;    // Calculate initial value of string   for (let i = 0; i < str.length; i++) {     initialValue += (str.charCodeAt(i) - 'a'.charCodeAt(0)) + 1;   }    let temp = str;    // Sort the string in decreasing order   str = str.split('').sort().reverse().join('');    // Store the deleted letters   let mpp = {};   let i = 0;    // Remove letters as long as the initial   // value is greater than K   while (initialValue > K) {     initialValue -= (str.charCodeAt(i) - 'a'.charCodeAt(0)) + 1;     if (mpp[str[i]]) {       mpp[str[i]]++;     } else {       mpp[str[i]] = 1;     }     i++;   }    // Store resultant string   let ans = '';   for (let i = 0; i < temp.length; i++) {     // If letter do exist in map, decrease     // frequency     if (mpp[temp[i]] > 0) {       mpp[temp[i]]--;     }     // Else store the letter in resultant     // string     else {       ans += temp[i];     }   }    // Return resultant string   return ans; }  // Driver code let str = 'geeksforgeeks'; let K = 15; //Function call document.write(minimiseStr(str, K));

Output

eee

Time Complexity: O(N * logN)
Auxiliary Space: O(N)

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Generate longest String with character sum at most K by deleting letters

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