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Minimum sum of squares of character counts in a given string after removing k characters
  • Difficulty Level : Medium
  • Last Updated : 19 Jan, 2021

Given a string of lowercase alphabets and a number k, the task is to print the minimum value of the string after removal of ‘k’ characters. The value of a string is defined as the sum of squares of the count of each distinct character. For example consider the string “saideep”, here frequencies of characters are s-1, a-1, i-1, e-2, d-1, p-1 and value of the string is 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 2^2 = 9.
Expected Time Complexity: O(k*logn)
Examples: 
 

Input :  str = abccc, K = 1
Output :  6
We remove c to get the value as 12 + 12 + 22

Input :  str = aaab, K = 2
Output :  2

Asked In : Amazon
 

One clear observation is that we need to remove character with highest frequency. One trick is the character ma
A Simple solution is to use sorting technique through all current highest frequency reduce up to k times. For After every reduce again sort frequency array. 
A Better Solution used to Priority Queue which has to the highest element on top. 
 

  1. Initialize empty priority queue.
  2. Count frequency of each character and Store into temp array.
  3. Remove K characters which have highest frequency from queue.
  4. Finally Count Sum of square of each element and return it.

Below is the implementation of the above idea. 
 

C++




// C++ program to find min sum of squares
// of characters after k removals
#include <bits/stdc++.h>
using namespace std;
 
const int MAX_CHAR = 26;
 
// Main Function to calculate min sum of
// squares of characters after k removals
int minStringValue(string str, int k)
{
    int l = str.length(); // find length of string
 
    // if K is greater than length of string
    // so reduced string will become 0
    if (k >= l)
        return 0;
 
    // Else find Frequency of each character and
    // store in an array
    int frequency[MAX_CHAR] = { 0 };
    for (int i = 0; i < l; i++)
        frequency[str[i] - 'a']++;
 
    // Push each char frequency into a priority_queue
    priority_queue<int> q;
    for (int i = 0; i < MAX_CHAR; i++)
        q.push(frequency[i]);
 
    // Removal of K characters
    while (k--) {
        // Get top element in priority_queue,
        // remove it. Decrement by 1 and again
        // push into priority_queue
        int temp = q.top();
        q.pop();
        temp = temp - 1;
        q.push(temp);
    }
 
    // After removal of K characters find sum
    // of squares of string Value
    int result = 0; // Initialize result
    while (!q.empty()) {
        int temp = q.top();
        result += temp * temp;
        q.pop();
    }
 
    return result;
}
 
// Driver Code
int main()
{
    string str = "abbccc"; // Input 1
    int k = 2;
    cout << minStringValue(str, k) << endl;
 
    str = "aaab"; // Input 2
    k = 2;
    cout << minStringValue(str, k);
 
    return 0;
}

Java




// Java program to find min sum of squares
// of characters after k removals
import java.util.Comparator;
import java.util.PriorityQueue;
import java.util.Collections;
public class GFG {
 
    static final int MAX_CHAR = 26;
 
    // Main Function to calculate min sum of
    // squares of characters after k removals
    static int minStringValue(String str, int k)
    {
        int l = str.length(); // find length of string
 
        // if K is greater than length of string
        // so reduced string will become 0
        if (k >= l)
            return 0;
 
        // Else find Frequency of each character and
        // store in an array
        int[] frequency = new int[MAX_CHAR];
        for (int i = 0; i < l; i++)
            frequency[str.charAt(i) - 'a']++;
 
        // creating a priority queue with comparator
        // such that elements in the queue are in
        // descending order.
        PriorityQueue<Integer> q = new PriorityQueue<>(Collections.reverseOrder());
 
        // Push each char frequency into a priority_queue
        for (int i = 0; i < MAX_CHAR; i++) {
            if (frequency[i] != 0)
                q.add(frequency[i]);
        }
 
        // Removal of K characters
        while (k != 0) {
            // Get top element in priority_queue,
            // remove it. Decrement by 1 and again
            // push into priority_queue
            q.add(q.poll() - 1);
            k--;
        }
 
        // After removal of K characters find sum
        // of squares of string Value
        int result = 0; // Initialize result
        while (!q.isEmpty()) {
            result += q.peek() * q.poll();
        }
 
        return result;
    }
 
    // Driver Code
    public static void main(String args[])
    {
        String str = "abbccc"; // Input 1
        int k = 2;
        System.out.println(minStringValue(str, k));
 
        str = "aaab"; // Input 2
        k = 2;
        System.out.println(minStringValue(str, k));
    }
}
// This code is contributed by Sumit Ghosh

Python 3




# Python3 program to find min sum of
# squares of characters after k removals
from queue import PriorityQueue
 
MAX_CHAR = 26
 
# Main Function to calculate min sum of
# squares of characters after k removals
def minStringValue(str, k):
    l = len(str) # find length of string
 
    # if K is greater than length of string
    # so reduced string will become 0
    if(k >= l):
        return 0
     
    # Else find Frequency of each
    # character and store in an array
    frequency = [0] * MAX_CHAR
    for i in range(0, l):
        frequency[ord(str[i]) - 97] += 1
 
    # Push each char frequency negative
    # into a priority_queue as the queue
    # by default is minheap
    q = PriorityQueue()
    for i in range(0, MAX_CHAR):
        q.put(-frequency[i])
 
    # Removal of K characters
    while(k > 0):
         
        # Get top element in priority_queue
        # multiply it by -1 as temp is negative
        # remove it. Increment by 1 and again
        # push into priority_queue
        temp = q.get()
        temp = temp + 1
        q.put(temp, temp)
        k = k - 1
 
    # After removal of K characters find
    # sum of squares of string Value    
    result = 0; # initialize result
    while not q.empty():
        temp = q.get()
        temp = temp * (-1)
        result += temp * temp
    return result
 
# Driver Code
if __name__ == "__main__":
    str = "abbccc"
    k = 2
    print(minStringValue(str, k))
 
    str = "aaab"
    k = 2
    print(minStringValue(str, k))
     
# This code is contributed
# by Sairahul Jella

C#




// C# program to find min sum of squares
// of characters after k removals
using System;
using System.Collections.Generic;
class GFG {
 
  static readonly int MAX_CHAR = 26;
 
  // Main Function to calculate min sum of
  // squares of characters after k removals
  static int minStringValue(String str, int k)
  {
    int l = str.Length; // find length of string
 
    // if K is greater than length of string
    // so reduced string will become 0
    if (k >= l)
      return 0;
 
    // Else find Frequency of each character and
    // store in an array
    int[] frequency = new int[MAX_CHAR];
    for (int i = 0; i < l; i++)
      frequency[str[i] - 'a']++;
 
    // creating a priority queue with comparator
    // such that elements in the queue are in
    // descending order.
    List<int> q = new List<int>();
 
    // Push each char frequency into a priority_queue
    for (int i = 0; i < MAX_CHAR; i++)
    {
      if (frequency[i] != 0)
        q.Add(frequency[i]);
    }
 
    // Removal of K characters
    while (k != 0)
    {
      q.Sort();
      q.Reverse();
 
      // Get top element in priority_queue,
      // remove it. Decrement by 1 and again
      // push into priority_queue
      q.Add(q[0] - 1);
      q.RemoveAt(0);
      k--;
    }
 
    // After removal of K characters find sum
    // of squares of string Value
    int result = 0; // Initialize result
    while (q.Count != 0)
    {
      result += q[0] * q[0];
      q.RemoveAt(0);
    }
    return result;
  }
 
  // Driver Code
  public static void Main(String []args)
  {
    String str = "abbccc"; // Input 1
    int k = 2;
    Console.WriteLine(minStringValue(str, k));
    str = "aaab"; // Input 2
    k = 2;
    Console.WriteLine(minStringValue(str, k));
  }
}
 
// This code is contributed by gauravrajput1

Output: 
 

6
2

Time Complexity: O(k*logn)
This article is contributed by Mr. Somesh Awasthi. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 




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