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 1^{2}+ 1^{2}+ 2^{2}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.

- Initialize empty priority queue.
- Count frequency of each character and Store into temp array.
- Remove K characters which have highest frequency from queue.
- 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.

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