# Minimum sum of squares of character counts in a given string after removing k characters

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 + 1^1 + 2^2 = 9.

Expected Time Complexity : O(n)

Examples:

```Input :  str = abccc, K = 1
Output :  6
We remove c to get the value as 11 + 11 + 22

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

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

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;
public class GFG {

static final int MAX_CHAR = 26;

// Defining a comparator class
static class IntCompare implements Comparator<Integer>{
@Override
public int compare(Integer arg0, Integer arg1) {
if(arg0 > arg1)
return -1;
else if(arg0 < arg1)
return 1;
else
return 0;
}
}

// 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 object for comparator
Comparator<Integer> c = new IntCompare();

// creating a priority queue with comparator
// such that elements in the queue are in
// descending order.
PriorityQueue<Integer> q = new PriorityQueue<>(c);

// Push each char frequency into a priority_queue
for (int i = 0; i < MAX_CHAR; i++){
if(frequency[i] != 0)
}

// Removal of K characters
while (k != 0)
{
// Get top element in priority_queue,
// remove it. Decrement by 1 and again
// push into priority_queue
int temp = q.peek();
q.poll();
temp = temp-1;
k--;
}

// After removal of K characters find sum
// of squares of string Value
int result = 0; // Initialize result
while (!q.isEmpty())
{
int temp = q.peek();
result += temp*temp;
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
```

Output:

```6
2
```

Time Complexity : O(n)

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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