Problem:

A string s is called good if there are no two different characters in s that have the same frequency.

Given a string s, return the minimum number of characters you need to delete to make s good.

The frequency of a character in a string is the number of times it appears in the string. For example, in the string "aab", the frequency of 'a' is 2, while the frequency of 'b' is 1.

Example 1:
Input: s = "aab"
Output: 0
Explanation: s is already good.

Example 2:
Input: s = "aaabbbcc"
Output: 2
Explanation: You can delete two 'b's resulting in the good string "aaabcc". Another way it to delete one 'b' and one 'c' resulting in the good string "aaabbc".

Example 3:
Input: s = "ceabaacb"
Output: 2
Explanation: You can delete both 'c's resulting in the good string "eabaab". Note that we only care about characters that are still in the string at the end (i.e. frequency of 0 is ignored).

Constraints:

• 1 <= s.length <= 105
• s contains only lowercase English letters.

Solution:

class Solution {
    public int minDeletions(String str) {
        // create array containing frequency of each character
        int[] freqArr = new int[26];
        for(char c: str.toCharArray()) {
            freqArr[c - 'a']++;
        }

        // sort frequency array
        Arrays.sort(freqArr);

        // traverse in reverse order and calculate deletions as freqArr[i] - maxAllowedFreq
        int deletions = 0;
        int maxAllowedFreq = str.length();

        for(int i=freqArr.length - 1; i>=0; i--) {
            if(freqArr[i] > maxAllowedFreq) {
                deletions += (freqArr[i] - maxAllowedFreq);
                freqArr[i] = maxAllowedFreq;
            }
            maxAllowedFreq = freqArr[i] - 1;
            if(maxAllowedFreq < 0) maxAllowedFreq = 0;
        }

        return deletions;

    }
}

Time Complexity: O(N) where N is the length of the given string. Sorting can be ignored as the array size will always be 26.
Space Complexity: O(1) as array size is always 26, hence it can be ignored.