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Leetcode 2716. Minimize String Length

Problem Statement

You are given a string s which contains only lowercase letters. You need to perform the following operation as many times as possible to minimize the length of the string:

• Find the most frequent character in the string and remove all occurrences of it.

Return the minimized length of the string after performing the operation as many times as possible.

Clarifying Questions

1. Input constraints:
   • What is the maximum length for the string s?
   • Are there any special characters, or is it strictly lowercase letters?
2. Output specifics:
   • If the string becomes empty after removing the most frequent characters, should we return 0?

Assuming standard constraints:

• The length of the string s is between 1 and 10^5.
• The string s contains only lowercase English letters.

Strategy

1. Frequency Calculation:
   • Use a frequency array or hash map to count occurrences of each character in the string.
2. Frequency Analysis:
   • Identify the character with the highest frequency and remove its count from the total length of the string.
3. Iterate and Minimize:
   • Repeat the process until all characters are removed or until no characters are left with non-zero frequency.

Code

import java.util.HashMap;
import java.util.Map;

class Solution {
    public int minimizeStringLength(String s) {
        // Step 1: Calculate frequency of each character
        Map<Character, Integer> frequencyMap = new HashMap<>();
        
        for (char c : s.toCharArray()) {
            frequencyMap.put(c, frequencyMap.getOrDefault(c, 0) + 1);
        }
        
        // Step 2: Identify most frequent character and reduce the string length
        int totalLength = s.length();
        
        while (!frequencyMap.isEmpty()) {
            // Find the character with the highest frequency
            char mostFrequentChar = ' ';
            int highestFrequency = 0;
            
            for (Map.Entry<Character, Integer> entry : frequencyMap.entrySet()) {
                if (entry.getValue() > highestFrequency) {
                    mostFrequentChar = entry.getKey();
                    highestFrequency = entry.getValue();
                }
            }

            // Remove all occurrences of the most frequent character
            totalLength -= highestFrequency;
            frequencyMap.remove(mostFrequentChar);
        }
        
        return totalLength;
    }
}

Time Complexity

1. Frequency Calculation (O(n)):
   • We iterate through the string once to calculate the frequency of each character.
2. Finding the Most Frequent Character (O(n log n)):
   • In the worst case, we may need to identify and remove characters log(n) times because the look-up and removal operation in a hash map for each character takes on average O(1) time, but the total number of characters can involve multiple iterations over the map entries.
3. Overall Time Complexity:
   • The overall time complexity will be O(n) for the initial iteration through the string plus O(n log n) for the repeated operations of finding and removing the most frequent character. Therefore, the combined time complexity is approximately O(n log n).

Summary

• This approach effectively uses a hash map to count character frequencies and iteratively removes the most frequent character to minimize the string length.
• The time complexity is efficient enough to handle the upper limit constraint of the input size.

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