LeetCode 3144: Minimum Substring Partition of Equal Character Frequency Solution

Master LeetCode problem 3144 (Minimum Substring Partition of Equal Character Frequency), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

3144. Minimum Substring Partition of Equal Character Frequency

Medium

Hash Table String Dynamic Programming Counting

Problem Explanation

Explanation

To solve this problem, we can iterate through the input string s and maintain a frequency map of characters in the current substring being considered. At each step, we check if the current substring can be partitioned into balanced substrings by checking if all character frequencies in the substring are equal. If they are, we increment the count of partitions and start a new substring.

We keep track of the frequency of characters in the current window and also maintain a set to keep track of the characters that are currently present in the window. When the frequencies of all characters in the window are the same, we increment the count of partitions and reset the frequency map and set for the next substring.

Time Complexity

The time complexity of this approach is O(n), where n is the length of the input string s. We only iterate through the string once.

Space Complexity

The space complexity of this approach is O(1) since we are using a fixed-size map and set to track character frequencies and characters in the current window.

Solution Code

class Solution {
    public int balancedStringSplit(String s) {
        int count = 0;
        int balance = 0;
        
        for (char c : s.toCharArray()) {
            if (c == 'L') {
                balance++;
            } else {
                balance--;
            }
            
            if (balance == 0) {
                count++;
            }
        }
        
        return count;
    }
}

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