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LeetCode 2405: Optimal Partition of String Solution

Master LeetCode problem 2405 (Optimal Partition of String), 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.

2405. Optimal Partition of String

Medium
Hash TableStringGreedy

Problem Explanation

Explanation:

To solve this problem, we can use a greedy approach. We will iterate through each character in the string and maintain a set to keep track of unique characters we have encountered so far in the current substring. Whenever we encounter a character that is already in the set, it means we need to start a new substring from that character.

Algorithm:

  1. Initialize an empty set to store unique characters.
  2. Initialize a variable count to keep track of the number of substrings.
  3. Iterate through each character in the string.
  4. If the character is not in the set, add it to the set.
  5. If the character is already in the set, increment count and clear the set.
  6. Return the final value of count.

Time Complexity:

The time complexity of this algorithm is O(n), where n is the length of the input string s.

Space Complexity:

The space complexity of this algorithm is O(26) = O(1) since we are using a set to store unique characters.

:

Solution Code

class Solution {
    public int minPartitions(String s) {
        Set<Character> uniqueChars = new HashSet<>();
        int count = 0;
        
        for (char c : s.toCharArray()) {
            if (!uniqueChars.contains(c)) {
                uniqueChars.add(c);
            } else {
                count++;
                uniqueChars.clear();
                uniqueChars.add(c);
            }
        }
        
        return count + 1;
    }
}

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What is the time complexity of LeetCode 2405 (Optimal Partition of String)?

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