LeetCode 763: Partition Labels Solution

Master LeetCode problem 763 (Partition Labels), 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.

763. Partition Labels

Medium

Hash Table Two Pointers String Greedy

Problem Explanation

Explanation

We can solve this problem using a greedy approach.
We iterate over the string and keep track of the last occurrence of each character.
As we iterate, we maintain a end variable that represents the end index of the current partition.
When we reach the end index of the current partition, we add the length of the current partition to the result and update the start index to the next character.

Time Complexity: O(n) where n is the length of the input string s
Space Complexity: O(1)

Solution Code

import java.util.*;

class Solution {
    public List<Integer> partitionLabels(String s) {
        List<Integer> result = new ArrayList<>();
        int[] lastOccurrence = new int[26];
        for (int i = 0; i < s.length(); i++) {
            lastOccurrence[s.charAt(i) - 'a'] = i;
        }

        int start = 0, end = 0;
        for (int i = 0; i < s.length(); i++) {
            end = Math.max(end, lastOccurrence[s.charAt(i) - 'a']);
            if (i == end) {
                result.add(end - start + 1);
                start = end + 1;
            }
        }

        return result;
    }
}

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