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Leetcode 522. Longest Uncommon Subsequence II

Problem Statement

Given a list of strings, you need to find the longest uncommon subsequence among them. The longest uncommon subsequence is defined as the longest subsequence which is not a subsequence of any other string in the list.

A subsequence can be derived from another string by deleting some characters without changing the order of the remaining characters.

Example:

Input: ["aba", "cdc", "eae"]
Output: 3

In this example, “aba” is not a subsequence of “cdc” or “eae”, and similarly for others. Therefore, any of the strings count as the longest uncommon subsequence and its length is 3.

If no such subsequence exists, return -1.

Clarifying Questions

  1. Input Size: What is the maximum size of the input list?
  2. String Length: What is the maximum length of each string in the list?
  3. Character Set: Can we assume the strings contain only lowercase English letters?
  4. Edge Cases: How should we handle edge cases like an empty list or list with one string?

Strategy

The key idea is to leverage the following approach:

  1. Sort the list of strings by length in descending order.
  2. Check each string if it is not a subsequence of any other string in the array.
  3. Return the length of the first string that meets this condition.

To determine if a string is a subsequence of another string, use two pointers for comparison.

Code

Let’s write the Java code to solve the problem:

import java.util.*;

public class Solution {
    // Helper function to determine if `s1` is a subsequence of `s2`
    private boolean isSubsequence(String s1, String s2) {
        int i = 0;
        for (int j = 0; j < s2.length() && i < s1.length(); j++) {
            if (s1.charAt(i) == s2.charAt(j)) {
                i++;
            }
        }
        return i == s1.length();
    }

    public int findLUSlength(String[] strs) {
        // Sort the strings by their lengths in descending order
        Arrays.sort(strs, (a, b) -> b.length() - a.length());

        for (int i = 0; i < strs.length; i++) {
            boolean isUncommon = true;
            for (int j = 0; j < strs.length; j++) {
                if (i != j && isSubsequence(strs[i], strs[j])) {
                    isUncommon = false;
                    break;
                }
            }
            if (isUncommon) {
                return strs[i].length();
            }
        }

        return -1;
    }

    public static void main(String[] args) {
        Solution solution = new Solution();
        String[] strs = {"aba", "cdc", "eae"};
        System.out.println(solution.findLUSlength(strs)); // Output: 3
    }
}

Time Complexity

Thus, the overall time complexity is O(n log n + n^2 * l).

Space Complexity

Space complexity is O(1) for extra variables, not considering the space used by the input data.

If you have any further questions or need clarifications, feel free to ask!

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