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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0501_0600.s0522_longest_uncommon_subsequence_ii;

// #Medium #Array #String #Hash_Table #Sorting #Two_Pointers
// #2022_07_28_Time_1_ms_(99.61%)_Space_41.7_MB_(40.70%)

/**
 * 522 - Longest Uncommon Subsequence II\.
 *
 * Medium
 *
 * Given an array of strings `strs`, return _the length of the **longest uncommon subsequence** between them_. If the longest uncommon subsequence does not exist, return `-1`.
 *
 * An **uncommon subsequence** between an array of strings is a string that is a **subsequence of one string but not the others**.
 *
 * A **subsequence** of a string `s` is a string that can be obtained after deleting any number of characters from `s`.
 *
 * *   For example, `"abc"` is a subsequence of `"aebdc"` because you can delete the underlined characters in `"aebdc"` to get `"abc"`. Other subsequences of `"aebdc"` include `"aebdc"`, `"aeb"`, and `""` (empty string).
 *
 * **Example 1:**
 *
 * **Input:** strs = ["aba","cdc","eae"]
 *
 * **Output:** 3
 *
 * **Example 2:**
 *
 * **Input:** strs = ["aaa","aaa","aa"]
 *
 * **Output:** -1
 *
 * **Constraints:**
 *
 * *   `2 <= strs.length <= 50`
 * *   `1 <= strs[i].length <= 10`
 * *   `strs[i]` consists of lowercase English letters.
**/
public class Solution {
    public int findLUSlength(String[] strs) {
        int maxUncommonLen = -1;
        for (int indx1 = 0; indx1 < strs.length; indx1++) {
            boolean isCommon = false;
            for (int indx2 = 0; indx2 < strs.length; indx2++) {
                if (indx2 != indx1 && isSubSequence(strs[indx1], strs[indx2])) {
                    isCommon = true;
                    break;
                }
            }
            if (!isCommon) {
                maxUncommonLen = Math.max(maxUncommonLen, strs[indx1].length());
            }
        }
        return maxUncommonLen;
    }

    private boolean isSubSequence(String str1, String str2) {
        int indx1 = 0;
        for (int indx2 = 0; indx2 < str2.length(); indx2++) {
            if (str1.charAt(indx1) == str2.charAt(indx2)) {
                indx1++;
            }
            if (indx1 == str1.length()) {
                return true;
            }
        }
        return indx1 == str1.length();
    }
}