LeetCode #522 — MEDIUM

Longest Uncommon Subsequence II

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

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.

Step 01

Brute Force Baseline

Problem summary: 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).

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map · Two Pointers

Example 1

["aba","cdc","eae"]

Example 2

["aaa","aaa","aa"]

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #522: Longest Uncommon Subsequence II
class Solution {
    public int findLUSlength(String[] strs) {
        int ans = -1;
        int n = strs.length;
        for (int i = 0, j; i < n; ++i) {
            int x = strs[i].length();
            for (j = 0; j < n; ++j) {
                if (i != j && check(strs[i], strs[j])) {
                    x = -1;
                    break;
                }
            }
            ans = Math.max(ans, x);
        }
        return ans;
    }

    private boolean check(String s, String t) {
        int m = s.length(), n = t.length();
        int i = 0;
        for (int j = 0; i < m && j < n; ++j) {
            if (s.charAt(i) == t.charAt(j)) {
                ++i;
            }
        }
        return i == m;
    }
}

// Accepted solution for LeetCode #522: Longest Uncommon Subsequence II
func findLUSlength(strs []string) int {
	ans := -1
	check := func(s, t string) bool {
		m, n := len(s), len(t)
		i := 0
		for j := 0; i < m && j < n; j++ {
			if s[i] == t[j] {
				i++
			}
		}
		return i == m
	}
	for i, s := range strs {
		x := len(s)
		for j, t := range strs {
			if i != j && check(s, t) {
				x = -1
				break
			}
		}
		ans = max(ans, x)
	}
	return ans
}

# Accepted solution for LeetCode #522: Longest Uncommon Subsequence II
class Solution:
    def findLUSlength(self, strs: List[str]) -> int:
        def check(s: str, t: str):
            i = j = 0
            while i < len(s) and j < len(t):
                if s[i] == t[j]:
                    i += 1
                j += 1
            return i == len(s)

        ans = -1
        for i, s in enumerate(strs):
            for j, t in enumerate(strs):
                if i != j and check(s, t):
                    break
            else:
                ans = max(ans, len(s))
        return ans

// Accepted solution for LeetCode #522: Longest Uncommon Subsequence II
struct Solution;
use std::cmp::Reverse;

trait IsSubSequenceOf {
    fn is_subsequence_of(&self, s: &str) -> bool;
}

impl IsSubSequenceOf for String {
    fn is_subsequence_of(&self, s: &str) -> bool {
        let mut it = self.chars().peekable();
        for c in s.chars() {
            if let Some(&first) = it.peek() {
                if first == c {
                    it.next();
                }
            }
        }
        it.next().is_none()
    }
}

impl Solution {
    fn find_lu_slength(mut strs: Vec<String>) -> i32 {
        let n = strs.len();
        strs.sort_by_key(|s| Reverse(s.len()));
        for i in 0..n {
            let mut count = 0;
            for j in 0..n {
                if i != j {
                    if !strs[i].is_subsequence_of(&strs[j]) {
                        count += 1;
                    }
                }
            }
            if count == n - 1 {
                return strs[i].len() as i32;
            }
        }
        -1
    }
}

#[test]
fn test() {
    let strs = vec_string!["aba", "cdc", "eae"];
    let res = 3;
    assert_eq!(Solution::find_lu_slength(strs), res);
}

// Accepted solution for LeetCode #522: Longest Uncommon Subsequence II
function findLUSlength(strs: string[]): number {
    const n = strs.length;
    let ans = -1;
    const check = (s: string, t: string): boolean => {
        const [m, n] = [s.length, t.length];
        let i = 0;
        for (let j = 0; i < m && j < n; ++j) {
            if (s[i] === t[j]) {
                ++i;
            }
        }
        return i === m;
    };
    for (let i = 0; i < n; ++i) {
        let x = strs[i].length;
        for (let j = 0; j < n; ++j) {
            if (i !== j && check(strs[i], strs[j])) {
                x = -1;
                break;
            }
        }
        ans = Math.max(ans, x);
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair of elements. The outer loop picks one element, the inner loop scans the rest. For n elements that is n × (n−1)/2 comparisons = O(n²). No extra memory — just two loop variables.

TWO POINTERS

O(n) time

O(1) space

Each pointer traverses the array at most once. With two pointers moving inward (or both moving right), the total number of steps is bounded by n. Each comparison is O(1), giving O(n) overall. No auxiliary data structures are needed — just two index variables.

Shortcut: Two converging pointers on sorted data → O(n) time, O(1) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Moving both pointers on every comparison

Wrong move: Advancing both pointers shrinks the search space too aggressively and skips candidates.

Usually fails on: A valid pair can be skipped when only one side should move.

Fix: Move exactly one pointer per decision branch based on invariant.