522. Longest Uncommon Subsequence II

Description

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.

Solutions

Solution 1: Subsequence Judgment

We define a function \(check(s, t)\) to determine whether string \(s\) is a subsequence of string \(t\). We can use a two-pointer approach, initializing two pointers \(i\) and \(j\) to point to the beginning of strings \(s\) and \(t\) respectively, then continuously move pointer \(j\). If \(s[i]\) equals \(t[j]\), then move pointer \(i\). Finally, check if \(i\) equals the length of \(s\). If \(i\) equals the length of \(s\), it means \(s\) is a subsequence of \(t\).

To determine if string \(s\) is unique, we only need to take string \(s\) itself and compare it with other strings in the list. If there exists a string for which \(s\) is a subsequence, then \(s\) is not unique. Otherwise, string \(s\) is unique. We take the longest string among all unique strings.

The time complexity is \(O(n^2 \times m)\), where \(n\) is the length of the list of strings, and \(m\) is the average length of the strings. The space complexity is \(O(1)\).

Python3JavaC++GoTypeScript

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

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;
    }
}

class Solution {
public:
    int findLUSlength(vector<string>& strs) {
        int ans = -1;
        int n = strs.size();
        auto check = [&](const string& s, const string& t) {
            int m = s.size(), n = t.size();
            int i = 0;
            for (int j = 0; i < m && j < n; ++j) {
                if (s[i] == t[j]) {
                    ++i;
                }
            }
            return i == m;
        };
        for (int i = 0, j; i < n; ++i) {
            int x = strs[i].size();
            for (j = 0; j < n; ++j) {
                if (i != j && check(strs[i], strs[j])) {
                    x = -1;
                    break;
                }
            }
            ans = max(ans, x);
        }
        return ans;
    }
};

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
}

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;
}

522. Longest Uncommon Subsequence II

Description

Solutions

Solution 1: Subsequence Judgment

Comments