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

• class Solution {
public int findLUSlength(String[] strs) {
int ans = -1;
for (int i = 0, j = 0, n = strs.length; i < n; ++i) {
for (j = 0; j < n; ++j) {
if (i == j) {
continue;
}
if (check(strs[j], strs[i])) {
break;
}
}
if (j == n) {
ans = Math.max(ans, strs[i].length());
}
}
return ans;
}

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

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

bool check(string a, string b) {
int j = 0;
for (int i = 0; i < a.size() && j < b.size(); ++i)
if (a[i] == b[j]) ++j;
return j == b.size();
}
};

• class Solution:
def findLUSlength(self, strs: List[str]) -> int:
def check(a, b):
i = j = 0
while i < len(a) and j < len(b):
if a[i] == b[j]:
j += 1
i += 1
return j == len(b)

n = len(strs)
ans = -1

for i in range(n):
j = 0
while j < n:
if i == j or not check(strs[j], strs[i]):
j += 1
else:
break
if j == n:
ans = max(ans, len(strs[i]))
return ans


• func findLUSlength(strs []string) int {
check := func(a, b string) bool {
j := 0
for i := 0; i < len(a) && j < len(b); i++ {
if a[i] == b[j] {
j++
}
}
return j == len(b)
}

ans := -1
for i, j, n := 0, 0, len(strs); i < n; i++ {
for j = 0; j < n; j++ {
if i == j {
continue
}
if check(strs[j], strs[i]) {
break
}
}
if j == n && ans < len(strs[i]) {
ans = len(strs[i])
}
}
return ans
}