Medium
Longest Uncommon Subsequence II — C++
Full explanation · Time O(l * n^2) · Space O(1)
// Time: O(l * n^2)
// Space: O(1)
class Solution {
public:
int findLUSlength(vector<string>& strs) {
sort(strs.begin(), strs.end(),
[](const string& a, const string& b) { return a.length() > b.length(); });
for (int i = 0; i < strs.size(); ++i) {
bool all_of = true;
for (int j = 0; j < strs.size() && strs[j].length() >= strs[i].length(); ++j) {
if (i != j && isSubsequence(strs[i], strs[j])) {
all_of = false;
break;
}
}
if (all_of) {
return strs[i].length();
}
}
return -1;
}
private:
bool isSubsequence(const string& a, const string& b) {
int i = 0;
for (int j = 0; j < b.length() && i < a.length(); ++j) {
if (a[i] == b[j]) {
++i;
}
}
return i == a.length();
}
};