This approach takes advantage of the fact that if two strings are identical, there are no uncommon subsequences. If they are different, the longest uncommon subsequence is the longest string itself.
a is equal to b, return -1 because all subsequences of a are subsequences of b and vice versa.a is not equal to b, return the maximum length of a or b since the longer string itself cannot be a subsequence of the other.Time Complexity: O(n) where n is the length of the strings, due to the string comparison.
Space Complexity: O(1), no additional space is needed.
1function findLUSlength(a, b) {
2 return a === b ? -1 : Math.max(a.length, b.length);
3}
4
5console.log(findLUSlength("aba", "cdc"));This JavaScript code compares strings using === and computes the length using Math.max.
Another approach is to analyze the problem by considering the entire strings as potential subsequences and determine their existence in the other string.
Time Complexity: O(n + m) where n and m are the lengths of the strings.
Space Complexity: O(1), since no additional structures are employed.
1function isSubsequence(s, t) {
2 let i = 0, j = 0;
3 while (i < s.length && j < t.length) {
4 if (s[i] === t[j]) i++;
5 j++;
6 }
7 return i === s.length;
8}
9
10function findLUSlength(a, b) {
11 if (isSubsequence(a, b) || isSubsequence(b, a)) return -1;
12 return Math.max(a.length, b.length);
13}
14
15console.log(findLUSlength("aba", "cdc"));The JavaScript solution defines a function isSubsequence to check if one string is a subsequence of another, then applies it accordingly to compute the result.