522. Longest Uncommon Subsequence II
Explanation:
To solve this problem, we can iterate through each pair of strings in the input array and check if one string is a subsequence of the other. If a string is not a subsequence of any other string, we can return its length as the longest uncommon subsequence. If all strings are subsequences of each other, we return -1.
- Create a helper method to check if one string is a subsequence of another string.
- Iterate through each pair of strings in the input array.
- For each pair, check if one string is a subsequence of the other.
- If a string is not a subsequence of any other string, return its length as the longest uncommon subsequence.
- If no such string is found, return -1.
Time Complexity: O(n^2 * m) where n is the number of strings and m is the maximum length of a string.
Space Complexity: O(1)
:
class Solution {
public int findLUSlength(String[] strs) {
int longestUncommon = -1;
for (int i = 0; i < strs.length; i++) {
boolean isUncommon = true;
for (int j = 0; j < strs.length; j++) {
if (i != j && isSubsequence(strs[i], strs[j])) {
isUncommon = false;
break;
}
}
if (isUncommon) {
longestUncommon = Math.max(longestUncommon, strs[i].length());
}
}
return longestUncommon;
}
private boolean isSubsequence(String a, String b) {
int i = 0, j = 0;
while (i < a.length() && j < b.length()) {
if (a.charAt(i) == b.charAt(j)) {
i++;
}
j++;
}
return i == a.length();
}
}Code Editor (Testing phase)
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