LeetCode 2901: Longest Unequal Adjacent Groups Subsequence II
Problem Description
Explanation:
To solve this problem, we can iterate through all possible subsequences of indices and check if they satisfy the given conditions. We can use backtracking to generate all possible subsequences, and for each subsequence, we can check if adjacent groups are unequal and if the hamming distance between adjacent words is 1.
- We will start with an empty subsequence and iterate through all possible indices.
- At each index, we will check if adding that index to the subsequence satisfies the conditions.
- If the conditions are met, we will continue building the subsequence by recursively calling the backtracking function with the updated subsequence.
- We will keep track of the longest valid subsequence found.
The time complexity of this approach is exponential as we are generating all possible subsequences. The space complexity is also exponential due to the recursive backtracking calls.
:
Solutions
class Solution {
List<String> result = new ArrayList<>();
public String[] longestUnequalAdjacentGroupsSubsequenceII(String[] words, int[] groups) {
backtrack(words, groups, new ArrayList<>(), 0);
return result.toArray(new String[0]);
}
private void backtrack(String[] words, int[] groups, List<String> current, int index) {
if (current.size() > result.size()) {
result = new ArrayList<>(current);
}
for (int i = index; i < words.length; i++) {
if (current.isEmpty() || (groups[i] != groups[current.get(current.size() - 1)]
&& hammingDistance(words[i], current.get(current.size() - 1)) == 1)) {
current.add(i);
backtrack(words, groups, current, i + 1);
current.remove(current.size() - 1);
}
}
}
private int hammingDistance(String word1, String word2) {
int count = 0;
for (int i = 0; i < word1.length(); i++) {
if (word1.charAt(i) != word2.charAt(i)) {
count++;
}
}
return count;
}
}Loading editor...