Leetcode 2900: Longest Unequal Adjacent Groups Subsequence I
You are given two arrays:
words (a list of distinct strings) and groups (a binary array where each element corresponds to an entry in words). Your task is to select the longest alternating subsequence of words, where each consecutive word has a different corresponding group value. For each pair of consecutive words in the subsequence, the corresponding values in groups must be different.Problem
Approach
Steps
Complexity
Input: The input consists of two arrays: `words` (a list of distinct strings) and `groups` (a binary array of integers corresponding to each word).
Example: ["apple","banana","cherry"]
Constraints:
• 1 <= n <= 100
• 1 <= words[i].length <= 10
• groups[i] is either 0 or 1
• words consists of distinct strings
Output: Return the longest alternating subsequence of words, where adjacent elements have non-matching corresponding group values in `groups`.
Example: ["apple","banana"]
Constraints:
• The output should be a valid subsequence from the given `words` array.
Goal: Select the longest alternating subsequence from `words` where consecutive words have different group values.
Steps:
• Initialize an empty list `ans` to store the subsequence.
• Initialize a variable `flag` to -1 to track the last added group's value.
• Iterate through the `words` array.
• For each word, if its group value is different from `flag`, add it to `ans` and update `flag`.
Goal: The problem constraints ensure that the input size is manageable and the values of `groups` are binary (0 or 1).
Steps:
• 1 <= words.length <= 100
• 1 <= words[i].length <= 10
• groups[i] is either 0 or 1
• words consists of distinct strings
Assumptions:
• The input contains distinct words.
• The group array has a corresponding entry for each word in `words`.
• Input: ["apple","banana","cherry"]
• Explanation: We start by picking the first word and alternating between different group values. We skip consecutive words with the same group.
Approach: Iterate through the `words` array, keeping track of the most recent group value and adding words to the result list if their group differs from the last added word.
Observations:
• The problem involves iterating through the list of words and checking the group values to maintain an alternating sequence.
• We can solve this efficiently in one pass by keeping track of the last group used.
Steps:
• Initialize `ans` as an empty list.
• Initialize `flag` to -1.
• For each word in `words`, check if its group differs from `flag`. If so, add it to `ans` and update `flag`.
Empty Inputs:
• The input is guaranteed to have at least one word.
Large Inputs:
• For large inputs, the algorithm should handle up to 100 elements in `words` efficiently.
Special Values:
• If all words belong to the same group, the result will contain only one word.
Constraints:
• The constraints guarantee that the solution will work within the limits provided.
vector<string> getWordsInLongestSubsequence(int n, vector<string>& words, vector<int>& groups) {
// Edges 0->1 / 1 -> 0, start has no conseq
vector<string>ans;
int flag=-1;
for(int i=0;i<n;i++){
if(groups[i]!=flag){
flag=groups[i];
ans.push_back(words[i]);
}
}
return ans;
}
1 : Function Definition
vector<string> getWordsInLongestSubsequence(int n, vector<string>& words, vector<int>& groups) {
This line defines a function that takes the number of words, a list of words, and a list of group identifiers. It returns a vector of strings representing the longest subsequence of words grouped together.
2 : Variable Initialization
vector<string>ans;
This line initializes an empty vector 'ans' to store the result, which will be the longest subsequence of words.
3 : Variable Initialization
int flag=-1;
This line initializes a variable 'flag' to -1. It will be used to track the current group identifier and detect changes between groups.
4 : Loop
for(int i=0;i<n;i++){
This loop iterates through all words in the input list. The index 'i' is used to access each word and group.
5 : Condition Check
if(groups[i]!=flag){
This condition checks if the current word belongs to a different group than the previous word (tracked by 'flag').
6 : Group Assignment
flag=groups[i];
If the condition is true, the 'flag' is updated to the current group's identifier.
7 : Action
ans.push_back(words[i]);
If the group changes, the current word is added to the result vector 'ans'.
8 : Return
return ans;
This line returns the result vector 'ans', which contains the longest subsequence of words grouped together.
Best Case: O(n)
Average Case: O(n)
Worst Case: O(n)
Description: The time complexity is O(n) since we process each word in the `words` array once.
Best Case: O(n)
Worst Case: O(n)
Description: The space complexity is O(n) as we store the result in the `ans` list.
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