2900. Longest Unequal Adjacent Groups Subsequence I
Problem Description
The goal is to find the longest subsequence from an array of indices [0, 1, ..., n - 1]
such that for any two consecutive indices in the subsequence i_j
and i_{j + 1}
, the elements in the binary array groups
at those indices are not the same, i.e., groups[i_j] != groups[i_{j + 1}]
. Each index in the subsequence corresponds to a word in the words
array. The task is to return an array of words that represents this longest subsequence.
A subsequence is a sequence that can be derived from another sequence by deleting some or no elements without changing the order of the remaining elements. Importantly, the words in the words
array may have different lengths, which doesn't impact the selection of the subsequence.
Intuition
The approach to finding the longest subsequence where consecutive elements in groups
are different is a greedy one. This means we can make local, optimal choices at each step without needing to consider the rest of the array.
For every element at index i
in groups
, we have two scenarios - either i
is the first index (i.e., i == 0
), or groups[i]
is different from the previous element groups[i - 1]
. If either of these conditions is met, we can include the corresponding word words[i]
in our subsequence.
By iterating over the entire groups
array and including words that meet the criteria, we ensure that no two consecutive selected words have their corresponding groups
elements equal, effectively giving us the longest subsequence by the definition provided.
The intuition comes from the fact that to maximize the length of the subsequence, we want to include as many words as possible as long as they meet the aforementioned condition. Since the condition only depends on the current and previous groups
elements, we only need a simple iteration to build our solution without needing to backtrack or look ahead.
Learn more about Greedy and Dynamic Programming patterns.
Solution Approach
The provided Python solution uses a list comprehension to create and return the subsequence of words, which is essentially a single-pass greedy algorithm. The algorithm iterates through the groups
array and applies a selection criteria to each element to determine if the corresponding element from the words
array should be included in our final output or not.
Algorithm:
- Initialize a list comprehension that will evaluate each element
x
ingroups
along with its corresponding indexi
. It uses Python’s built-inenumerate
function to obtain each element ingroups
and its index simultaneously. - In this comprehension, for every element
x
and indexi
, the following condition is checked:i == 0 or x != groups[i - 1]
. This condition says that the first element (i == 0
) should always be included, and then every subsequent element should be included only if it is different than the one preceding it (x != groups[i - 1]
). This ensures that no two adjacent elements in the subsequence have the samegroups
value. - If the condition is true for a given
i
, we selectwords[i]
for inclusion in the final output. - Once the list comprehension finishes iterating over all elements in
groups
, it will have produced a list of words that constitutes the longest subsequence satisfying the problem's constraints. - The final step is to return this list of words.
Data Structures:
- We utilize Python's list data structure to store the
words
andgroups
.
Patterns used:
- The solution applies a greedy approach to the problem: at each step, it makes a local optimum choice (whether to include a word) based only on the current and immediate previous elements from
groups
, which guarantees the finding of the global optimum (the longest subsequence under the given conditions). - List comprehension, a concise way to create lists, is used for its readability and efficiency in selecting the appropriate words.
This simple yet effective approach leverages the characteristics of the problem's constraints to arrive at an optimal solution without needing a complex algorithm.
Ready to land your dream job?
Unlock your dream job with a 2-minute evaluator for a personalized learning plan!
Start EvaluatorExample Walkthrough
Let's walk through a small example to illustrate the solution approach described above. Suppose we have the following groups
and words
arrays:
groups = [1, 0, 0, 1, 0, 1]
words = ["apple", "banana", "grape", "cherry", "mango", "peach"]
Our task is to find the longest subsequence such that no two consecutive indices in the subsequence correspond to equal elements in the groups
array. Let's apply the algorithm step by step:
-
Start iterating through the
groups
array, comparing the element at indexi
with the one at indexi - 1
. -
The first element in
groups
is1
(at index0
). Sincei == 0
, we don't have a previous element to compare with, so we include "apple" from thewords
array in our subsequence. -
The next element in
groups
is0
(at index1
). Sincegroups[1] != groups[0]
, we include "banana" from thewords
array in our subsequence. -
At index
2
,groups
has another0
. This time,groups[2] == groups[1]
, so "grape" does not get included in our subsequence since consecutive elements must be different. -
Now at index
3
, we have a1
ingroups
. Sincegroups[3] != groups[2]
, "cherry" gets included in our subsequence. -
Moving to index
4
, the element ingroups
is0
. Becausegroups[4] != groups[3]
, we include "mango" in our subsequence. -
Lastly, at index
5
,groups
contains a1
. Asgroups[5] != groups[4]
, we include "peach" in our subsequence.
Following the steps of our algorithm, the final subsequence of words
is:
["apple", "banana", "cherry", "mango", "peach"]
Thus, by iterating through each element of the groups
array and checking our defined condition, we successfully construct the longest subsequence of words without having identical consecutive elements from groups
. This illustrates the effectiveness of the greedy approach to solve the problem, as we made local optimal selections to achieve a global optimal solution.
Solution Implementation
1# Import the List type from typing module for type hints
2from typing import List
3
4class Solution:
5 def getWordsInLongestSubsequence(self, n: int, words: List[str], groups: List[int]) -> List[str]:
6 # Initialize an empty list to store the words in the longest subsequence
7 longest_subsequence_words = []
8
9 # Iterate through each index and corresponding group number in the groups list
10 for i, group_number in enumerate(groups):
11 # Check if it's the first word or if the current group number is different from the previous one
12 if i == 0 or group_number != groups[i - 1]:
13 # If yes, append the corresponding word to the longest_subsequence_words list
14 longest_subsequence_words.append(words[i])
15
16 # Return the list of words that form the longest subsequence
17 return longest_subsequence_words
18
1import java.util.ArrayList;
2import java.util.List;
3
4class Solution {
5
6 /**
7 * Finds the words in the longest subsequence with alternating groups.
8 *
9 * @param n the number of words.
10 * @param words an array of words.
11 * @param groups an array of group identifiers corresponding to each word.
12 * @return a list of words in the longest subsequence with alternating groups.
13 */
14 public List<String> getWordsInLongestSubsequence(int n, String[] words, int[] groups) {
15 // Initialize an ArrayList to store the resulting words.
16 List<String> result = new ArrayList<>();
17
18 // Iterate over the words to find the longest subsequence.
19 for (int i = 0; i < n; ++i) {
20 // Add the first word and any word that starts a new group (compared to the previous word).
21 if (i == 0 || groups[i] != groups[i - 1]) {
22 result.add(words[i]);
23 }
24 }
25 // Return the list of words in the longest subsequence.
26 return result;
27 }
28}
29
1#include <vector>
2#include <string>
3
4class Solution {
5public:
6 // Function that generates a vector of strings, which consists of the words
7 // in the longest non-repeating subsequence based on the given groups.
8 // Parameters:
9 // n - the number of elements in the words and groups vectors.
10 // words - a vector of strings representing the words.
11 // groups - a vector of integers, where each integer corresponds to the group of the word at the same index.
12 std::vector<std::string> getWordsInLongestSubsequence(int n, std::vector<std::string>& words, std::vector<int>& groups) {
13 // Answer vector to store the resulting words.
14 std::vector<std::string> answer;
15
16 // Iterate through each group by index.
17 for (int index = 0; index < n; ++index) {
18 // If we are at the first word, or the current word belongs to a different group than the previous one,
19 // then it is a part of the longest non-repeating subsequence.
20 if (index == 0 || groups[index] != groups[index - 1]) {
21 // Add the current word to the answer vector.
22 answer.emplace_back(words[index]);
23 }
24 }
25 // Return the answer vector containing the words in the longest non-repeating subsequence.
26 return answer;
27 }
28};
29
1function getWordsInLongestSubsequence(totalWords: number, wordsArray: string[], wordGroups: number[]): string[] {
2 // Initialize an array to hold the resulting longest subsequence of words.
3 const longestSubsequence: string[] = [];
4
5 // Iterate through the array of words to identify the longest subsequence.
6 for (let index = 0; index < totalWords; ++index) {
7 // If we are at the first word or the current word's group is different from the previous word's group,
8 // it is a part of the longest subsequence, so we add it to the result array.
9 if (index === 0 || wordGroups[index] !== wordGroups[index - 1]) {
10 longestSubsequence.push(wordsArray[index]);
11 }
12 }
13
14 // Return the longest subsequence found.
15 return longestSubsequence;
16}
17
Time and Space Complexity
The time complexity of the code is O(n)
, where n
is the length of the list groups
. This is because the code iterates over the list groups
once and performs a constant time check for each element.
The space complexity is O(k)
, where k
is the number of unique subsequences identified, which in the worst case could be equal to n
. This would happen if no two consecutive elements in groups
are the same, resulting in each word from words
being added to the output list. Thus, the space used for the output list is proportional to the number of selected words.
Learn more about how to find time and space complexity quickly using problem constraints.
How does quick sort divide the problem into subproblems?
Recommended Readings
Greedy Introduction div class responsive iframe iframe src https www youtube com embed WTslqPbj7I title YouTube video player frameborder 0 allow accelerometer autoplay clipboard write encrypted media gyroscope picture in picture web share allowfullscreen iframe div When do we use greedy Greedy algorithms tend to solve optimization problems Typically they will ask you to calculate the max min of some value Commonly you may see this phrased in the problem as max min longest shortest largest smallest etc These keywords can be identified by just scanning
What is Dynamic Programming Prerequisite DFS problems dfs_intro Backtracking problems backtracking Memoization problems memoization_intro Pruning problems backtracking_pruning Dynamic programming is an algorithmic optimization technique that breaks down a complicated problem into smaller overlapping sub problems in a recursive manner and uses solutions to the sub problems to construct a solution
LeetCode Patterns Your Personal Dijkstra's Algorithm to Landing Your Dream Job The goal of AlgoMonster is to help you get a job in the shortest amount of time possible in a data driven way We compiled datasets of tech interview problems and broke them down by patterns This way we
Got a question? Ask the Monster Assistant anything you don't understand.
Still not clear?  Submit the part you don't understand to our editors. Or join our Discord and ask the community.