Off-by-one on range boundaries
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.
Move from brute-force thinking to an efficient approach using array strategy.
You are given a string array words, and an array groups, both arrays having length n.
The hamming distance between two strings of equal length is the number of positions at which the corresponding characters are different.
You need to select the longest subsequence from an array of indices [0, 1, ..., n - 1], such that for the subsequence denoted as [i0, i1, ..., ik-1] having length k, the following holds:
groups[ij] != groups[ij+1], for each j where 0 < j + 1 < k.words[ij] and words[ij+1] are equal in length, and the hamming distance between them is 1, where 0 < j + 1 < k, for all indices in the subsequence.Return a string array containing the words corresponding to the indices (in order) in the selected subsequence. If there are multiple answers, return any of them.
Note: strings in words may be unequal in length.
Example 1:
Input: words = ["bab","dab","cab"], groups = [1,2,2]
Output: ["bab","cab"]
Explanation: A subsequence that can be selected is [0,2].
groups[0] != groups[2]words[0].length == words[2].length, and the hamming distance between them is 1.So, a valid answer is [words[0],words[2]] = ["bab","cab"].
Another subsequence that can be selected is [0,1].
groups[0] != groups[1]words[0].length == words[1].length, and the hamming distance between them is 1.So, another valid answer is [words[0],words[1]] = ["bab","dab"].
It can be shown that the length of the longest subsequence of indices that satisfies the conditions is 2.
Example 2:
Input: words = ["a","b","c","d"], groups = [1,2,3,4]
Output: ["a","b","c","d"]
Explanation: We can select the subsequence [0,1,2,3].
It satisfies both conditions.
Hence, the answer is [words[0],words[1],words[2],words[3]] = ["a","b","c","d"].
It has the longest length among all subsequences of indices that satisfy the conditions.
Hence, it is the only answer.
Constraints:
1 <= n == words.length == groups.length <= 10001 <= words[i].length <= 101 <= groups[i] <= nwords consists of distinct strings.words[i] consists of lowercase English letters.Problem summary: You are given a string array words, and an array groups, both arrays having length n. The hamming distance between two strings of equal length is the number of positions at which the corresponding characters are different. You need to select the longest subsequence from an array of indices [0, 1, ..., n - 1], such that for the subsequence denoted as [i0, i1, ..., ik-1] having length k, the following holds: For adjacent indices in the subsequence, their corresponding groups are unequal, i.e., groups[ij] != groups[ij+1], for each j where 0 < j + 1 < k. words[ij] and words[ij+1] are equal in length, and the hamming distance between them is 1, where 0 < j + 1 < k, for all indices in the subsequence. Return a string array containing the words corresponding to the indices (in order) in the selected subsequence. If there are multiple answers, return any of them. Note: strings in words may be
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Array · Dynamic Programming
["bab","dab","cab"] [1,2,2]
["a","b","c","d"] [1,2,3,4]
Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #2901: Longest Unequal Adjacent Groups Subsequence II
class Solution {
public List<String> getWordsInLongestSubsequence(String[] words, int[] groups) {
int n = groups.length;
int[] f = new int[n];
int[] g = new int[n];
Arrays.fill(f, 1);
Arrays.fill(g, -1);
int mx = 1;
for (int i = 0; i < n; ++i) {
for (int j = 0; j < i; ++j) {
if (groups[i] != groups[j] && f[i] < f[j] + 1 && check(words[i], words[j])) {
f[i] = f[j] + 1;
g[i] = j;
mx = Math.max(mx, f[i]);
}
}
}
List<String> ans = new ArrayList<>();
for (int i = 0; i < n; ++i) {
if (f[i] == mx) {
for (int j = i; j >= 0; j = g[j]) {
ans.add(words[j]);
}
break;
}
}
Collections.reverse(ans);
return ans;
}
private boolean check(String s, String t) {
if (s.length() != t.length()) {
return false;
}
int cnt = 0;
for (int i = 0; i < s.length(); ++i) {
if (s.charAt(i) != t.charAt(i)) {
++cnt;
}
}
return cnt == 1;
}
}
// Accepted solution for LeetCode #2901: Longest Unequal Adjacent Groups Subsequence II
func getWordsInLongestSubsequence(words []string, groups []int) []string {
check := func(s, t string) bool {
if len(s) != len(t) {
return false
}
cnt := 0
for i := range s {
if s[i] != t[i] {
cnt++
}
}
return cnt == 1
}
n := len(groups)
f := make([]int, n)
g := make([]int, n)
for i := range f {
f[i] = 1
g[i] = -1
}
mx := 1
for i, x := range groups {
for j, y := range groups[:i] {
if x != y && f[i] < f[j]+1 && check(words[i], words[j]) {
f[i] = f[j] + 1
g[i] = j
if mx < f[i] {
mx = f[i]
}
}
}
}
ans := make([]string, 0, mx)
for i, x := range f {
if x == mx {
for j := i; j >= 0; j = g[j] {
ans = append(ans, words[j])
}
break
}
}
slices.Reverse(ans)
return ans
}
# Accepted solution for LeetCode #2901: Longest Unequal Adjacent Groups Subsequence II
class Solution:
def getWordsInLongestSubsequence(
self, words: List[str], groups: List[int]
) -> List[str]:
def check(s: str, t: str) -> bool:
return len(s) == len(t) and sum(a != b for a, b in zip(s, t)) == 1
n = len(groups)
f = [1] * n
g = [-1] * n
mx = 1
for i, x in enumerate(groups):
for j, y in enumerate(groups[:i]):
if x != y and f[i] < f[j] + 1 and check(words[i], words[j]):
f[i] = f[j] + 1
g[i] = j
mx = max(mx, f[i])
ans = []
for i in range(n):
if f[i] == mx:
j = i
while j >= 0:
ans.append(words[j])
j = g[j]
break
return ans[::-1]
// Accepted solution for LeetCode #2901: Longest Unequal Adjacent Groups Subsequence II
impl Solution {
pub fn get_words_in_longest_subsequence(words: Vec<String>, groups: Vec<i32>) -> Vec<String> {
fn check(s: &str, t: &str) -> bool {
s.len() == t.len() && s.chars().zip(t.chars()).filter(|(a, b)| a != b).count() == 1
}
let n = groups.len();
let mut f = vec![1; n];
let mut g = vec![-1; n];
let mut mx = 1;
for i in 0..n {
let x = groups[i] as usize;
for j in 0..i {
let y = groups[j] as usize;
if x != y && f[i] < f[j] + 1 && check(&words[i], &words[j]) {
f[i] = f[j] + 1;
g[i] = j as i32;
mx = mx.max(f[i]);
}
}
}
let mut ans = vec![];
let mut i = n - 1;
while f[i] != mx {
i -= 1;
}
let mut j = i as i32;
while j >= 0 {
ans.push(words[j as usize].clone());
j = g[j as usize];
}
ans.reverse();
ans
}
}
// Accepted solution for LeetCode #2901: Longest Unequal Adjacent Groups Subsequence II
function getWordsInLongestSubsequence(words: string[], groups: number[]): string[] {
const n = groups.length;
const f: number[] = Array(n).fill(1);
const g: number[] = Array(n).fill(-1);
let mx = 1;
const check = (s: string, t: string) => {
if (s.length !== t.length) {
return false;
}
let cnt = 0;
for (let i = 0; i < s.length; ++i) {
if (s[i] !== t[i]) {
++cnt;
}
}
return cnt === 1;
};
for (let i = 0; i < n; ++i) {
for (let j = 0; j < i; ++j) {
if (groups[i] !== groups[j] && f[i] < f[j] + 1 && check(words[i], words[j])) {
f[i] = f[j] + 1;
g[i] = j;
mx = Math.max(mx, f[i]);
}
}
}
const ans: string[] = [];
for (let i = 0; i < n; ++i) {
if (f[i] === mx) {
for (let j = i; ~j; j = g[j]) {
ans.push(words[j]);
}
break;
}
}
return ans.reverse();
}
Use this to step through a reusable interview workflow for this problem.
Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.
Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.
Review these before coding to avoid predictable interview regressions.
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.
Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.
Usually fails on: Correctness breaks on cases that differ only in hidden state.
Fix: Define state so each unique subproblem maps to one DP cell.