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.
Given a list of strings words and a string pattern, return a list of words[i] that match pattern. You may return the answer in any order.
A word matches the pattern if there exists a permutation of letters p so that after replacing every letter x in the pattern with p(x), we get the desired word.
Recall that a permutation of letters is a bijection from letters to letters: every letter maps to another letter, and no two letters map to the same letter.
Example 1:
Input: words = ["abc","deq","mee","aqq","dkd","ccc"], pattern = "abb"
Output: ["mee","aqq"]
Explanation: "mee" matches the pattern because there is a permutation {a -> m, b -> e, ...}.
"ccc" does not match the pattern because {a -> c, b -> c, ...} is not a permutation, since a and b map to the same letter.
Example 2:
Input: words = ["a","b","c"], pattern = "a" Output: ["a","b","c"]
Constraints:
1 <= pattern.length <= 201 <= words.length <= 50words[i].length == pattern.lengthpattern and words[i] are lowercase English letters.Problem summary: Given a list of strings words and a string pattern, return a list of words[i] that match pattern. You may return the answer in any order. A word matches the pattern if there exists a permutation of letters p so that after replacing every letter x in the pattern with p(x), we get the desired word. Recall that a permutation of letters is a bijection from letters to letters: every letter maps to another letter, and no two letters map to the same letter.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Array · Hash Map
["abc","deq","mee","aqq","dkd","ccc"] "abb"
["a","b","c"] "a"
isomorphic-strings)word-pattern)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #890: Find and Replace Pattern
class Solution {
public List<String> findAndReplacePattern(String[] words, String pattern) {
List<String> ans = new ArrayList<>();
for (String word : words) {
if (match(word, pattern)) {
ans.add(word);
}
}
return ans;
}
private boolean match(String s, String t) {
int[] m1 = new int[128];
int[] m2 = new int[128];
for (int i = 0; i < s.length(); ++i) {
char c1 = s.charAt(i);
char c2 = t.charAt(i);
if (m1[c1] != m2[c2]) {
return false;
}
m1[c1] = i + 1;
m2[c2] = i + 1;
}
return true;
}
}
// Accepted solution for LeetCode #890: Find and Replace Pattern
func findAndReplacePattern(words []string, pattern string) []string {
match := func(s, t string) bool {
m1, m2 := make([]int, 128), make([]int, 128)
for i := 0; i < len(s); i++ {
if m1[s[i]] != m2[t[i]] {
return false
}
m1[s[i]] = i + 1
m2[t[i]] = i + 1
}
return true
}
var ans []string
for _, word := range words {
if match(word, pattern) {
ans = append(ans, word)
}
}
return ans
}
# Accepted solution for LeetCode #890: Find and Replace Pattern
class Solution:
def findAndReplacePattern(self, words: List[str], pattern: str) -> List[str]:
def match(s, t):
m1, m2 = [0] * 128, [0] * 128
for i, (a, b) in enumerate(zip(s, t), 1):
if m1[ord(a)] != m2[ord(b)]:
return False
m1[ord(a)] = m2[ord(b)] = i
return True
return [word for word in words if match(word, pattern)]
// Accepted solution for LeetCode #890: Find and Replace Pattern
use std::collections::HashMap;
impl Solution {
pub fn find_and_replace_pattern(words: Vec<String>, pattern: String) -> Vec<String> {
let pattern = pattern.as_bytes();
let n = pattern.len();
words
.into_iter()
.filter(|word| {
let word = word.as_bytes();
let mut map1 = HashMap::new();
let mut map2 = HashMap::new();
for i in 0..n {
if map1.get(&word[i]).unwrap_or(&n) != map2.get(&pattern[i]).unwrap_or(&n) {
return false;
}
map1.insert(word[i], i);
map2.insert(pattern[i], i);
}
true
})
.collect()
}
}
// Accepted solution for LeetCode #890: Find and Replace Pattern
function findAndReplacePattern(words: string[], pattern: string): string[] {
return words.filter(word => {
const map1 = new Map<string, number>();
const map2 = new Map<string, number>();
for (let i = 0; i < word.length; i++) {
if (map1.get(word[i]) !== map2.get(pattern[i])) {
return false;
}
map1.set(word[i], i);
map2.set(pattern[i], i);
}
return true;
});
}
Use this to step through a reusable interview workflow for this problem.
Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.
Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.
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: Zero-count keys stay in map and break distinct/count constraints.
Usually fails on: Window/map size checks are consistently off by one.
Fix: Delete keys when count reaches zero.