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.
Build confidence with an intuition-first walkthrough focused on array fundamentals.
In an alien language, surprisingly, they also use English lowercase letters, but possibly in a different order. The order of the alphabet is some permutation of lowercase letters.
Given a sequence of words written in the alien language, and the order of the alphabet, return true if and only if the given words are sorted lexicographically in this alien language.
Example 1:
Input: words = ["hello","leetcode"], order = "hlabcdefgijkmnopqrstuvwxyz" Output: true Explanation: As 'h' comes before 'l' in this language, then the sequence is sorted.
Example 2:
Input: words = ["word","world","row"], order = "worldabcefghijkmnpqstuvxyz" Output: false Explanation: As 'd' comes after 'l' in this language, then words[0] > words[1], hence the sequence is unsorted.
Example 3:
Input: words = ["apple","app"], order = "abcdefghijklmnopqrstuvwxyz" Output: false Explanation: The first three characters "app" match, and the second string is shorter (in size.) According to lexicographical rules "apple" > "app", because 'l' > '∅', where '∅' is defined as the blank character which is less than any other character (More info).
Constraints:
1 <= words.length <= 1001 <= words[i].length <= 20order.length == 26words[i] and order are English lowercase letters.Problem summary: In an alien language, surprisingly, they also use English lowercase letters, but possibly in a different order. The order of the alphabet is some permutation of lowercase letters. Given a sequence of words written in the alien language, and the order of the alphabet, return true if and only if the given words are sorted lexicographically in this alien language.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Array · Hash Map
["hello","leetcode"] "hlabcdefgijkmnopqrstuvwxyz"
["word","world","row"] "worldabcefghijkmnpqstuvxyz"
["apple","app"] "abcdefghijklmnopqrstuvwxyz"
Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #953: Verifying an Alien Dictionary
class Solution {
public boolean isAlienSorted(String[] words, String order) {
int[] m = new int[26];
for (int i = 0; i < 26; ++i) {
m[order.charAt(i) - 'a'] = i;
}
for (int i = 0; i < 20; ++i) {
int prev = -1;
boolean valid = true;
for (String x : words) {
int curr = i >= x.length() ? -1 : m[x.charAt(i) - 'a'];
if (prev > curr) {
return false;
}
if (prev == curr) {
valid = false;
}
prev = curr;
}
if (valid) {
break;
}
}
return true;
}
}
// Accepted solution for LeetCode #953: Verifying an Alien Dictionary
func isAlienSorted(words []string, order string) bool {
m := make([]int, 26)
for i, c := range order {
m[c-'a'] = i
}
for i := 0; i < 20; i++ {
prev := -1
valid := true
for _, x := range words {
curr := -1
if i < len(x) {
curr = m[x[i]-'a']
}
if prev > curr {
return false
}
if prev == curr {
valid = false
}
prev = curr
}
if valid {
break
}
}
return true
}
# Accepted solution for LeetCode #953: Verifying an Alien Dictionary
class Solution:
def isAlienSorted(self, words: List[str], order: str) -> bool:
m = {c: i for i, c in enumerate(order)}
for i in range(20):
prev = -1
valid = True
for x in words:
curr = -1 if i >= len(x) else m[x[i]]
if prev > curr:
return False
if prev == curr:
valid = False
prev = curr
if valid:
return True
return True
// Accepted solution for LeetCode #953: Verifying an Alien Dictionary
use std::collections::HashMap;
impl Solution {
pub fn is_alien_sorted(words: Vec<String>, order: String) -> bool {
let n = words.len();
let mut map = HashMap::new();
order.as_bytes().iter().enumerate().for_each(|(i, &v)| {
map.insert(v, i);
});
for i in 1..n {
let s1 = words[i - 1].as_bytes();
let s2 = words[i].as_bytes();
let mut is_equal = true;
for i in 0..s1.len().min(s2.len()) {
if map.get(&s1[i]) > map.get(&s2[i]) {
return false;
}
if map.get(&s1[i]) < map.get(&s2[i]) {
is_equal = false;
break;
}
}
if is_equal && s1.len() > s2.len() {
return false;
}
}
true
}
}
// Accepted solution for LeetCode #953: Verifying an Alien Dictionary
function isAlienSorted(words: string[], order: string): boolean {
const map = new Map();
for (const c of order) {
map.set(c, map.size);
}
const n = words.length;
for (let i = 1; i < n; i++) {
const s1 = words[i - 1];
const s2 = words[i];
const m = Math.min(s1.length, s2.length);
let isEqual = false;
for (let j = 0; j < m; j++) {
if (map.get(s1[j]) > map.get(s2[j])) {
return false;
}
if (map.get(s1[j]) < map.get(s2[j])) {
isEqual = true;
break;
}
}
if (!isEqual && s1.length > s2.length) {
return false;
}
}
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.