LeetCode-in-Rust
438. Find All Anagrams in a String
Medium
Given two strings s and p, return an array of all the start indices of p’s anagrams in s. You may return the answer in any order.
An Anagram is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once.
Example 1:
Input: s = “cbaebabacd”, p = “abc”
Output: [0,6]
Explanation:
The substring with start index = 0 is “cba”, which is an anagram of “abc”.
The substring with start index = 6 is “bac”, which is an anagram of “abc”.
Example 2:
Input: s = “abab”, p = “ab”
Output: [0,1,2]
Explanation:
The substring with start index = 0 is “ab”, which is an anagram of “ab”.
The substring with start index = 1 is “ba”, which is an anagram of “ab”.
The substring with start index = 2 is “ab”, which is an anagram of “ab”.
Constraints:
1 <= s.length, p.length <= 3 * 104sandpconsist of lowercase English letters.
Solution
use std::collections::VecDeque;
impl Solution {
pub fn find_anagrams(s: String, p: String) -> Vec<i32> {
let mut map = vec![0; 26];
let p_len = p.len();
let s_len = s.len();
// Fill the map with the character frequencies of string `p`
for ch in p.chars() {
map[ch as usize - 'a' as usize] += 1;
}
let mut res = Vec::new();
let mut j = 0;
// Sliding window
for i in 0..s_len {
let idx = s.as_bytes()[i] as usize - 'a' as usize;
map[idx] -= 1;
// If window size exceeds p_len, remove the left character
if i >= p_len {
let left_idx = s.as_bytes()[j] as usize - 'a' as usize;
map[left_idx] += 1;
j += 1;
}
// Check if the window is an anagram of `p`
if i >= p_len - 1 && map.iter().all(|&x| x == 0) {
res.push(j as i32);
}
}
res
}
}