LeetCode-in-All
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
#include <vector>
#include <string>
using namespace std;
class Solution {
public:
vector<int> findAnagrams(string s, string p) {
vector<int> map(26, 0);
for (char c : p) {
map[c - 'a']++;
}
vector<int> res;
int i = 0, j = 0;
while (i < s.length()) {
int idx = s[i] - 'a';
// Add the new character
map[idx]--;
// If the length is greater than window's length, pop the left character in the window
if (i >= p.length()) {
map[s[j++] - 'a']++;
}
bool finish = true;
for (int k = 0; k < 26; k++) {
// If it is not an anagram of string p
if (map[k] != 0) {
finish = false;
break;
}
}
if (i >= p.length() - 1 && finish) {
res.push_back(j);
}
i++;
}
return res;
}
};