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This approach utilizes a sliding window technique combined with character count comparisons to efficiently find all anagrams of the string p within the string s. By maintaining a window of length equal to p as we iterate through s, and by checking the frequency of characters within this window with the frequency of characters in p, we can determine if the substring is an anagram.
Time Complexity: O(n) where n is the length of s, because we iterate over `s` once.
Space Complexity: O(1) since the size of the character frequency array is fixed with respect to the alphabet.
1function findAnagrams(s, p) {
2 const pCount = new Array(26).fill(0);
3 const sCount = new Array(26).fill(0);
4 const result = [];
5 const aCharCode = 'a'.charCodeAt(0);
6
7 for (const char of p) {
8 pCount[char.charCodeAt(0) - aCharCode]++;
9 }
10
11 for (let i = 0; i < s.length; i++) {
12 sCount[s[i].charCodeAt(0) - aCharCode]++;
13
14 if (i >= p.length) {
15 sCount[s[i - p.length].charCodeAt(0) - aCharCode]--;
16 }
17
18 if (arraysEqual(pCount, sCount)) {
19 result.push(i - p.length + 1);
20 }
21 }
22 return result;
23}
24
25function arraysEqual(arr1, arr2) {
26 for (let i = 0; i < 26; i++) {
27 if (arr1[i] !== arr2[i]) return false;
28 }
29 return true;
30}This JavaScript solution uses two frequency arrays to keep count of the characters in p and a sliding window in s. The window size is maintained to be the length of p. As the window slides through s, we check if the two frequency arrays (for the window and p) are equal, indicating an anagram found.
This approach leverages hashmaps (or dictionaries) and a two-pointer technique to find the anagrams of p in s. We maintain counts of characters using hashmaps and adjust them as the two pointers traverse through s.
Time Complexity: O(n) where n is the length of `s`.
Space Complexity: O(1) if considering character frequency hashmaps to have at most 26 entries.
1import java.util.*;
2
3
This Java approach employs two arrays to store the frequency of each character. We iterate over s with a sliding window of size equal to p, updating these arrays accordingly. If both frequency arrays match at any point, it indicates the presence of an anagram.