# Find All Anagrams in a String

438. Find All Anagrams in a String

Given a string s and a non-empty string p, find all the start indices of p's anagrams in s.

Strings consists of lowercase English letters only and the length of both strings s and p will not be larger than 20,100.

The order of output does not matter.

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".
``````
``````public class Solution {
public List<Integer> findAnagrams(String s, String p) {
List<Integer> list = new ArrayList<>();
if (s == null || s.length() == 0
|| p == null || p.length() == 0) {
return list;
}

for (int i = 0; i <= s.length() - p.length();i++) {
String str = s.substring(i, i + p.length());

if (isAnagram(str, p)) {
}
}

return list;
}

private boolean isAnagram(String s, String t) {
if (s == null || t == null) {
return false;
}

if (s.length() != t.length()) {
return false;
}

int[] charCount = new int[256];
for (int i = 0; i < s.length(); i++) {
charCount[s.charAt(i)]++;
}

for (int i = 0; i < t.length(); i++) {
if (--charCount[t.charAt(i)] < 0) {
return false;
}
}

return true;
}
}
``````
``````public class Solution {
public List<Integer> findAnagrams(String s, String p) {
List<Integer> result = new ArrayList<>();
if (s == null || s.length() == 0
|| p == null || p.length() == 0) {
return result;
}

// record each character in p
int[] charCount = new int[256];
for (char c : p.toCharArray()) {
charCount[c]++;
}

int left = 0, right = 0; // two pointers
int count = p.length(); // initialize count to p's length

while (right < s.length()) {
// move right everytime, if the character exists in p's hash, decrease the count
// current hash value >= 1 means the character is existing in p
if (charCount[s.charAt(right)] >= 1) {
count--;
}

charCount[s.charAt(right)]--;
right++;

// when the count is down to 0, means we found the right anagram
// then add window's left to result list
if (count == 0) {
}

// if we find the window's size equals to p, then we have to move left
// to find the new match window
// ++ to reset the hash because we kicked out the left
// only increase the count if the character is in p
// the count >= 0 indicate it was original in the hash, cuz it won't go below 0
if (right - left == p.length()) {
if (charCount[s.charAt(left)] >= 0) {
count++;
}

charCount[s.charAt(left)]++;
left++;
}
}

return result;
}
}
``````

Hope this helps,
Michael