Find All Anagrams in a String

String, Two Pointers

Easy? ... ~ Medium if O(n) required

Given a stringsand 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".

Solution & Analysis

Approach 2：Flexible Size Sliding Window

public class Solution {
public List<Integer> findAnagrams(String s, String t) {
List<Integer> result = new LinkedList<>();
if(t.length()> s.length()) return result;
Map<Character, Integer> map = new HashMap<>();
for(char c : t.toCharArray()){
map.put(c, map.getOrDefault(c, 0) + 1);
}
int counter = map.size();

int begin = 0, end = 0;

while(end < s.length()){
char c = s.charAt(end);
if( map.containsKey(c) ){
map.put(c, map.get(c)-1);
if(map.get(c) == 0) counter--;
}
end++;

while(counter == 0){
char tempc = s.charAt(begin);
if(map.containsKey(tempc)){
map.put(tempc, map.get(tempc) + 1);
if(map.get(tempc) > 0){
counter++;
}
}
if(end-begin == t.length()){
}
begin++;
}

}
return result;
}
}

Other Approach:

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;

int[] hash = new int; //character hash

//record each character in p to hash
for (char c : p.toCharArray()) {
hash[c]++;
}
//two points, initialize count to p's length
int left = 0, right = 0, count = p.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 (hash[s.charAt(right)] >= 1) {
count--;
}
hash[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 (narrow the window) 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 (hash[s.charAt(left)] >= 0) {
count++;
}
hash[s.charAt(left)]++;
left++;

}

}
return list;
}
}