Massive Algorithms: LeetCode 438 - Find All Anagrams in a String

LeetCode 438 - Find All Anagrams in a String

https://leetcode.com/problems/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".

X. use map
https://discuss.leetcode.com/topic/64423/java-o-n-using-hashmap-easy-understanding
https://discuss.leetcode.com/topic/64447/java-o-n-solution-hashmap-sliding-window
https://discuss.leetcode.com/topic/71138/sliding-window-and-hashmap-o-n

    public List<Integer> findAnagrams(String s, String p) {
        List<Integer> result = new ArrayList<>();
        if (s == null || s.length() == 0) {
            return result;
        }
        if (p.length() > s.length()) {
            return result;
        }
        Map<Character, Integer> map = new HashMap<>();
        for (int i = 0; i < p.length(); i++) {
            char c = p.charAt(i);
            if (map.containsKey(c)) {
                map.put(c, map.get(c) + 1);
            } else {
                map.put(c, 1);
            }
        }
        int match = 0;
        for (int i = 0; i < s.length(); i++) {
            char c = s.charAt(i);
            if (map.containsKey(c)) {
                map.put(c, map.get(c) - 1);
                if (map.get(c) == 0) {
                    match++;
                }
            }
            if (i >= p.length()) {
                c = s.charAt(i - p.length());
                if (map.containsKey(c)) {
                    map.put(c, map.get(c) + 1);
                    if (map.get(c) == 1) {
                        match--;
                    }
                }
            }
            if (match == map.size()) {
                result.add(i - p.length() + 1);
            }
        }
        return result;
    }

https://leetcode.com/problems/find-all-anagrams-in-a-string/discuss/92015/ShortestConcise-JAVA-O(n)-Sliding-Window-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[256]; //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--; 
        
        //when the count is down to 0, means we found the right anagram
        //then add window's left to result list
        if (count == 0) list.add(left);
    
        //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() && hash[s.charAt(left++)]++ >= 0) count++;
    }
    return list;
}

https://discuss.leetcode.com/topic/87897/sliding-window-in-java-very-similar-to-find-all-anagrams-in-a-string
https://discuss.leetcode.com/topic/64423/java-o-n-using-hashmap-easy-understanding

    public List<Integer> slidingWindowTemplateByHarryChaoyangHe(String s, String t) {
        //init a collection or int value to save the result according the question.
        List<Integer> result = new LinkedList<>();
        if(t.length()> s.length()) return result;
        
        //create a hashmap to save the Characters of the target substring.
        //(K, V) = (Character, Frequence of the Characters)
        Map<Character, Integer> map = new HashMap<>();
        for(char c : t.toCharArray()){
            map.put(c, map.getOrDefault(c, 0) + 1);
        }
        //maintain a counter to check whether match the target string.
        int counter = map.size();//must be the map size, NOT the string size because the char may be duplicate.
        
        //Two Pointers: begin - left pointer of the window; end - right pointer of the window
        int begin = 0, end = 0;
        
        //the length of the substring which match the target string.
        int len = Integer.MAX_VALUE; 
        
        //loop at the begining of the source string
        while(end < s.length()){
            
            char c = s.charAt(end);//get a character
            
            if( map.containsKey(c) ){
                map.put(c, map.get(c)-1);// plus or minus one
                if(map.get(c) == 0) counter--;//modify the counter according the requirement(different condition).
            }
            end++;
            
            //increase begin pointer to make it invalid/valid again
            while(counter == 0 /* counter condition. different question may have different condition */){
                
                char tempc = s.charAt(begin);//***be careful here: choose the char at begin pointer, NOT the end pointer
                if(map.containsKey(tempc)){
                    map.put(tempc, map.get(tempc) + 1);//plus or minus one
                    if(map.get(tempc) > 0) counter++;//modify the counter according the requirement(different condition).
                }
                
                /* save / update(min/max) the result if find a target*/
                // result collections or result int value
                
                begin++;
            }
        }
        return result;
    }

https://discuss.leetcode.com/topic/64434/shortest-concise-java-o-n-sliding-window-solution
https://tech.liuchao.me/2016/11/leetcode-solution-438/

思路：直接的方式是暴力用hash，这样时间复杂度O(m*n)．可以用hash+slide window来优化到O(n)．

Same idea from a fantastic sliding window template, please refer:
https://discuss.leetcode.com/topic/30941/here-is-a-10-line-template-that-can-solve-most-substring-problems

Time Complexity will be O(n) because the "start" and "end" points will only move from left to right once.

Basically, we are interested only when every hash[i] becomes 0. There are a number of ways of doing it. To understand OP's approach, we observe that:

the sum of all hash[i] is always >=0;
count is the sum of all positive hash[i];
therefore, every hash[i] is zero if and only if count is 0.

The genius of this approach is that the code is shorter, compared to our instinctive approach of maintaining the count of hash[i]==0.

eed to convert a char to int, 256 is good..
Oh, wait. only lower case letters, 26 is sufficient but don't want to type - 'a'
Well, could find out what 'z' is and use that, but probably not worth the effort
Yeah, 256 is good...

The same goes for 128, except the other author probably roughly knows 'z' is just over 100

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[256]; //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--; 
        
        //when the count is down to 0, means we found the right anagram
        //then add window's left to result list
        if (count == 0) list.add(left);
    
        //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() && hash[s.charAt(left++)]++ >= 0) count++;
    }
    return list;
}

https://discuss.leetcode.com/topic/64622/17ms-java-sliding-window

public List<Integer> findAnagrams(String s, String p) {
    int[] chars = new int[26];//\\
    List<Integer> result = new ArrayList<>();

    if (s == null || p == null || s.length() < p.length())
        return result;
    for (char c : p.toCharArray())
        chars[c-'a']++;

    int start = 0, end = 0, count = p.length();
    // Go over the string
    while (end < s.length()) {
        // If the char at start appeared in p, we increase count
        if (end - start == p.length() && chars[s.charAt(start++)-'a']++ >= 0)
            count++;
        // If the char at end appeared in p (since it's not -1 after decreasing), we decrease count
        if (--chars[s.charAt(end++)-'a'] >= 0)
            count--;
        if (count == 0)
            result.add(start);
    }
    
    return result;
}

X.
https://discuss.leetcode.com/topic/64491/java-using-isanagram-helper-function-easy-to-understand

   public List<Integer> findAnagrams(String s, String p) {
        List<Integer> res = new ArrayList<>();
        if (p == null || s == null || s.length() < p.length()) return res;
        int m = s.length(), n = p.length();
        for (int i = 0; i < m-n+1; i++) {
            String cur = s.substring(i, i+n);
            if (helper(cur, p)) res.add(i);
        }
        return res;
    }
    public boolean helper(String a, String b) {
        if (a == null || b == null || a.length() != b.length()) return false;
        int[] dict = new int[26];
        for (int i = 0; i < a.length(); i++) {
            char ch = a.charAt(i);
            dict[ch-'a']++;
        }
        for (int i = 0; i < b.length(); i++) {
            char ch = b.charAt(i);
            dict[ch-'a']--;
            if (dict[ch-'a'] < 0) return false;
        }
        return true;
    }

X. Not efficient
https://dyang2016.wordpress.com/2016/10/26/438-find-all-anagrams-in-a-string/
https://discuss.leetcode.com/topic/64401/java-sliding-window
https://discuss.leetcode.com/topic/64417/commented-java-solution

    public List<Integer> findAnagrams(String s, String p) {
        List<Integer> res = new ArrayList<>();
        if(p.length() > s.length())
            return res;
        char[] sStr = s.toCharArray();
        int[]map = new int[26];
        for(char ch:p.toCharArray())
            map[ch - 'a']++;
        int n = s.length(), m = p.length();;
        int j = 0;
        for(j=0; j<m-1; j++)
            map[sStr[j] - 'a']--;
        for(int i=0; j<n; i++, j++){
            map[sStr[j] - 'a']--;
            if(check(map))
                res.add(i);
            map[sStr[i] - 'a']++;
        }
        return res;
    }
    public boolean check(int[]map){
        for(int n:map)
            if(n > 0)   return false;
        return true;
    }

X. Using rolling hash -> convert to long
https://www.dailycodingproblem.com/blog/anagram-indices/
1. Rolling hash

One way to solve this problem is to use the Rabin-Karp algorithm. The basic idea is that we could make a frequency-based hash of the target word and check if any window under s hashes to the same value. That is, the hash would be the sum of char * prime_num ** char_freq for each character and their frequency. If the hash of the word and window matches, we could do a manual == of the two strings. Since collisions are expected to be rare, this would be O(S). However, there’s an easier way to solve this problem:

Count difference

Notice that moving along the window seems to mean recomputing the frequency counts of the entire window, when only a little bit of it actually updated. This insight leads us to the following strategy:

Make a frequency dictionary of the target word
Continuously diff against it as we go along the string
When the dict is empty, the window and the word matches

We diff in our frequency dict by incrementing the new character in the window and removing
old one.

http://blog.gainlo.co/index.php/2016/04/08/if-a-string-contains-an-anagram-of-another-string/

How to check if a string contains an anagram of another string?

LeetCode 438 - Find All Anagrams in a String

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