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 1:Fixed Size Sliding Window

一种思路是不断移动大小为t.length()的window,如果当前window中的字符统计和target一样,就放入结果中。

Approach 2:Flexible Size Sliding Window

通用Sliding Window 模板可以解决:counter记录map中还需要match的字符个数,如果counter==0,说明当前window中substring包含了要找的target string anagram,但是可以进一步检测是否可以缩短这个window。通过调整begin来实现。所以这个sliding window其实window size是在变化的,只有当counter==0,并且当前window size: end - begine == t.length()时,才记录结果。

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()){
                    result.add(begin);
                }
                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[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--;
        }
        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) {
            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() ) {

            if (hash[s.charAt(left)] >= 0) {
                count++;
            }
            hash[s.charAt(left)]++;
            left++;

        }


    }
        return list;
    }
}

Reference

results matching ""

    No results matching ""