438. Find All Anagrams in a String (Medium)

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".

Solutions

class Solution {

    // The key idea for this problem is that anagrams are of same length and the frequence of different
    // chars are identical.

    // We may image a window of the same length as p and work out the character
    // frequence in this window.

    // We define two arrays of length 256 of which one keeps the char frequence of p and the other keeps s's.

    // we update the char frequence in the window when scanning the given string. We hit the anagram string
    // as long as the frequence matches.

    public List<Integer> findAnagrams(String s, String p) {
        List<Integer> ans = new ArrayList<>();

        if (s == null || p == null || s.length() < p.length()) {
            return ans;
        }

        // window size
        int ws = p.length();

        // frequence map for chars in pattern string
        int[] freqP = new int[26];
        Arrays.fill(freqP, 0);

        for (int i = 0; i < p.length(); i++) {
            freqP[p.charAt(i) - 'a']++;
        }

        // frequence map for chars in test string of moving window
        int[] freqS = new int[26];
        Arrays.fill(freqS, 0);

        for (int i = 0; i < s.length(); i++) {
            // The new element incorporated when window moving forward
            int cRight = s.charAt(i) - 'a';

            freqS[cRight]++;

            if (i < ws - 1) {
                continue;
            }

            if (i > ws - 1) {
                // the leftmost element of previous window. It is edged off when window
                // moving forward.
                int cLeft = s.charAt(i - ws) - 'a';

                freqS[cLeft]--;
            }

            boolean isAnagram = true;

            for (int j = 0; j < 26; j++) {
                if (freqS[j] == freqP[j]) {
                    continue;
                }

                isAnagram = false;
                break;
            }

            if (isAnagram) {
                ans.add(i - ws + 1);
            }
        }

        return ans;
    }
}

Incorrect Solutions

References

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