LeetCode--567--medium--PermutationInString

package com.app.main.LeetCode.slidingwindow;

import java.util.Arrays;
import java.util.Collections;

/**
 *567
 *
 * medium
 *
 * https://leetcode.com/problems/permutation-in-string/
 *
 * Given two strings s1 and s2, write a function to return true if s2 contains the permutation of s1. In other words, one of the first string's permutations is the substring of the second string.
 *
 *
 *
 * Example 1:
 *
 * Input: s1 = "ab" s2 = "eidbaooo"
 * Output: True
 * Explanation: s2 contains one permutation of s1 ("ba").
 * Example 2:
 *
 * Input:s1= "ab" s2 = "eidboaoo"
 * Output: False
 *
 *
 * Note:
 *
 * The input strings only contain lower case letters.
 * The length of both given strings is in range [1, 10,000].
 *
 *
 * Created with IDEA
 * author:Dingsheng Huang
 * Date:2020/1/20
 * Time:下午8:35
 */
public class PermutationInString {

    // sliding window  O(n)
    public boolean checkInclusion(String s1, String s2) {

        int cn = s1.length();
        int[] count = new int[26];
        for (int i = 0; i < s1.length(); i++) {
            count[s1.charAt(i) - 'a']++;
        }
        int[] count2 = new int[26];

        int l = 0;
        for (int r = 0; r < s2.length(); r++) {
            if (count[s2.charAt(r) - 'a'] == 0) {
                // optimize , needn't new int[26], just transfer l and backtrack data
//                l = r + 1;
//                count2 = new int[26];
//                cn = s1.length();
                while (l < r) {
                    count2[s2.charAt(l) - 'a']--;
                    cn++;
                    l++;
                }
                l++;
                continue;
            }
            if (count2[s2.charAt(r) - 'a'] < count[s2.charAt(r) - 'a']) {
                count2[s2.charAt(r) - 'a']++;
                cn--;
                if (cn == 0) {
                    return true;
                }
                continue;
            }
            // count2[s2.charAt(r) - 'a'] == count[s2.charAt(r) - 'a'] ， curr ch is full, transfer l and backtrack data
            while (l < r && s2.charAt(l) != s2.charAt(r)) {
                count2[s2.charAt(l) - 'a']--;
                cn++;
                l++;
            }
            if (l < r) {
                l++;
            } else {
                count2[s2.charAt(r) - 'a']++;
                cn--;
            }
        }

        return false;
    }

}