You have a list of words and a pattern, and you want to know which words in words matches the pattern.

A word matches the pattern if there exists a permutation of letters p so that after replacing every letter x in the pattern with p(x), we get the desired word.

(Recall that a permutation of letters is a bijection from letters to letters: every letter maps to another letter, and no two letters map to the same letter.)

Return a list of the words in words that match the given pattern.

You may return the answer in any order.

Example 1:

Input: words = ["abc","deq","mee","aqq","dkd","ccc"], pattern = "abb"
Output: ["mee","aqq"]
Explanation: "mee" matches the pattern because there is a permutation {a -> m, b -> e, ...}. 
"ccc" does not match the pattern because {a -> c, b -> c, ...} is not a permutation,
since a and b map to the same letter.

Note:

• 1 <= words.length <= 50
• 1 <= pattern.length = words[i].length <= 20

Solution: Remember the last pos of each char.

Time complexity: O(n*l)

Space complexity: O(128) -> O(26)

C++

// Author: Huahua
// Running time: 0 ms
class Solution {
public:
  vector<string> findAndReplacePattern(vector<string>& words, string pattern) {    
    vector<string> ans;
    for (const string& w : words)
      if (match(w, pattern)) ans.push_back(w);
    return ans;
  }
private:
  bool match(const string& w, const string& p) {
    vector<int> last_pos_w(128, INT_MIN); // last pos of x in w
    vector<int> last_pos_p(128, INT_MIN); // last pos of x in p
    for (int i = 0; i < w.size(); ++i)
      if (last_pos_w[w[i]] != last_pos_p[p[i]]) {
        return false;
      } else {
        last_pos_w[w[i]] = last_pos_p[p[i]] = i;
      }
    return true;
  }
};

The post 花花酱 LeetCode 890. Find and Replace Pattern appeared first on Huahua's Tech Road.

Problem

题目大意：给你s1, s2，问你s2的子串中是否存在s1的一个排列。

https://leetcode.com/problems/permutation-in-string/description/

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:

1. The input strings only contain lower case letters.
2. The length of both given strings is in range [1, 10,000].

Solution: Sliding Window

Time Complexity: O(l1 + l2 * 26) = O(l1 + l2)

Space Complexity: O(26 * 2) = O(1)

C++

// Author: Huahua
// Running time: 12 ms
class Solution {
public:
  bool checkInclusion(string s1, string s2) {
    int l1 = s1.length();
    int l2 = s2.length();    
    vector<int> c1(26);
    vector<int> c2(26);
    for (const char c : s1)
      ++c1[c - 'a'];
    for (int i = 0; i < l2; ++i) {
      if (i >= l1)
        --c2[s2[i - l1] - 'a'];
      ++c2[s2[i] - 'a'];      
      if (c1 == c2) return true;      
    }
    return false;
  }
};

Related Problems

• 花花酱 LeetCode 480. Sliding Window Median
• 花花酱 LeetCode 239. Sliding Window Maximum
• 花花酱 LeetCode 438. Find All Anagrams in a String

The post 花花酱 LeetCode 567. Permutation in String appeared first on Huahua's Tech Road.