Problem

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;

}

};

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。

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].

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;

}

};

Posts tagged as “permuation”

花花酱 LeetCode 890. Find and Replace Pattern

Problem

Solution: Remember the last pos of each char.

花花酱 LeetCode 567. Permutation in String

Problem

Solution: Sliding Window

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