logseq_notes/pages/OJ notes/pages/Leetcode Permutation-In-String.md

Leetcode Permutation-In-String

2022-07-14 10:29

Algorithms:

#algorithm #sliding_window

Data structures:

#DS #string

Difficulty:

#coding_problems #difficulty_medium

Additional tags:

#leetcode #CS_list_need_understanding

Revisions:

N/A

Related topics:

Links:

• Link to problem

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Problem

Given two strings s1 and s2, return true if s2 contains a permutation of s1, or false otherwise.

In other words, return true if one of s1's permutations is the substring of s2.

Examples

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

Constraints

• 1 <= s1.length, s2.length <= 104
• s1 and s2 consist of lowercase English letters.

Thoughts

[!summary] This is a #sliding_window problem.

I tried to use kadane's algorithm, but the problem is a premature string, not set. So I gave up.

Then I found out this is sliding window problem, similar to Leetcode Longest-Substring-Without-Repeating-Characters

Solution

class Solution {
public:
bool checkInclusion(string s1, string s2) {
  // Sliding windows problem.

  // First, register the freqs of s1
  vector<int> freq(26, 0);
  for (char ch : s1) {
    freq[ch - 'a']++;
  }

  // Second, start constructing window and window freq;
  int len = s1.length();
  int l = 0, r = 0;
  vector<int> window(26, 0);

  // Start expanding and sliding!
  while (r < s2.length()) {
    // I add value to window here, to avoid r OOB when r = s2.length
    window[s2[r] - 'a']++;
    if (r - l + 1 < len) {
      // Expand right
      r++;
    } else if (window == freq) { // note that I use else if, to make sure
                                 // every time r is incremented once.
      return true;
    } else {
      // Remove l from the window, slide right
      window[s2[l] - 'a']--;
      l++;
      r++;
    }
  }

  return false;
}
};