Leetcode

Permutation in String

May 26, 2022 by aAshish

Approach 1: Classic sliding window

  • Time:O(|\texttt{s1}| + |\texttt{s2}|)
  • Space:O(128) = O(1)

C++

 
class Solution {
 public:
  bool checkInclusion(string s1, string s2) {
    vector<int> count(128);
    int required = s1.length();

    for (const char c : s1)
      ++count[c];

    for (int l = 0, r = 0; r < s2.length(); ++r) {
      if (--count[s2[r]] >= 0)
        --required;
      while (required == 0) {
        if (r - l + 1 == s1.length())
          return true;
        if (++count[s2[l++]] > 0)
          ++required;
      }
    }

    return false;
  }
};

JAVA

 
class Solution {
  public boolean checkInclusion(String s1, String s2) {
    int[] count = new int[128];
    int required = s1.length();

    for (final char c : s1.toCharArray())
      ++count[c];

    for (int l = 0, r = 0; r < s2.length(); ++r) {
      if (--count[s2.charAt(r)] >= 0)
        --required;
      while (required == 0) {
        if (r - l + 1 == s1.length())
          return true;
        if (++count[s2.charAt(l++)] > 0)
          ++required;
      }
    }

    return false;
  }
}

Approach 2: Constant-sized moving window

  • Time:O(|\texttt{s1}| + |\texttt{s2}|)
  • Space:O(128) = O(1)

C++

 
class Solution {
 public:
  bool checkInclusion(string s1, string s2) {
    vector<int> count(128);
    int required = s1.length();

    for (const char c : s1)
      ++count[c];

    for (int r = 0; r < s2.length(); ++r) {
      if (--count[s2[r]] >= 0)
        --required;
      if (r >= s1.length())  // the window is oversized
        if (++count[s2[r - s1.length()]] > 0)
          ++required;
      if (required == 0)
        return true;
    }

    return false;
  }
};

JAVA

 
class Solution {
  public boolean checkInclusion(String s1, String s2) {
    int[] count = new int[128];
    int required = s1.length();

    for (final char c : s1.toCharArray())
      ++count[c];

    for (int r = 0; r < s2.length(); ++r) {
      if (--count[s2.charAt(r)] >= 0)
        --required;
      if (r >= s1.length()) // the window is oversized
        if (++count[s2.charAt(r - s1.length())] > 0)
          ++required;
      if (required == 0)
        return true;
    }

    return false;
  }
}

Python

 
class Solution:
  def checkInclusion(self, s1: str, s2: str) -> bool:
    count = Counter(s1)
    required = len(s1)

    for r, c in enumerate(s2):
      count[c] -= 1
      if count[c] >= 0:
        required -= 1
      if r >= len(s1):
        count[s2[r - len(s1)]] += 1
        if count[s2[r - len(s1)]] > 0:
          required += 1
      if required == 0:
        return True

    return False