Question

567. Permutation in String

中文文档

Description

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.

 

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.

Solutions

Solution 1

class Solution:
  def checkInclusion(self, s1: str, s2: str) -> bool:
    n = len(s1)
    cnt1 = Counter(s1)
    cnt2 = Counter(s2[:n])
    if cnt1 == cnt2:
      return True
    for i in range(n, len(s2)):
      cnt2[s2[i]] += 1
      cnt2[s2[i - n]] -= 1
      if cnt1 == cnt2:
        return True
    return False

class Solution {
  public boolean checkInclusion(String s1, String s2) {
    int n = s1.length();
    int m = s2.length();
    if (n > m) {
      return false;
    }
    int[] cnt1 = new int[26];
    int[] cnt2 = new int[26];
    for (int i = 0; i < n; ++i) {
      ++cnt1[s1.charAt(i) - 'a'];
      ++cnt2[s2.charAt(i) - 'a'];
    }
    if (Arrays.equals(cnt1, cnt2)) {
      return true;
    }
    for (int i = n; i < m; ++i) {
      ++cnt2[s2.charAt(i) - 'a'];
      --cnt2[s2.charAt(i - n) - 'a'];
      if (Arrays.equals(cnt1, cnt2)) {
        return true;
      }
    }
    return false;
  }
}

class Solution {
public:
  bool checkInclusion(string s1, string s2) {
    int n = s1.size(), m = s2.size();
    if (n > m) {
      return false;
    }
    vector<int> cnt1(26), cnt2(26);
    for (int i = 0; i < n; ++i) {
      ++cnt1[s1[i] - 'a'];
      ++cnt2[s2[i] - 'a'];
    }
    if (cnt1 == cnt2) {
      return true;
    }
    for (int i = n; i < m; ++i) {
      ++cnt2[s2[i] - 'a'];
      --cnt2[s2[i - n] - 'a'];
      if (cnt1 == cnt2) {
        return true;
      }
    }
    return false;
  }
};

func checkInclusion(s1 string, s2 string) bool {
  n, m := len(s1), len(s2)
  if n > m {
    return false
  }
  cnt1 := [26]int{}
  cnt2 := [26]int{}
  for i := range s1 {
    cnt1[s1[i]-'a']++
    cnt2[s2[i]-'a']++
  }
  if cnt1 == cnt2 {
    return true
  }
  for i := n; i < m; i++ {
    cnt2[s2[i]-'a']++
    cnt2[s2[i-n]-'a']--
    if cnt1 == cnt2 {
      return true
    }
  }
  return false
}

function checkInclusion(s1: string, s2: string): boolean {
  // 滑动窗口方案
  if (s1.length > s2.length) {
    return false;
  }

  const n = s1.length;
  const m = s2.length;

  const toCode = (s: string) => s.charCodeAt(0) - 97;
  const isMatch = () => {
    for (let i = 0; i < 26; i++) {
      if (arr1[i] !== arr2[i]) {
        return false;
      }
    }
    return true;
  };

  const arr1 = new Array(26).fill(0);
  for (const s of s1) {
    const index = toCode(s);
    arr1[index]++;
  }

  const arr2 = new Array(26).fill(0);
  for (let i = 0; i < n; i++) {
    const index = toCode(s2[i]);
    arr2[index]++;
  }

  for (let l = 0, r = n; r < m; l++, r++) {
    if (isMatch()) {
      return true;
    }

    const i = toCode(s2[l]);
    const j = toCode(s2[r]);
    arr2[i]--;
    arr2[j]++;
  }
  return isMatch();
}

use std::collections::HashMap;

impl Solution {
  // 测试两个哈希表是否匹配
  fn is_match(m1: &HashMap<char, i32>, m2: &HashMap<char, i32>) -> bool {
    for (k, v) in m1.iter() {
      if m2.get(k).unwrap_or(&0) != v {
        return false;
      }
    }
    true
  }
  pub fn check_inclusion(s1: String, s2: String) -> bool {
    if s1.len() > s2.len() {
      return false;
    }
    let mut m1 = HashMap::new();
    let mut m2 = HashMap::new();
    // 初始化表 1
    for c in s1.chars() {
      m1.insert(c, m1.get(&c).unwrap_or(&0) + 1);
    }
    let cs: Vec<char> = s2.chars().collect();
    // 初始化窗口
    let mut i = 0;
    while i < s1.len() {
      m2.insert(cs[i], m2.get(&cs[i]).unwrap_or(&0) + 1);
      i += 1;
    }
    if Self::is_match(&m1, &m2) {
      return true;
    }
    // 持续滑动窗口，直到匹配或超出边界
    let mut j = 0;
    while i < cs.len() {
      m2.insert(cs[j], m2.get(&cs[j]).unwrap_or(&1) - 1);
      m2.insert(cs[i], m2.get(&cs[i]).unwrap_or(&0) + 1);
      j += 1;
      i += 1;
      if Self::is_match(&m1, &m2) {
        return true;
      }
    }
    false
  }
}

Solution 2

class Solution:
  def checkInclusion(self, s1: str, s2: str) -> bool:
    n, m = len(s1), len(s2)
    if n > m:
      return False
    cnt = Counter()
    for a, b in zip(s1, s2):
      cnt[a] -= 1
      cnt[b] += 1
    diff = sum(x != 0 for x in cnt.values())
    if diff == 0:
      return True
    for i in range(n, m):
      a, b = s2[i - n], s2[i]

      if cnt[b] == 0:
        diff += 1
      cnt[b] += 1
      if cnt[b] == 0:
        diff -= 1

      if cnt[a] == 0:
        diff += 1
      cnt[a] -= 1
      if cnt[a] == 0:
        diff -= 1

      if diff == 0:
        return True
    return False

class Solution {
  public boolean checkInclusion(String s1, String s2) {
    int n = s1.length();
    int m = s2.length();
    if (n > m) {
      return false;
    }
    int[] cnt = new int[26];
    for (int i = 0; i < n; ++i) {
      --cnt[s1.charAt(i) - 'a'];
      ++cnt[s2.charAt(i) - 'a'];
    }
    int diff = 0;
    for (int x : cnt) {
      if (x != 0) {
        ++diff;
      }
    }
    if (diff == 0) {
      return true;
    }
    for (int i = n; i < m; ++i) {
      int a = s2.charAt(i - n) - 'a';
      int b = s2.charAt(i) - 'a';
      if (cnt[b] == 0) {
        ++diff;
      }
      if (++cnt[b] == 0) {
        --diff;
      }
      if (cnt[a] == 0) {
        ++diff;
      }
      if (--cnt[a] == 0) {
        --diff;
      }
      if (diff == 0) {
        return true;
      }
    }
    return false;
  }
}

class Solution {
public:
  bool checkInclusion(string s1, string s2) {
    int n = s1.size(), m = s2.size();
    if (n > m) {
      return false;
    }
    vector<int> cnt(26);
    for (int i = 0; i < n; ++i) {
      --cnt[s1[i] - 'a'];
      ++cnt[s2[i] - 'a'];
    }
    int diff = 0;
    for (int x : cnt) {
      if (x) {
        ++diff;
      }
    }
    if (diff == 0) {
      return true;
    }
    for (int i = n; i < m; ++i) {
      int a = s2[i - n] - 'a';
      int b = s2[i] - 'a';
      if (cnt[b] == 0) {
        ++diff;
      }
      if (++cnt[b] == 0) {
        --diff;
      }
      if (cnt[a] == 0) {
        ++diff;
      }
      if (--cnt[a] == 0) {
        --diff;
      }
      if (diff == 0) {
        return true;
      }
    }
    return false;
  }
};

func checkInclusion(s1 string, s2 string) bool {
  n, m := len(s1), len(s2)
  if n > m {
    return false
  }
  cnt := [26]int{}
  for i := range s1 {
    cnt[s1[i]-'a']--
    cnt[s2[i]-'a']++
  }
  diff := 0
  for _, x := range cnt {
    if x != 0 {
      diff++
    }
  }
  if diff == 0 {
    return true
  }
  for i := n; i < m; i++ {
    a, b := s2[i-n]-'a', s2[i]-'a'
    if cnt[b] == 0 {
      diff++
    }
    cnt[b]++
    if cnt[b] == 0 {
      diff--
    }
    if cnt[a] == 0 {
      diff++
    }
    cnt[a]--
    if cnt[a] == 0 {
      diff--
    }
    if diff == 0 {
      return true
    }
  }
  return false
}

Solution 3

func checkInclusion(s1 string, s2 string) bool {
  need, window := make(map[byte]int), make(map[byte]int)
  validate, left, right := 0, 0, 0
  for i := range s1 {
    need[s1[i]] += 1
  }
  for ; right < len(s2); right++ {
    c := s2[right]
    window[c] += 1
    if need[c] == window[c] {
      validate++
    }
    for right-left+1 >= len(s1) {
      if validate == len(need) {
        return true
      }
      d := s2[left]
      if need[d] == window[d] {
        validate--
      }
      window[d] -= 1
      left++
    }
  }
  return false
}