567. Permutation in String

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

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;
    }
};

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

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
}

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
    }
}

567 - Permutation in String

567. Permutation in String

Description

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567. Permutation in String

Description

Solutions

All Problems

All Solutions