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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

Solution 1: Sliding Window

We use an array $\textit{cnt}$ to record the characters and their counts that need to be matched, and a variable $\textit{need}$ to record the number of different characters that still need to be matched. Initially, $\textit{cnt}$ contains the character counts from the string $\textit{s1}$, and $\textit{need}$ is the number of different characters in $\textit{s1}$.

Then we traverse the string $\textit{s2}$. For each character, we decrement its corresponding value in $\textit{cnt}$. If the decremented value equals $0$, it means the current character's count in $\textit{s1}$ is satisfied, and we decrement $\textit{need}$. If the current index $i$ is greater than or equal to the length of $\textit{s1}$, we need to increment the corresponding value in $\textit{cnt}$ for $\textit{s2}[i-\textit{s1}]$. If the incremented value equals $1$, it means the current character's count in $\textit{s1}$ is no longer satisfied, and we increment $\textit{need}$. During the traversal, if the value of $\textit{need}$ equals $0$, it means all character counts are satisfied, and we have found a valid substring, so we return $\text{true}$.

Otherwise, if the traversal ends without finding a valid substring, we return $\text{false}$.

The time complexity is $O(m + n)$, where $m$ and $n$ are the lengths of strings $\textit{s1}$ and $\textit{s2}$, respectively. The space complexity is $O(|\Sigma|)$, where $\Sigma$ is the character set. In this problem, the character set is lowercase letters, so the space complexity is constant.

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class Solution:
    def checkInclusion(self, s1: str, s2: str) -> bool:
        cnt = Counter(s1)
        need = len(cnt)
        m = len(s1)
        for i, c in enumerate(s2):
            cnt[c] -= 1
            if cnt[c] == 0:
                need -= 1
            if i >= m:
                cnt[s2[i - m]] += 1
                if cnt[s2[i - m]] == 1:
                    need += 1
            if need == 0:
                return True
        return False

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class Solution {
    public boolean checkInclusion(String s1, String s2) {
        int need = 0;
        int[] cnt = new int[26];
        for (char c : s1.toCharArray()) {
            if (++cnt[c - 'a'] == 1) {
                ++need;
            }
        }
        int m = s1.length(), n = s2.length();
        for (int i = 0; i < n; ++i) {
            int c = s2.charAt(i) - 'a';
            if (--cnt[c] == 0) {
                --need;
            }
            if (i >= m) {
                c = s2.charAt(i - m) - 'a';
                if (++cnt[c] == 1) {
                    ++need;
                }
            }
            if (need == 0) {
                return true;
            }
        }
        return false;
    }
}

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class Solution {
public:
    bool checkInclusion(string s1, string s2) {
        int need = 0;
        int cnt[26]{};
        for (char c : s1) {
            if (++cnt[c - 'a'] == 1) {
                ++need;
            }
        }
        int m = s1.size(), n = s2.size();
        for (int i = 0; i < n; ++i) {
            int c = s2[i] - 'a';
            if (--cnt[c] == 0) {
                --need;
            }
            if (i >= m) {
                c = s2[i - m] - 'a';
                if (++cnt[c] == 1) {
                    ++need;
                }
            }
            if (need == 0) {
                return true;
            }
        }
        return false;
    }
};

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func checkInclusion(s1 string, s2 string) bool {
    need := 0
    cnt := [26]int{}

    for _, c := range s1 {
        if cnt[c-'a']++; cnt[c-'a'] == 1 {
            need++
        }
    }

    m, n := len(s1), len(s2)
    for i := 0; i < n; i++ {
        c := s2[i] - 'a'
        if cnt[c]--; cnt[c] == 0 {
            need--
        }
        if i >= m {
            c = s2[i-m] - 'a'
            if cnt[c]++; cnt[c] == 1 {
                need++
            }
        }
        if need == 0 {
            return true
        }
    }
    return false
}

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function checkInclusion(s1: string, s2: string): boolean {
    let need = 0;
    const cnt: number[] = Array(26).fill(0);
    const a = 'a'.charCodeAt(0);
    for (const c of s1) {
        if (++cnt[c.charCodeAt(0) - a] === 1) {
            need++;
        }
    }

    const [m, n] = [s1.length, s2.length];
    for (let i = 0; i < n; i++) {
        let c = s2.charCodeAt(i) - a;
        if (--cnt[c] === 0) {
            need--;
        }
        if (i >= m) {
            c = s2.charCodeAt(i - m) - a;
            if (++cnt[c] === 1) {
                need++;
            }
        }
        if (need === 0) {
            return true;
        }
    }
    return false;
}

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impl Solution {
    pub fn check_inclusion(s1: String, s2: String) -> bool {
        let mut need = 0;
        let mut cnt = vec![0; 26];

        for c in s1.chars() {
            let index = (c as u8 - b'a') as usize;
            if cnt[index] == 0 {
                need += 1;
            }
            cnt[index] += 1;
        }

        let m = s1.len();
        let n = s2.len();
        let s2_bytes = s2.as_bytes();

        for i in 0..n {
            let c = (s2_bytes[i] - b'a') as usize;
            cnt[c] -= 1;
            if cnt[c] == 0 {
                need -= 1;
            }

            if i >= m {
                let c = (s2_bytes[i - m] - b'a') as usize;
                cnt[c] += 1;
                if cnt[c] == 1 {
                    need += 1;
                }
            }

            if need == 0 {
                return true;
            }
        }

        false
    }
}

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public class Solution {
    public bool CheckInclusion(string s1, string s2) {
        int need = 0;
        int[] cnt = new int[26];

        foreach (char c in s1) {
            if (++cnt[c - 'a'] == 1) {
                need++;
            }
        }

        int m = s1.Length, n = s2.Length;
        for (int i = 0; i < n; i++) {
            int c = s2[i] - 'a';
            if (--cnt[c] == 0) {
                need--;
            }

            if (i >= m) {
                c = s2[i - m] - 'a';
                if (++cnt[c] == 1) {
                    need++;
                }
            }

            if (need == 0) {
                return true;
            }
        }
        return false;
    }
}

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class Solution {
    /**
     * @param String $s1
     * @param String $s2
     * @return Boolean
     */
    function checkInclusion($s1, $s2) {
        $need = 0;
        $cnt = array_fill(0, 26, 0);

        for ($i = 0; $i < strlen($s1); $i++) {
            $index = ord($s1[$i]) - ord('a');
            if (++$cnt[$index] == 1) {
                $need++;
            }
        }

        $m = strlen($s1);
        $n = strlen($s2);

        for ($i = 0; $i < $n; $i++) {
            $c = ord($s2[$i]) - ord('a');
            if (--$cnt[$c] == 0) {
                $need--;
            }

            if ($i >= $m) {
                $c = ord($s2[$i - $m]) - ord('a');
                if (++$cnt[$c] == 1) {
                    $need++;
                }
            }

            if ($need == 0) {
                return true;
            }
        }

        return false;
    }
}

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