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1461. Check If a String Contains All Binary Codes of Size K

Description

Given a binary string s and an integer k, return true if every binary code of length k is a substring of s. Otherwise, return false.

 

Example 1:

Input: s = "00110110", k = 2
Output: true
Explanation: The binary codes of length 2 are "00", "01", "10" and "11". They can be all found as substrings at indices 0, 1, 3 and 2 respectively.

Example 2:

Input: s = "0110", k = 1
Output: true
Explanation: The binary codes of length 1 are "0" and "1", it is clear that both exist as a substring.

Example 3:

Input: s = "0110", k = 2
Output: false
Explanation: The binary code "00" is of length 2 and does not exist in the array.

 

Constraints:

  • 1 <= s.length <= 5 * 105
  • s[i] is either '0' or '1'.
  • 1 <= k <= 20

Solutions

Solution 1: Hash Table

First, for a string $s$ of length $n$, the number of substrings of length $k$ is $n - k + 1$. If $n - k + 1 < 2^k$, then there must exist a binary string of length $k$ that is not a substring of $s$, so we return false.

Next, we traverse the string $s$ and store all substrings of length $k$ in a set $ss$. Finally, we check if the size of the set $ss$ is equal to $2^k$.

The time complexity is $O(n \times k)$, and the space complexity is $O(n)$. Here, $n$ is the length of the string $s$.

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class Solution:
    def hasAllCodes(self, s: str, k: int) -> bool:
        n = len(s)
        m = 1 << k
        if n - k + 1 < m:
            return False
        ss = {s[i : i + k] for i in range(n - k + 1)}
        return len(ss) == m

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class Solution {
    public boolean hasAllCodes(String s, int k) {
        int n = s.length();
        int m = 1 << k;
        if (n - k + 1 < m) {
            return false;
        }
        Set<String> ss = new HashSet<>();
        for (int i = 0; i < n - k + 1; ++i) {
            ss.add(s.substring(i, i + k));
        }
        return ss.size() == m;
    }
}

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class Solution {
public:
    bool hasAllCodes(string s, int k) {
        int n = s.size();
        int m = 1 << k;
        if (n - k + 1 < m) {
            return false;
        }
        unordered_set<string> ss;
        for (int i = 0; i + k <= n; ++i) {
            ss.insert(move(s.substr(i, k)));
        }
        return ss.size() == m;
    }
};

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func hasAllCodes(s string, k int) bool {
    n, m := len(s), 1<<k
    if n-k+1 < m {
        return false
    }
    ss := map[string]bool{}
    for i := 0; i+k <= n; i++ {
        ss[s[i:i+k]] = true
    }
    return len(ss) == m
}

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function hasAllCodes(s: string, k: number): boolean {
    const n = s.length;
    const m = 1 << k;
    if (n - k + 1 < m) {
        return false;
    }
    const ss = new Set<string>();
    for (let i = 0; i + k <= n; ++i) {
        ss.add(s.slice(i, i + k));
    }
    return ss.size === m;
}

Solution 2: Sliding Window

In Solution 1, we stored all distinct substrings of length $k$, and processing each substring requires $O(k)$ time. We can instead use a sliding window, where each time we add the latest character, we remove the leftmost character from the window. During this process, we use an integer $x$ to store the substring.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the string $s$.

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class Solution:
    def hasAllCodes(self, s: str, k: int) -> bool:
        n = len(s)
        m = 1 << k
        if n - k + 1 < m:
            return False
        ss = set()
        x