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

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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:

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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:

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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:

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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 * 10^5
  • s[i] is either '0' or '1'.
  • 1 <= k <= 20

Hints/Notes

  • 2024/10/13
  • sliding window
  • No solution from 0x3F

Solution

Language: C++

Easier solution

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class Solution {
public:
    bool hasAllCodes(string s, int k) {
        unordered_set<int> nums;
        int n = s.size(), numbits = sizeof(int);
        int right = 0, digits = 0;
        string cur;
        while (right < n) {
            cur.push_back(s[right]);
            if (right >= k - 1) {
                nums.insert(bitset<32>(cur).to_ulong());
                cur.erase(cur.begin());
            }
            right++;
        }
        return nums.size() == (2 << k - 1);
    }
};
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class Solution {
public:
    bool hasAllCodes(string s, int k) {
        unordered_set<int> nums;
        int n = s.size(), numbits = sizeof(int);
        int right = 0, cur = 0, digits = 0;
        while (right < n) {
            cur += s[right] - '0';
            if (right >= k - 1) {
                nums.insert(cur);
                cur &= (1 << k - 1) - 1;
            }
            cur <<= 1;
            right++;
        }
        return nums.size() == (2 << k - 1);
    }
};