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

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 indicies 0, 1, 3 and 2 respectively.

Example 2:

Input: s = "00110", k = 2
Output: true

Example 3:

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 4:

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

Example 5:

Input: s = "0000000001011100", k = 4
Output: false

Constraints:

• 1 <= s.length <= 5 * 10^5
• s consists of 0's and 1's only.
• 1 <= k <= 20

检查一个字符串是否包含所有长度为 K 的二进制子串。

给你一个二进制字符串 s 和一个整数 k 。

如果所有长度为 k 的二进制字符串都是 s 的子串，请返回 True ，否则请返回 False 。

这道题问的是在字符串中是否能找到所有长度为 K 的二进制子串，应该不难想到需要用滑动窗口做，但是这里有一个很巧妙的处理，能快速知道你到底需要找多少个子串。因为子串的长度是 K，所以一共需要找 Math.pow(2, k) 个子串。所以这里我们利用一个hashset，如果遍历完input字符串之后set.size() == k则证明找全了，否则就是有缺失。

时间O(n)

空间O(n)

Java实现

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

LeetCode 题目总结