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.

Constraints

• 1 <= s.length <= 5 * 10^5

• s consists of 0's and 1's only.

• 1 <= k <= 20

Approach

Links

• GeeksforGeeks

• Leetcode

• ProgramCreek

• YouTube

Examples

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

Solutions

Solution 1

/**
 * Time complexity : O(N*K). Where N is length of s. We need to iterate the 
 *    string, and use O(K) to calculate the hash of each substring.
 * Space complexity : O(N*K). There are at most N strings with length K in the set.
 */

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

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