Check If a String Contains All Binary Codes of Size K - Problem

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.

A binary code of length k is a string of exactly k characters, where each character is either '0' or '1'.

Note: There are exactly 2^k different binary codes of length k. For example, when k=2, the binary codes are "00", "01", "10", and "11".

Input & Output

Example 1 — All Codes Present

$ Input: s = "00110110", k = 2

› Output: true

💡 Note: For k=2, we need all 4 binary codes: "00", "01", "10", "11". String contains: "00" at index 0, "01" at index 1, "11" at index 2-3, "10" at index 4-5. All codes are present.

Example 2 — Missing Code

$ Input: s = "0110", k = 1

› Output: true

💡 Note: For k=1, we need codes "0" and "1". String "0110" contains both: "0" at index 0 and "1" at indices 1,2. All codes present.

Example 3 — String Too Short

$ Input: s = "0110", k = 2

› Output: false

💡 Note: For k=2, we need 4 codes: "00", "01", "10", "11". String contains "01" at index 0, "11" at index 1, "10" at index 2. Missing "00", so return false.

Constraints

1 ≤ s.length ≤ 5 × 10⁴
1 ≤ k ≤ 20
s consists of only '0' and '1' characters

Visualization

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Asked in

F Facebook 35 G Google 28 a Amazon 22

The key insight is that we need exactly 2^k different binary codes of length k. The optimal approach uses rolling hash to efficiently convert binary substrings to integers, avoiding string operations. Best approach is rolling hash with early termination. Time: O(n), Space: O(2^k)

Common Approaches

✓ Frequency Count

⏱️ Time: N/A Space: N/A

Generate All Codes and Search

⏱️ Time: O(2^k × n × k) Space: O(2^k × k)

First generate all possible binary codes of length k (there are 2^k of them), then search for each code in the string s using substring search.

Rolling Hash with Early Termination

⏱️ Time: O(n) Space: O(2^k)

Convert each k-length substring to an integer using rolling hash technique. This avoids string operations and provides O(1) hash computation for each window position.

Sliding Window with Hash Set

⏱️ Time: O(n × k) Space: O(2^k × k)

Slide a window of size k through the string, collecting all unique k-length substrings in a hash set. If the set size equals 2^k, then all possible binary codes are present.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stdbool.h>

bool hasAllBinaryCodes(char* s, int k) {
    int len = strlen(s);
    int totalCodes = 1 << k; // 2^k
    
    // If string is too short to contain all codes
    if (len < k + totalCodes - 1) {
        return false;
    }
    
    // Use array to mark which codes we've seen
    static bool seen[1024]; // Max 2^10 = 1024 for reasonable k values
    memset(seen, false, sizeof(seen));
    
    int foundCount = 0;
    
    // Check all substrings of length k
    for (int i = 0; i <= len - k; i++) {
        // Convert k-length substring to integer
        int code = 0;
        for (int j = 0; j < k; j++) {
            code = code * 2 + (s[i + j] - '0');
        }
        
        // If we haven't seen this code before
        if (!seen[code]) {
            seen[code] = true;
            foundCount++;
            
            // Early termination if we found all codes
            if (foundCount == totalCodes) {
                return true;
            }
        }
    }
    
    return foundCount == totalCodes;
}

int main() {
    char s[100001];
    int k;
    
    // Read binary string
    fgets(s, sizeof(s), stdin);
    // Remove newline if present
    int len = strlen(s);
    if (len > 0 && s[len-1] == '\n') {
        s[len-1] = '\0';
    }
    
    // Read k
    scanf("%d", &k);
    
    bool result = hasAllBinaryCodes(s, k);
    printf(result ? "true\n" : "false\n");
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

✓ Linear Growth

Space Complexity

✓ Linear Space

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Medium Frequency

~25 min Avg. Time

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Smart Actions

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