LeetCode #1461 — MEDIUM

Check If a String Contains All Binary Codes of Size K

Move from brute-force thinking to an efficient approach using hash map strategy.

The Problem

Problem Statement

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Hash Map · Bit Manipulation

Example 1

"00110110"
2

Example 2

"0110"
1

Example 3

"0110"
2

Step 02

Core Insight

What unlocks the optimal approach

We need only to check all sub-strings of length k.
The number of distinct sub-strings should be exactly 2^k.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1461: Check If a String Contains All Binary Codes of Size K
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;
    }
}

// Accepted solution for LeetCode #1461: Check If a String Contains All Binary Codes of Size K
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
}

# Accepted solution for LeetCode #1461: Check If a String Contains All Binary Codes of Size K
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

// Accepted solution for LeetCode #1461: Check If a String Contains All Binary Codes of Size K
struct Solution;

impl Solution {
    fn has_all_codes(s: String, k: i32) -> bool {
        let k = k as usize;
        let m = 1 << k;
        let mut set = vec![false; m];
        let mut x = 0;
        let n = s.len();
        let mask = m as u32 - 1;
        for (i, c) in s.char_indices().rev() {
            x <<= 1;
            x |= (c as u8 - b'0') as u32;
            x &= mask;
            if i + k <= n {
                set[x as usize] = true;
            }
        }
        set.iter().all(|&b| b)
    }
}

#[test]
fn test() {
    let s = "00110110".to_string();
    let k = 2;
    let res = true;
    assert_eq!(Solution::has_all_codes(s, k), res);
    let s = "00110".to_string();
    let k = 2;
    let res = true;
    assert_eq!(Solution::has_all_codes(s, k), res);
    let s = "0110".to_string();
    let k = 1;
    let res = true;
    assert_eq!(Solution::has_all_codes(s, k), res);
    let s = "0110".to_string();
    let k = 2;
    let res = false;
    assert_eq!(Solution::has_all_codes(s, k), res);
    let s = "0000000001011100".to_string();
    let k = 4;
    let res = false;
    assert_eq!(Solution::has_all_codes(s, k), res);
}

// Accepted solution for LeetCode #1461: Check If a String Contains All Binary Codes of Size K
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;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × k)

Space

O(n)

Approach Breakdown

SORT + SCAN

O(n log n) time

O(n) space

Sort the array in O(n log n), then scan for the missing or unique element by comparing adjacent pairs. Sorting requires O(n) auxiliary space (or O(1) with in-place sort but O(n log n) time remains). The sort step dominates.

BIT MANIPULATION

O(n) time

O(1) space

Bitwise operations (AND, OR, XOR, shifts) are O(1) per operation on fixed-width integers. A single pass through the input with bit operations gives O(n) time. The key insight: XOR of a number with itself is 0, which eliminates duplicates without extra space.

Shortcut: Bit operations are O(1). XOR cancels duplicates. Single pass → O(n) time, O(1) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.