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:
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 * 105s[i] is either '0' or '1'.1 <= k <= 20
Solutions
Solution 1: Hash Table
First, for a string s of length n, the number of substrings of length k is n - k + 1. If n - k + 1 < 2k, then there must exist a binary string of length k that is not a substring of s, so we return false.
Next, we traverse the string s and store all substrings of length k in a set ss. Finally, we check if the size of the set ss is equal to 2k.
The time complexity is O(n × k), and the space complexity is O(n). Here, n is the length of the string s.
PythonJavaC++GoTypeScript
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(code-box)
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;
}
}(code-box)
class Solution {
public:
bool hasAllCodes(string s, int k) {
int n = s.size();
int m = 1 << k;
if (n - k + 1 < m) {
return false;
}
unordered_set<string> ss;
for (int i = 0; i + k <= n; ++i) {
ss.insert(move(s.substr(i, k)));
}
return ss.size() == m;
}
};(code-box)
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
}(code-box)
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;
}(code-box)
Solution 2: Sliding Window
In Solution 1, we stored all distinct substrings of length k, and processing each substring requires O(k) time. We can instead use a sliding window, where each time we add the latest character, we remove the leftmost character from the window. During this process, we use an integer x to store the substring.
The time complexity is O(n), and the space complexity is O(n). Here, n is the length of the string s.
PythonJavaC++GoTypeScript
class Solution:
def hasAllCodes(self, s: str, k: int) -> bool:
n = len(s)
m = 1 << k
if n - k + 1 < m:
return False
ss = set()
x = int(s[:k], 2)
ss.add(x)
for i in range(k, n):
a = int(s[i - k]) << (k - 1)
b = int(s[i])
x = (x - a) << 1 | b
ss.add(x)
return len(ss) == m(code-box)
class Solution {
public boolean hasAllCodes(String s, int k) {
int n = s.length();
int m = 1 << k;
if (n - k + 1 < m) {
return false;
}
boolean[] ss = new boolean[m];
int x = Integer.parseInt(s.substring(0, k), 2);
ss[x] = true;
for (int i = k; i < n; ++i) {
int a = (s.charAt(i - k) - '0') << (k - 1);
int b = s.charAt(i) - '0';
x = (x - a) << 1 | b;
ss[x] = true;
}
for (boolean v : ss) {
if (!v) {
return false;
}
}
return true;
}
}(code-box)
class Solution {
public:
bool hasAllCodes(string s, int k) {
int n = s.size();
int m = 1 << k;
if (n - k + 1 < m) {
return false;
}
bool ss[m];
memset(ss, false, sizeof(ss));
int x = stoi(s.substr(0, k), nullptr, 2);
ss[x] = true;
for (int i = k; i < n; ++i) {
int a = (s[i - k] - '0') << (k - 1);
int b = s[i] - '0';
x = (x - a) << 1 | b;
ss[x] = true;
}
return all_of(ss, ss + m, [](bool v) { return v; });
}
};(code-box)
func hasAllCodes(s string, k int) bool {
n, m := len(s), 1<<k
if n-k+1 < m {
return false
}
ss := make([]bool, m)
x, _ := strconv.ParseInt(s[:k], 2, 64)
ss[x] = true
for i := k; i < n; i++ {
a := int64(s[i-k]-'0') << (k - 1)
b := int64(s[i] - '0')
x = (x-a)<<1 | b
ss[x] = true
}
for _, v := range ss {
if !v {
return false
}
}
return true
}(code-box)
function hasAllCodes(s: string, k: number): boolean {
const n = s.length;
const m = 1 << k;
if (n - k + 1 < m) {
return false;
}
let x = +`0b${s.slice(0, k)}`;
const ss = new Set<number>();
ss.add(x);
for (let i = k; i < n; ++i) {
const a = +s[i - k] << (k - 1);
const b = +s[i];
x = ((x - a) << 1) | b;
ss.add(x);
}
return ss.size === m;
}(code-box)
