Description
For a string sequence, a string word is k-repeating if word concatenated k times is a substring of sequence. The word's maximum k-repeating value is the highest value k where word is k-repeating in sequence. If word is not a substring of sequence, word's maximum k-repeating value is 0.
Given strings sequence and word, return the maximum k-repeating value of word in sequence.
Example 1:
Input: sequence = "ababc", word = "ab"
Output: 2
Explanation: "abab" is a substring in "ababc".
Example 2:
Input: sequence = "ababc", word = "ba"
Output: 1
Explanation: "ba" is a substring in "ababc". "baba" is not a substring in "ababc".
Example 3:
Input: sequence = "ababc", word = "ac"
Output: 0
Explanation: "ac" is not a substring in "ababc".
Constraints:
1 <= sequence.length <= 1001 <= word.length <= 100sequence and word contains only lowercase English letters.
Solutions
Solution 1
PythonJavaC++GoTypeScriptRustC
class Solution:
def maxRepeating(self, sequence: str, word: str) -> int:
for k in range(len(sequence) // len(word), -1, -1):
if word * k in sequence:
return k(code-box)
class Solution {
public int maxRepeating(String sequence, String word) {
for (int k = sequence.length() / word.length(); k > 0; --k) {
if (sequence.contains(word.repeat(k))) {
return k;
}
}
return 0;
}
}(code-box)
class Solution {
public:
int maxRepeating(string sequence, string word) {
int ans = 0;
string t = word;
int x = sequence.size() / word.size();
for (int k = 1; k <= x; ++k) {
// C++ 这里从小到大枚举重复值
if (sequence.find(t) != string::npos) {
ans = k;
}
t += word;
}
return ans;
}
};(code-box)
func maxRepeating(sequence string, word string) int {
for k := len(sequence) / len(word); k > 0; k-- {
if strings.Contains(sequence, strings.Repeat(word, k)) {
return k
}
}
return 0
}(code-box)
function maxRepeating(sequence: string, word: string): number {
let n = sequence.length;
let m = word.length;
for (let k = Math.floor(n / m); k > 0; k--) {
if (sequence.includes(word.repeat(k))) {
return k;
}
}
return 0;
}(code-box)
impl Solution {
pub fn max_repeating(sequence: String, word: String) -> i32 {
let n = sequence.len();
let m = word.len();
if n < m {
return 0;
}
let mut dp = vec![0; n - m + 1];
for i in 0..=n - m {
let s = &sequence[i..i + m];
if s == word {
dp[i] = (if (i as i32) - (m as i32) < 0 {
0
} else {
dp[i - m]
}) + 1;
}
}
*dp.iter().max().unwrap()
}
}(code-box)
#define max(a, b) (((a) > (b)) ? (a) : (b))
int findWord(int i, char* sequence, char* word) {
int n = strlen(word);
for (int j = 0; j < n; j++) {
if (sequence[j + i] != word[j]) {
return 0;
}
}
return 1 + findWord(i + n, sequence, word);
}
int maxRepeating(char* sequence, char* word) {
int n = strlen(sequence);
int m = strlen(word);
int ans = 0;
for (int i = 0; i <= n - m; i++) {
ans = max(ans, findWord(i, sequence, word));
}
return ans;
}(code-box)
