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Formatted question description: https://leetcode.ca/all/1668.html

# 1668. Maximum Repeating Substring (Easy)

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 <= 100
• 1 <= word.length <= 100
• sequence and word contains only lowercase English letters.

Related Topics:
String

Similar Questions:

## Solution 1. Brute force

Note that find takes O(N^2) instead of O(N).

// OJ: https://leetcode.com/problems/maximum-repeating-substring/
// Time: O(N^3)
// Space: O(N)
class Solution {
public:
int maxRepeating(string sequence, string word) {
string f = word;
int k = 1;
for (; sequence.find(f) != string::npos; ++k, f += word);
return k - 1;
}
};

• class Solution {
public int maxRepeating(String sequence, String word) {
int maxRepeating = 0;
StringBuffer sb = new StringBuffer();
int k = 0;
while (sb.length() <= sequence.length()) {
sb.append(word);
k++;
if (sequence.indexOf(sb.toString()) >= 0)
maxRepeating = k;
}
return maxRepeating;
}
}

############

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;
}
}

• // OJ: https://leetcode.com/problems/maximum-repeating-substring/
// Time: O(N^3)
// Space: O(N)
class Solution {
public:
int maxRepeating(string sequence, string word) {
string f = word;
int k = 1;
for (; sequence.find(f) != string::npos; ++k, f += word);
return k - 1;
}
};

• 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

############

# 1668. Maximum Repeating Substring
# https://leetcode.com/problems/maximum-repeating-substring/

class Solution:
def maxRepeating(self, seq: str, w: str) -> int:

res = 0
s = set()
times = len(seq) // len(w)
p = [w * i for i in range(1, times+1)]

for c in p:
if c not in s and c in seq:
res += 1

return res

• 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
}

• 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;
}


• 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()
}
}