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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;
    }
};

Java

  • 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;
    }
}

  • // 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:
                s.add(c)
                res += 1
        
        return res

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