Welcome to Subscribe On Youtube

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:
                s.add(c)
                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()
    }
}

All Problems

All Solutions