Question

1668. Maximum Repeating Substring

中文文档

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

Solutions

Solution 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

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;
  }
}

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;
  }
};

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()
  }
}

#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;
}