1668. Maximum Repeating Substring

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

Python3JavaC++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

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;
}

1668. Maximum Repeating Substring

Description

Solutions

Solution 1

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