Skip to content

2730. Find the Longest Semi-Repetitive Substring

Description

You are given a 0-indexed string s that consists of digits from 0 to 9.

A string t is called a semi-repetitive if there is at most one consecutive pair of the same digits inside t. For example, 0010, 002020, 0123, 2002, and 54944 are semi-repetitive while 00101022, and 1101234883 are not.

Return the length of the longest semi-repetitive substring inside s.

A substring is a contiguous non-empty sequence of characters within a string.

 

Example 1:

Input: s = "52233"
Output: 4
Explanation: The longest semi-repetitive substring is "5223", which starts at i = 0 and ends at j = 3.

Example 2:

Input: s = "5494"
Output: 4
Explanation: s is a semi-reptitive string, so the answer is 4.

Example 3:

Input: s = "1111111"
Output: 2
Explanation: The longest semi-repetitive substring is "11", which starts at i = 0 and ends at j = 1.

 

Constraints:

  • 1 <= s.length <= 50
  • '0' <= s[i] <= '9'

Solutions

Solution 1: Two Pointers

We use two pointers to maintain a range $s[j..i]$ such that there is at most one pair of adjacent characters that are equal, initially $j = 0$, $i = 1$. Initialize the answer $ans = 1$.

We use $cnt$ to record the number of pairs of adjacent characters that are equal in the range. If $cnt > 1$, then we need to move the left pointer $j$ until $cnt \le 1$. Each time, we update the answer as $ans = \max(ans, i - j + 1)$.

The time complexity is $O(n)$, where $n$ is the length of the string. The space complexity is $O(1)$.

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
class Solution:
    def longestSemiRepetitiveSubstring(self, s: str) -> int:
        ans, n = 1, len(s)
        cnt = j = 0
        for i in range(1, n):
            cnt += s[i] == s[i - 1]
            while cnt > 1:
                cnt -= s[j] == s[j + 1]
                j += 1
            ans = max(ans, i - j + 1)
        return ans

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
class Solution {
    public int longestSemiRepetitiveSubstring(String s) {
        int ans = 1, n = s.length();
        for (int i = 1, j = 0, cnt = 0; i < n; ++i) {
            cnt += s.charAt(i) == s.charAt(i - 1) ? 1 : 0;
            for (; cnt > 1; ++j) {
                cnt -= s.charAt(j) == s.charAt(j + 1) ? 1 : 0;
            }
            ans = Math.max(ans, i - j + 1);
        }
        return ans;
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
class Solution {
public:
    int longestSemiRepetitiveSubstring(string s) {
        int ans = 1, n = s.size();
        for (int i = 1, j = 0, cnt = 0; i < n; ++i) {
            cnt += s[i] == s[i - 1] ? 1 : 0;
            for (; cnt > 1; ++j) {
                cnt -= s[j] == s[j + 1] ? 1 : 0;
            }
            ans = max(ans, i - j + 1);
        }
        return ans;
    }
};

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
func longestSemiRepetitiveSubstring(s string) (ans int) {
    ans = 1
    for i, j, cnt := 1, 0, 0; i < len(s); i++ {
        if s[i] == s[i-1] {
            cnt++
        }
        for ; cnt > 1; j++ {
            if s[j] == s[j+1] {
                cnt--
            }
        }
        ans = max(ans, i-j+1)
    }
    return
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
function longestSemiRepetitiveSubstring(s: string): number {
    const n = s.length;
    let ans = 1;
    for (let i = 1, j = 0, cnt = 0; i < n; ++i) {
        cnt += s[i] === s[i - 1] ? 1 : 0;
        for (; cnt > 1; ++j) {
            cnt -= s[j] === s[j + 1] ? 1 : 0;
        }
        ans = Math.max(ans, i - j + 1);
    }
    return ans;
}

Comments