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3090. Maximum Length Substring With Two Occurrences

Description

Given a string s, return the maximum length of a substring such that it contains at most two occurrences of each character.

 

Example 1:

Input: s = "bcbbbcba"

Output: 4

Explanation:

The following substring has a length of 4 and contains at most two occurrences of each character: "bcbbbcba".

Example 2:

Input: s = "aaaa"

Output: 2

Explanation:

The following substring has a length of 2 and contains at most two occurrences of each character: "aaaa".

 

Constraints:

  • 2 <= s.length <= 100
  • s consists only of lowercase English letters.

Solutions

Solution 1: Two Pointers

We use two pointers $i$ and $j$ to maintain a sliding window, and an array $cnt$ to record the occurrence times of each character in the window.

In each iteration, we add the character $c$ at the pointer $j$ into the window, then check if $cnt[c]$ is greater than $2$. If it is, we move the pointer $i$ to the right until $cnt[c]$ is less than or equal to $2$. At this point, we update the answer $ans = \max(ans, j - i + 1)$.

Finally, we return the answer $ans$.

The time complexity is $O(n)$, where $n$ is the length of the string $s$. The space complexity is $O(|\Sigma|)$, where $\Sigma$ is the character set, and in this problem, $\Sigma = 26$.

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class Solution:
    def maximumLengthSubstring(self, s: str) -> int:
        cnt = Counter()
        ans = i = 0
        for j, c in enumerate(s):
            cnt[c] += 1
            while cnt[c] > 2:
                cnt[s[i]] -= 1
                i += 1
            ans = max(ans, j - i + 1)
        return ans

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class Solution {
    public int maximumLengthSubstring(String s) {
        int[] cnt = new int[26];
        int ans = 0;
        for (int i = 0, j = 0; j < s.length(); ++j) {
            int idx = s.charAt(j) - 'a';
            ++cnt[idx];
            while (cnt[idx] > 2) {
                --cnt[s.charAt(i++) - 'a'];
            }
            ans = Math.max(ans, j - i + 1);
        }
        return ans;
    }
}

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class Solution {
public:
    int maximumLengthSubstring(string s) {
        int cnt[26]{};
        int ans = 0;
        for (int i = 0, j = 0; j < s.length(); ++j) {
            int idx = s[j] - 'a';
            ++cnt[idx];
            while (cnt[idx] > 2) {
                --cnt[s[i++] - 'a'];
            }
            ans = max(ans, j - i + 1);
        }
        return ans;
    }
};

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func maximumLengthSubstring(s string) (ans int) {
    cnt := [26]int{}
    i := 0
    for j, c := range s {
        idx := c - 'a'
        cnt[idx]++
        for cnt[idx] > 2 {
            cnt[s[i]-'a']--
            i++
        }
        ans = max(ans, j-i+1)
    }
    return
}

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function maximumLengthSubstring(s: string): number {
    let ans = 0;
    const cnt: number[] = Array(26).fill(0);
    for (let i = 0, j = 0; j < s.length; ++j) {
        const idx = s[j].charCodeAt(0) - 'a'.charCodeAt(0);
        ++cnt[idx];
        while (cnt[idx] > 2) {
            --cnt[s[i++].charCodeAt(0) - 'a'.charCodeAt(0)];
        }
        ans = Math.max(ans, j - i + 1);
    }
    return ans;
}

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