# 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, $\Sigma = 26$.

• 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;
}
}

• 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;
}
};

• 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


• 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
}

• 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;
}