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1156. Swap For Longest Repeated Character Substring

Description

You are given a string text. You can swap two of the characters in the text.

Return the length of the longest substring with repeated characters.

 

Example 1:

Input: text = "ababa"
Output: 3
Explanation: We can swap the first 'b' with the last 'a', or the last 'b' with the first 'a'. Then, the longest repeated character substring is "aaa" with length 3.

Example 2:

Input: text = "aaabaaa"
Output: 6
Explanation: Swap 'b' with the last 'a' (or the first 'a'), and we get longest repeated character substring "aaaaaa" with length 6.

Example 3:

Input: text = "aaaaa"
Output: 5
Explanation: No need to swap, longest repeated character substring is "aaaaa" with length is 5.

 

Constraints:

  • 1 <= text.length <= 2 * 104
  • text consist of lowercase English characters only.

Solutions

Solution 1: Two Pointers

First, we use a hash table or array $cnt$ to count the occurrence of each character in the string $text$.

Next, we define a pointer $i$, initially $i = 0$. Each time, we set the pointer $j$ to $i$, and continuously move $j$ to the right until the character pointed by $j$ is different from the character pointed by $i$. At this time, we get a substring $text[i..j-1]$ of length $l = j - i$, where all characters are the same.

Then we skip the character pointed by the pointer $j$, and continue to move the pointer $k$ to the right until the character pointed by $k$ is different from the character pointed by $i$. At this time, we get a substring $text[j+1..k-1]$ of length $r = k - j - 1$, where all characters are the same. So the longest single-character repeated substring we can get by at most one swap operation is $\min(l + r + 1, cnt[text[i]])$. Next, we move the pointer $i$ to $j$ and continue to find the next substring. We take the maximum length of all substrings that meet the conditions.

The time complexity is $O(n)$, and the space complexity is $O(C)$. Here, $n$ is the length of the string, and $C$ is the size of the character set. In this problem, $C = 26$.

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class Solution:
    def maxRepOpt1(self, text: str) -> int:
        cnt = Counter(text)
        n = len(text)
        ans = i = 0
        while i < n:
            j = i
            while j < n and text[j] == text[i]:
                j += 1
            l = j - i
            k = j + 1
            while k < n and text[k] == text[i]:
                k += 1
            r = k - j - 1
            ans = max(ans, min(l + r + 1, cnt[text[i]]))
            i = j
        return ans
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class Solution {
    public int maxRepOpt1(String text) {
        int[] cnt = new int[26];
        int n = text.length();
        for (int i = 0; i < n; ++i) {
            ++cnt[text.charAt(i) - 'a'];
        }
        int ans = 0, i = 0;
        while (i < n) {
            int j = i;
            while (j < n && text.charAt(j) == text.charAt(i)) {
                ++j;
            }
            int l = j - i;
            int k = j + 1;
            while (k < n && text.charAt(k) == text.charAt(i)) {
                ++k;
            }
            int r = k - j - 1;
            ans = Math.max(ans, Math.min(l + r + 1, cnt[text.charAt(i) - 'a']));
            i = j;
        }
        return ans;
    }
}
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class Solution {
public:
    int maxRepOpt1(string text) {
        int cnt[26] = {0};
        for (char& c : text) {
            ++cnt[c - 'a'];
        }
        int n = text.size();
        int ans = 0, i = 0;
        while (i < n) {
            int j = i;
            while (j < n && text[j] == text[i]) {
                ++j;
            }
            int l = j - i;
            int k = j + 1;
            while (k < n && text[k] == text[i]) {
                ++k;
            }
            int r = k - j - 1;
            ans = max(ans, min(l + r<