424 Longest Repeating Character Replacement

424. Longest Repeating Character Replacement

1. Question

Given a string that consists of only uppercase English letters, you can replace any letter in the string with another letter at mostktimes. Find the length of a longest substring containing all repeating letters you can get after performing the above operations.

Note: Both the string's length andkwill not exceed 104.

Example 1:

Input: s = "ABAB", k = 2

Output: 4

Explanation: Replace the two 'A's with two 'B's or vice versa.

Example 2:

Input: s = "AABABBA", k = 1

Output: 4

Explanation: 
Replace the one 'A' in the middle with 'B' and form "AABBBBA".
The substring "BBBB" has the longest repeating letters, which is 4.

2. Implementation

(1) Two Pointer + Hash

思路: 这道题的本质是问找到一个最长的substring满足，该substring的长度减substring中最多重复的character的个数是小于等于k的。所以我们维护一个变量maxCount, 从左向右扫s， maxCount记录目前最多重复的character的个数，当substring的长度减去maxCount的值大于k时，我们需要更新substring的起点。

class Solution {
    public int characterReplacement(String s, int k) {
        if (s.length() <= k) {
            return s.length();
        }

        int start = 0, end = 0, maxCount = 0, maxLen = 0;
        int[] map = new int[256];

        while (end < s.length()) {
            ++map[s.charAt(end)];
            maxCount = Math.max(maxCount, map[s.charAt(end)]);
            ++end;

            if (end - start - maxCount > k) {
                --map[s.charAt(start)];
                ++start;
            }
            maxLen = Math.max(maxLen, end - start);
        }
        return maxLen;
    }
}

3. Time & Space Complexity

Two Pointer + Hash: 时间复杂度O(n), 空间复杂度O(1)