424. Longest Repeating Character Replacement

Description

You are given a string s and an integer k. You can choose any character of the string and change it to any other uppercase English character. You can perform this operation at most k times.

Return the length of the longest substring containing the same letter you can get after performing the above operations.

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.
There may exists other ways to achieve this answer too.

Constraints:

1 <= s.length <= 10⁵
s consists of only uppercase English letters.
0 <= k <= s.length

Solutions

Solution 1: Two Pointers

We use a hash table cnt to count the occurrence of each character in the string, and two pointers l and r to maintain a sliding window, such that the size of the window minus the count of the most frequent character does not exceed $k$.

We iterate through the string, updating the right boundary r of the window each time, updating the count of characters within the window, and updating the maximum count mx of the characters that have appeared. When the size of the window minus mx is greater than $k$, we need to shrink the left boundary l of the window, updating the count of characters within the window, until the size of the window minus mx is no longer greater than $k$.

Finally, the answer is $n - l$, where $n$ is the length of the string.

Python3JavaC++GoTypeScript

class Solution:
    def characterReplacement(self, s: str, k: int) -> int:
        cnt = Counter()
        l = mx = 0
        for r, c in enumerate(s):
            cnt[c] += 1
            mx = max(mx, cnt[c])
            if r - l + 1 - mx > k:
                cnt[s[l]] -= 1
                l += 1
        return len(s) - l

class Solution {
    public int characterReplacement(String s, int k) {
        int[] cnt = new int[26];
        int l = 0, mx = 0;
        int n = s.length();
        for (int r = 0; r < n; ++r) {
            mx = Math.max(mx, ++cnt[s.charAt(r) - 'A']);
            if (r - l + 1 - mx > k) {
                --cnt[s.charAt(l++) - 'A'];
            }
        }
        return n - l;
    }
}

class Solution {
public:
    int characterReplacement(string s, int k) {
        int cnt[26]{};
        int l = 0, mx = 0;
        int n = s.length();
        for (int r = 0; r < n; ++r) {
            mx = max(mx, ++cnt[s[r] - 'A']);
            if (r - l + 1 - mx > k) {
                --cnt[s[l++] - 'A'];
            }
        }
        return n - l;
    }
};

func characterReplacement(s string, k int) int {
    cnt := [26]int{}
    l, mx := 0, 0
    for r, c := range s {
        cnt[c-'A']++
        mx = max(mx, cnt[c-'A'])
        if r-l+1-mx > k {
            cnt[s[l]-'A']--
            l++
        }
    }
    return len(s) - l
}

function characterReplacement(s: string, k: number): number {
    const idx = (c: string) => c.charCodeAt(0) - 65;
    const cnt: number[] = Array(26).fill(0);
    const n = s.length;
    let [l, mx] = [0, 0];
    for (let r = 0; r < n; ++r) {
        mx = Math.max(mx, ++cnt[idx(s[r])]);
        if (r - l + 1 - mx > k) {
            --cnt[idx(s[l++])];
        }
    }
    return n - l;
}

424. Longest Repeating Character Replacement

Description

Solutions

Solution 1: Two Pointers

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