LeetCode //C - 541. Reverse String II

541. Reverse String II

Given a string s and an integer k, reverse the first k characters for every 2k characters counting from the start of the string.

If there are fewer than k characters left, reverse all of them. If there are less than 2k but greater than or equal to k characters, then reverse the first k characters and leave the other as original.

Example 1:

Input: s = "abcdefg", k = 2
Output: "bacdfeg"

Example 2:

Input: s = "abcd", k = 2
Output: "bacd"

Constraints:

• 1 < = s . l e n g t h < = 1 0 4 1 <= s.length <= 10^4 1<=s.length<=104
• s consists of only lowercase English letters.
• 1 < = k < = 1 0 4 1 <= k <= 10^4 1<=k<=104

From: LeetCode

Link: 541. Reverse String II

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Solution:

Ideas:

1. Helper Function (reverse):

• This function reverses the characters in the substring of s from index start to end.

2. Main Function (reverseStr):

• It iterates through the string in segments of 2k.
• For every 2k segment, the first k characters are reversed. The rest remain unchanged.
• If there are fewer than k characters left, reverse all of them.
• If there are between k and 2k characters, reverse the first k and leave the rest unchanged.

3. Edge Cases:

• When k is greater than the remaining length of the string, it handles it by only reversing up to the string's end.
• The function is efficient and adheres to the constraints, as the operations are performed in linear time relative to the string length.

Code:

void reverse(char* s, int start, int end) {
    while (start < end) {
        char temp = s[start];
        s[start] = s[end];
        s[end] = temp;
        start++;
        end--;
    }
}

char* reverseStr(char* s, int k) {
    int len = strlen(s);
    for (int i = 0; i < len; i += 2 * k) {
        // Reverse the first k characters in the current segment
        int end = (i + k - 1 < len) ? i + k - 1 : len - 1;
        reverse(s, i, end);
    }
    return s;
}