Reverse First K Characters in Every 2K Block

Given a string s and a positive integer k, process s from left to right in consecutive blocks of 2 * k characters.
For every complete block of 2 * k characters, reverse its first k characters and leave its next k characters unchanged.
If fewer than k characters remain, reverse all remaining characters.
If at least k but fewer than 2 * k characters remain, reverse the first k remaining characters and leave the rest unchanged.
Return the transformed string.

Function
reverseFirstKInEvery2KBlock(s: String, k: int) → String

Examples
Example 1
s = "abcdefgh"
k = 2
return = "bacdfegh"
In abcd, reverse ab to obtain bacd. In efgh, reverse ef to obtain fegh. Combining the blocks gives bacdfegh.
Example 2
s = "abcdefg"
k = 3
return = "cbadefg"
Reverse abc in the first six-character block, producing cbadef. Only g remains, so reversing the final one-character suffix leaves it unchanged.

The trick: loop i from 0 to n in steps of 2k. For each i, reverse the slice from i to min(i + k, n) - 1. That single min() covers both tail rules. If fewer than k remain, you reverse everything left. If between k and 2k remain, you reverse only the first k. No separate branches needed. Convert the string to a char array first, since strings are immutable in most languages and slicing repeatedly costs extra. Use two pointers, left at i and right at min(i + k, n) - 1, and swap inward until they cross. Time is O(n), space is O(n) for the array. The common pitfall is stepping by k instead of 2k, or forgetting to clamp the right bound and running off the end. Trace Example 2 by hand before submitting. If the live OA freezes you, StealthCoder is the hedge that gives you the loop instantly.

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