1class Solution {
2    /**
3     * Splits the array into blocks of size (length / k) and reverses
4     * each block in place using a two-pointer approach.
5     *
6     * @param nums the input array to be processed
7     * @param k    divisor used to determine the block size
8     * @return the array after reversing each block in place
9     */
10    public int[] reverseSubarrays(int[] nums, int k) {
11        int length = nums.length;
12        // Size of each block to reverse
13        int blockSize = length / k;
14
15        // Iterate over the array one block at a time
16        for (int start = 0; start < length; start += blockSize) {
17            // Set the two pointers at the boundaries of the current block
18            int left = start;
19            int right = start + blockSize - 1;
20
21            // Reverse the current block by swapping elements
22            // from the outside toward the center
23            while (left < right) {
24                int temp = nums[left];
25                nums[left++] = nums[right];
26                nums[right--] = temp;
27            }
28        }
29
30        return nums;
31    }
32}
33

1class Solution {
2public:
3    vector<int> reverseSubarrays(vector<int>& nums, int k) {
4        // Total number of elements in the input array
5        int n = nums.size();
6
7        // Size of each block to be reversed (integer division)
8        int blockSize = n / k;
9
10        // Iterate over the array, advancing one block at a time
11        for (int start = 0; start < n; start += blockSize) {
12            // Set two pointers at the ends of the current block
13            int left = start;
14            int right = start + blockSize - 1;
15
16            // Reverse the elements within this block in place
17            while (left < right) {
18                swap(nums[left++], nums[right--]);
19            }
20        }
21
22        // Return the modified array
23        return nums;
24    }
25};
26```
27
28**A few notes on potential concerns:**
29
301. **Edge case (`k == 0` or `blockSize == 0`):** If `k > n`, then `blockSize` becomes `0`, and `start += blockSize` would create an infinite loop. The original code shares this risk—worth guarding if such input is possible.
31
322. **Remainder handling:** When `n` is not divisible by `m`, the last partial block (size `n % blockSize`) is still reversed. The reverse loop naturally handles a partial block since `right` is bounded by the iteration's `start`, but note it does **not** clamp `right` to `n - 1`. If `start + blockSize - 1` could exceed `n - 1` on the final block, you'd want `right = min(start + blockSize - 1, n - 1)` for safety. In the divisible case this never triggers.
33
34A defensive variant would add:
35```cpp
36if (blockSize == 0) return nums;          // guard against infinite loop
37int right = min(start + blockSize - 1, n - 1);  // clamp last block
38

1/**
2 * Divides the array into consecutive subarrays (each of size Math.floor(n / k))
3 * and reverses each subarray in place, then returns the modified array.
4 *
5 * @param nums - The input array of numbers to be processed.
6 * @param k - Used to determine the size of each subarray (size = floor(n / k)).
7 * @returns The same array reference after in-place reversal of each subarray.
8 */
9function reverseSubarrays(nums: number[], k: number): number[] {
10    // Total number of elements in the array.
11    const n: number = nums.length;
12
13    // Size of each subarray group to be reversed.
14    const groupSize: number = Math.floor(n / k);
15
16    // Iterate over the array one group at a time, stepping by groupSize.
17    for (let start = 0; start < n; start += groupSize) {
18        // Two pointers: left at the group's start, right at the group's end.
19        let left: number = start;
20        let right: number = start + groupSize - 1;
21
22        // Reverse the current group in place by swapping inward.
23        while (left < right) {
24            // Swap the elements at the left and right pointers.
25            const temp: number = nums[left];
26            nums[left++] = nums[right];
27            nums[right--] = temp;
28        }
29    }
30
31    // Return the modified array (same reference as the input).
32    return nums;
33}
34