Let me first explain what the provided code does, including...

Let me first explain what the provided code does, including how the k = k % nums.size() line affects the result.

Code Explanation

The code is designed to rotate an array nums to the right by k steps. Rotating an array to the right by k steps means that the last k elements will move to the front of the array, and the rest will shift to the right. For example, rotating [1, 2, 3, 4, 5] by k = 2 results in [4, 5, 1, 2, 3].

Key Steps in the Code

1. Reverse the Entire Array:

   int i = 0;
   int j = nums.size() - 1;
   while (i < j) {
       swap(nums[i], nums[j]);
       i++;
       j--;
   }
   

   This reverses the entire array. For example:

   • Input: nums = [1, 2, 3, 4, 5]
   • After reversing: nums = [5, 4, 3, 2, 1]

2. Reverse the First k Elements:

   int s = 0;
   int e = k - 1;
   while (s <= e) {
       swap(nums[s], nums[e]);
       s++;
       e--;
   }
   

   This reverses the first k elements of the array (where k is the number of steps to rotate). For example:

   • Before: nums = [5, 4, 3, 2, 1]
   • After reversing the first k elements (say, k = 2): nums = [4, 5, 3, 2, 1]

3. Reverse the Remaining Last (size - k) Elements:

   int p = k;
   int t = nums.size() - 1;
   while (p <= t) {
       swap(nums[p], nums[t]);
       p++;
       t--;
   }
   

   This reverses the remaining portion of the array (i.e., from index k to the end). For example:

   • Before: nums = [4, 5, 3, 2, 1]
   • After reversing the last (size - k) elements: nums = [4, 5, 1, 2, 3]

The Issue with k = k % nums.size() and Example

Why is k = k % nums.size() important?

If k is greater than the size of the array (nums.size()), performing the rotation would produce the same result as k % nums.size() steps. For example, rotating an array of size 7 by 10 steps is equivalent to rotating it by 10 % 7 = 3 steps.

However, your statement about k = 0 is incorrect. Here's why:

• If k = k % nums.size() and k = 3 initially with size = 7:

  k = 3 % 7; // k is still 3
  

• Only if k = nums.size() (e.g., k = 7), then:

  k = 7 % 7; // k becomes 0
  

When k = 0, no rotation is needed, and the array remains unchanged. This is the correct expected behavior because performing 7 (or any multiple of 7) rotations on an array of size 7 results in the same array.

Example:

1. Input: nums = [1, 2, 3, 4, 5, 6, 7], k = 3

   • After k = k % nums.size(): k = 3 % 7 = 3
   • Final output: [5, 6, 7, 1, 2, 3, 4]

2. Input: nums = [1, 2, 3, 4, 5, 6, 7], k = 10

   • After k = k % nums.size(): k = 10 % 7 = 3
   • Final output: [5, 6, 7, 1, 2, 3, 4]

3. Input: nums = [1, 2, 3, 4, 5, 6, 7], k = 7

   • After k = k % nums.size(): k = 7 % 7 = 0
   • No rotation happens. Final output: [1, 2, 3, 4, 5, 6, 7]

Why Your Concern About k = 0 Being "Wrong" Is Incorrect

When k = 0 after k = k % nums.size(), the code correctly does not perform the rotation since the array remains unchanged. This behavior is correct, and it is not a wrong result.

Summary

• The code rotates the array to the right by k steps using a three-reverse approach.
• Adding k = k % nums.size() prevents unnecessary rotations when k is greater than the array size or exactly equal to it.
• If k = 0 after the modulo operation, the array remains unchanged, which is the correct behavior.