The provided code defines a solution to rotate an array...

The provided code defines a solution to rotate an array (nums) to the right by k steps. It utilizes a three-step process that involves reversing parts of the array in order to achieve the rotation.

Step-by-step Explanation:

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 so that the last element moves to the beginning, and so on.

2. Reverse the first k elements:

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

   After reversing the entire array, this reverses the first k elements (from index 0 to k-1).

3. Reverse the remaining elements:

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

   This step reverses the rest of the array (from index k to the end).

By combining these three steps, the array gets effectively rotated k steps to the right.

Flaw in the Code:

Although the logic works, there is one key edge case that needs to be addressed:

• If k is greater than or equal to the size of the array (nums.size()), the rotation is redundant since rotating an array by its length results in the same array. Thus, k needs to be adjusted using k = k % nums.size().

Optimized Version of the Code:

The optimized version incorporates the modulo operation to avoid unnecessary rotations and simplifies the code slightly.

class Solution {
public:
    void rotate(vector<int>& nums, int k) {
        int n = nums.size();
        k = k % n;  // Handle cases where k >= nums.size()

        // Helper function to reverse part of the array
        auto reverse = [&](int start, int end) {
            while (start < end) {
                swap(nums[start], nums[end]);
                start++;
                end--;
            }
        };

        // Reverse the three segments
        reverse(0, n - 1);      // Reverse the entire array
        reverse(0, k - 1);      // Reverse the first k elements
        reverse(k, n - 1);      // Reverse the remaining n-k elements
    }
};

Explanation of the Changes:

1. Modulo Operation:

   • The line k = k % n; ensures that if k is larger than the length of the array, only the relevant rotation is performed. For example, rotating an array of size 5 by 7 steps is the same as rotating it by 7 % 5 = 2 steps.

2. Helper Function:

   • The reverse lambda function simplifies the reversing logic to avoid duplicating similar code.

3. Simplified Reverse Steps:

   • The three steps (entire array, first k, and remaining part) are executed in a clear and concise manner using the helper function.

Time and Space Complexity:

• Time Complexity: O(n) (Each element is swapped at most once in the three reversals.)
• Space Complexity: O(1) (In-place rotation.)