array rotation using modulus

array rotation using modulus

class Solution {
public:
    void rotate(vector<int>& nums, int k) {

        // Get the size of the array
        int n = nums.size();

        // Create a temporary array of the same size to store rotated values
        vector<int> temp(n);

        // Loop through each element in the original array
        for(int i = 0; i < n; i++) {
            // Calculate the new position using (i + k) % n
            // This ensures that the index wraps around when it exceeds the array size
            temp[(i + k) % n] = nums[i];
        }

        // Copy the values from the temporary array back to the original array
        for(int i = 0; i < n; i++) {
            nums[i] = temp[i];
        }
    }
};

Logic:

• Modulus Operator: The key idea is to use the modulus operator to compute the new index after the rotation. (i + k) % n ensures that the index wraps around to the beginning if it exceeds the array size n.
• Temporary Array: A temporary array temp is used to store the rearranged elements based on their new positions.
• Final Step: After calculating the rotated positions, the values in temp are copied back into the original nums array.

Example:

vector<int> nums = {1, 2, 3, 4, 5, 6, 7};
int k = 3;

1. Initial Array: nums[] = {1, 2, 3, 4, 5, 6, 7}
k = 3 (rotate the array 3 positions to the right)

2. Step-by-Step Execution:

   • Array size n = 7.

   • Create a temporary array temp of size n. Loop to Rotate:

   • For i = 0:
     temp[(0 + 3) % 7] = nums[0] → temp[3] = 1 temp[] = {_, _, _, 1, _, _, _}

   • For i = 1:
     temp[(1 + 3) % 7] = nums[1]
     → temp[4] = 2
temp[] = {_, _, _, 1, 2, _, _}

   • For i = 2:
     temp[(2 + 3) % 7] = nums[2]
     → temp[5] = 3
temp[] = {_, _, _, 1, 2, 3, _}

   • For i = 3:
     temp[(3 + 3) % 7] = nums[3]
     → temp[6] = 4
temp[] = {_, _, _, 1, 2, 3, 4}

   • For i = 4:
     temp[(4 + 3) % 7] = nums[4]
     → temp[0] = 5
temp[] = {5, _, _, 1, 2, 3, 4}

   • For i = 5:
     temp[(5 + 3) % 7] = nums[5]
     → temp[1] = 6
temp[] = {5, 6, _, 1, 2, 3, 4}

   • For i = 6:
     temp[(6 + 3) % 7] = nums[6]
     → temp[2] = 7
temp[] = {5, 6, 7, 1, 2, 3, 4}

3. Copy Back to Original Array:

   • Now, copy temp[] back to nums[]:
   • nums[] = {5, 6, 7, 1, 2, 3, 4}

Final Output:

After rotating the array 3 positions to the right, the result is:

nums[] = {5, 6, 7, 1, 2, 3, 4}