Solution Approach
Solution Overview
There are 3 main approaches to solve this problem:
- Using Extra Array - Complex approach
- Using Cyclic Replacements - Complex approach
- Using Array Reversal - Complex approach
Approach 1: Using Extra Array
Let's start by understanding the problem: we need to rotate an array to the right by k steps, which means each element should be moved k positions to the right, with elements that go beyond the end of the array wrapping around to the beginning.
Thinking Process: The most straightforward approach is to use an extra array to store the rotated elements. We can calculate the new position of each element after rotation and place it in the correct position in the new array.
For an element at index i in the original array, its new position after rotating k steps to the right would be (i + k) % n, where n is the length of the array. We can use this formula to place each element in its correct position in the new array.
Intuition: This approach directly follows the problem statement by calculating the new position of each element after rotation. It's simple to understand and implement, but it requires O(n) extra space.
Algorithm:
- Create a new array of the same length as the original array.
- For each element at index i in the original array, calculate its new position as (i + k) % n, where n is the length of the array.
- Place the element in the new array at the calculated position.
- Copy the elements from the new array back to the original array.
Time Complexity:
We need to iterate through all n elements of the array once.
Space Complexity:
We need an extra array of size n to store the rotated elements.
Approach 2: Using Cyclic Replacements
We can optimize the space complexity by performing the rotation in-place using cyclic replacements.
Thinking Process: Instead of using an extra array, we can directly place each element in its final position in the original array. We start with the first element and keep replacing elements in a cyclic manner until we've processed all elements.
However, there's a complication: if we simply replace elements one by one, we might overwrite some elements before we've had a chance to move them. To avoid this, we need to keep track of the element we're currently replacing and the element we're replacing it with.
Intuition: This approach is more complex than the extra array approach, but it allows us to perform the rotation in-place with O(1) extra space. The key insight is to use cyclic replacements to move each element to its final position without losing any information.
Algorithm:
- Calculate the effective number of rotations as k % n, where n is the length of the array.
- Initialize a count variable to 0 to keep track of the number of elements we've processed.
- Iterate through the array until we've processed all elements:
- a. Start with the current element and its position.
- b. Calculate the next position as (current_position + k) % n.
- c. Save the element at the next position.
- d. Replace the element at the next position with the current element.
- e. Update the current element and position for the next iteration.
- f. Increment the count variable.
- g. If we return to the starting position, move to the next position and continue.
- Repeat until all elements have been processed.
Time Complexity:
We need to process all n elements of the array once.
Space Complexity:
We only need a constant amount of extra space to store temporary variables.
Approach 3: Using Array Reversal
Another elegant approach to solve this problem in-place is by using array reversal.
Thinking Process: The key insight is that rotating an array by k steps is equivalent to:
- Reversing the entire array.
- Reversing the first k elements.
- Reversing the remaining n-k elements.
For example, if we have the array [1, 2, 3, 4, 5, 6, 7] and k = 3:
- After reversing the entire array, we get [7, 6, 5, 4, 3, 2, 1].
- After reversing the first k elements, we get [5, 6, 7, 4, 3, 2, 1].
- After reversing the remaining n-k elements, we get [5, 6, 7, 1, 2, 3, 4], which is the correct result.
Intuition: This approach is elegant and easy to implement. It allows us to perform the rotation in-place with O(1) extra space. The key insight is to use array reversal to achieve the rotation effect.
Algorithm:
- Calculate the effective number of rotations as k % n, where n is the length of the array.
- Reverse the entire array.
- Reverse the first k elements.
- Reverse the remaining n-k elements.
Time Complexity:
We need to reverse the array three times, each taking O(n) time.
Space Complexity:
We only need a constant amount of extra space to store temporary variables during the reversal.
Pseudocode
1234567891011121314function rotate(nums, k): n = length of nums k = k % n // Calculate effective number of rotations // Create a new array to store the rotated elements result = new array of size n // Place each element in its correct position after rotation for i from 0 to n-1: result[(i + k) % n] = nums[i] // Copy the elements back to the original array for i from 0 to n-1: nums[i] = result[i]String Problems
Learning Objective
Understand different approaches to solve the array rotator problem and analyze their efficiency.
Solution Approaches
Approach 1: Using Extra Array
Let's start by understanding the problem: we need to rotate an array to the right by k steps, which means each element should be moved k positions to the right, with elements that go beyond the end of the array wrapping around to the beginning.
Thinking Process: The most straightforward approach is to use an extra array to store the rotated elements. We can calculate the new position of each element after rotation and place it in the correct position in the new array.
For an element at index i in the original array, its new position after rotating k steps to the right would be (i + k) % n, where n is the length of the array. We can use this formula to place each element in its correct position in the new array.
Intuition: This approach directly follows the problem statement by calculating the new position of each element after rotation. It's simple to understand and implement, but it requires O(n) extra space.
Algorithm:
- Create a new array of the same length as the original array.
- For each element at index i in the original array, calculate its new position as (i + k) % n, where n is the length of the array.
- Place the element in the new array at the calculated position.
- Copy the elements from the new array back to the original array.
Approach 2: Using Cyclic Replacements
We can optimize the space complexity by performing the rotation in-place using cyclic replacements.
Thinking Process: Instead of using an extra array, we can directly place each element in its final position in the original array. We start with the first element and keep replacing elements in a cyclic manner until we've processed all elements.
However, there's a complication: if we simply replace elements one by one, we might overwrite some elements before we've had a chance to move them. To avoid this, we need to keep track of the element we're currently replacing and the element we're replacing it with.
Intuition: This approach is more complex than the extra array approach, but it allows us to perform the rotation in-place with O(1) extra space. The key insight is to use cyclic replacements to move each element to its final position without losing any information.
Algorithm:
- Calculate the effective number of rotations as k % n, where n is the length of the array.
- Initialize a count variable to 0 to keep track of the number of elements we've processed.
- Iterate through the array until we've processed all elements:
- a. Start with the current element and its position.
- b. Calculate the next position as (current_position + k) % n.
- c. Save the element at the next position.
- d. Replace the element at the next position with the current element.
- e. Update the current element and position for the next iteration.
- f. Increment the count variable.
- g. If we return to the starting position, move to the next position and continue.
- Repeat until all elements have been processed.
Approach 3: Using Array Reversal
Another elegant approach to solve this problem in-place is by using array reversal.
Thinking Process: The key insight is that rotating an array by k steps is equivalent to:
- Reversing the entire array.
- Reversing the first k elements.
- Reversing the remaining n-k elements.
For example, if we have the array [1, 2, 3, 4, 5, 6, 7] and k = 3:
- After reversing the entire array, we get [7, 6, 5, 4, 3, 2, 1].
- After reversing the first k elements, we get [5, 6, 7, 4, 3, 2, 1].
- After reversing the remaining n-k elements, we get [5, 6, 7, 1, 2, 3, 4], which is the correct result.
Intuition: This approach is elegant and easy to implement. It allows us to perform the rotation in-place with O(1) extra space. The key insight is to use array reversal to achieve the rotation effect.
Algorithm:
- Calculate the effective number of rotations as k % n, where n is the length of the array.
- Reverse the entire array.
- Reverse the first k elements.
- Reverse the remaining n-k elements.
Complexity Analysis
Using Extra Array Approach
Time Complexity
We need to iterate through all n elements of the array once.
Space Complexity
We need an extra array of size n to store the rotated elements.
Using Cyclic Replacements Approach
Time Complexity
We need to process all n elements of the array once.
Space Complexity
We only need a constant amount of extra space to store temporary variables.
Using Array Reversal Approach
Time Complexity
We need to reverse the array three times, each taking O(n) time.
Space Complexity
We only need a constant amount of extra space to store temporary variables during the reversal.
Pseudocode
Using Extra Array Approach
1234567891011121314function rotate(nums, k): n = length of nums k = k % n // Calculate effective number of rotations // Create a new array to store the rotated elements result = new array of size n // Place each element in its correct position after rotation for i from 0 to n-1: result[(i + k) % n] = nums[i] // Copy the elements back to the original array for i from 0 to n-1: nums[i] = result[i]Using Cyclic Replacements Approach
123456789101112131415161718192021222324function rotate(nums, k): n = length of nums k = k % n // Calculate effective number of rotations if n == 1 or k == 0: return // No rotation needed count = 0 // Count of elements processed for start from 0 to n-1: if count == n: break // All elements processed current = start prev = nums[start] do: next = (current + k) % n // Calculate next position temp = nums[next] // Save element at next position nums[next] = prev // Replace element at next position prev = temp // Update prev for next iteration current = next // Update current position count++ // Increment count of processed elements while current != start // Continue until we return to startUsing Array Reversal Approach
12345678910111213141516171819function rotate(nums, k): n = length of nums k = k % n // Calculate effective number of rotations // Helper function to reverse a portion of the array function reverse(start, end): while start < end: swap nums[start] and nums[end] start++ end-- // Reverse the entire array reverse(0, n-1) // Reverse the first k elements reverse(0, k-1) // Reverse the remaining n-k elements reverse(k, n-1)