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Rotate Function

Medium

Asked by:

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Topics:

Arrays

You are given an integer array nums of length n.

Assume arr_k to be an array obtained by rotating nums by k positions clock-wise. We define the rotation function F on nums as follow:

F(k) = 0 * arr_k[0] + 1 * arr_k[1] + ... + (n - 1) * arr_k[n - 1].

Return the maximum value of F(0), F(1), ..., F(n-1).

The test cases are generated so that the answer fits in a 32-bit integer.

Example 1:

Input: nums = [4,3,2,6]
Output: 26
Explanation:
F(0) = (0 * 4) + (1 * 3) + (2 * 2) + (3 * 6) = 0 + 3 + 4 + 18 = 25
F(1) = (0 * 6) + (1 * 4) + (2 * 3) + (3 * 2) = 0 + 4 + 6 + 6 = 16
F(2) = (0 * 2) + (1 * 6) + (2 * 4) + (3 * 3) = 0 + 6 + 8 + 9 = 23
F(3) = (0 * 3) + (1 * 2) + (2 * 6) + (3 * 4) = 0 + 2 + 12 + 12 = 26
So the maximum value of F(0), F(1), F(2), F(3) is F(3) = 26.

Example 2:

Input: nums = [100]
Output: 0

Constraints:

n == nums.length
1 <= n <= 10⁵
-100 <= nums[i] <= 100

Try coding it in LeetCode

Rotate Function

Medium

Asked by:

113 views

Topics:

Arrays

You are given an integer array nums of length n.

Assume arr_k to be an array obtained by rotating nums by k positions clock-wise. We define the rotation function F on nums as follow:

F(k) = 0 * arr_k[0] + 1 * arr_k[1] + ... + (n - 1) * arr_k[n - 1].

Return the maximum value of F(0), F(1), ..., F(n-1).

The test cases are generated so that the answer fits in a 32-bit integer.

Example 1:

Input: nums = [4,3,2,6]
Output: 26
Explanation:
F(0) = (0 * 4) + (1 * 3) + (2 * 2) + (3 * 6) = 0 + 3 + 4 + 18 = 25
F(1) = (0 * 6) + (1 * 4) + (2 * 3) + (3 * 2) = 0 + 4 + 6 + 6 = 16
F(2) = (0 * 2) + (1 * 6) + (2 * 4) + (3 * 3) = 0 + 6 + 8 + 9 = 23
F(3) = (0 * 3) + (1 * 2) + (2 * 6) + (3 * 4) = 0 + 2 + 12 + 12 = 26
So the maximum value of F(0), F(1), F(2), F(3) is F(3) = 26.

Example 2:

Input: nums = [100]
Output: 0

Constraints:

n == nums.length
1 <= n <= 10⁵
-100 <= nums[i] <= 100

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the range of values for the integers in the input array?
Can the input array be empty or null? If so, what should the function return?
Are there any constraints on the size of the input array?
If there are multiple rotations that produce the same maximum value, is any one of them acceptable?
Can you provide a small example input and the corresponding expected output?

Brute Force Solution

Approach

The brute force method for the rotate function problem involves trying every possible rotation of the input. We calculate a score for each rotation and then find the best score. It's like manually shifting a set of values around and checking their value at each shift.

Here's how the algorithm would work step-by-step:

Imagine the set of numbers sitting in a circle.
First, don't rotate the numbers at all, calculate the score based on their current positions.
Next, rotate the circle one position. Calculate the score again.
Keep rotating the circle by one position each time, calculating the score for each rotation.
Do this until you've tried every possible rotation, going all the way around the circle.
Compare the scores from all the different rotations.
The highest score is the answer.

Code Implementation

def rotate_function_brute_force(numbers):
    array_length = len(numbers)
    max_function_value = float('-inf')

    # Iterate through all possible rotations
    for rotation_index in range(array_length):
        function_value = 0

        # Calculate the function value for the current rotation
        for element_index in range(array_length):
            rotated_index = (element_index + rotation_index) % array_length
            function_value += element_index * numbers[rotated_index]

        # Update max_function_value if current rotation results in a higher value
        if function_value > max_function_value:
            max_function_value = function_value

    #Handle the case of empty array
    if array_length == 0:
        return 0

    return max_function_value

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through all possible rotations of the array, which amounts to n rotations. For each rotation, the algorithm calculates the function's value by iterating through all n elements of the array to compute the sum of index * value. Therefore, the time complexity is driven by nested loops, with the outer loop iterating n times for rotations and the inner loop iterating n times for calculation. The total number of operations approximates n * n, simplifying to O(n²).

Space Complexity

O(1) – The brute force method calculates the score for each rotation by directly manipulating the input array. It does not use any auxiliary data structures like new arrays, lists, or hash maps to store intermediate results. The algorithm only requires a few integer variables to track the current rotation and the maximum score calculated so far. Therefore, the space used is constant and independent of the input size N, where N is the number of elements in the input array.

Optimal Solution

Approach

The core idea is to avoid recomputing the entire 'rotation score' for each shifted version of the input. Instead, we leverage the relationship between consecutive rotations to update the score efficiently. This significantly reduces redundant calculations.

Here's how the algorithm would work step-by-step:

First, calculate the score of the original (unrotated) input.
Next, find the total sum of all the numbers in the input. This will be used repeatedly.
Now, use the following clever trick: The score of the next rotation can be easily calculated from the current rotation's score, using the total sum and the last element of the input. It is not required to recalculate it from scratch.
Repeat the process, rotating the input and calculating the score from the previous score until all rotations have been considered.
Keep track of the highest score found during this process.
Finally, return the highest score.

Code Implementation

def max_rotate_function(numbers):
    array_length = len(numbers)
    if array_length == 0:
        return 0

    # Calculate initial rotation score.
    initial_rotation_score = 0
    array_sum = 0
    for i in range(array_length):
        initial_rotation_score += i * numbers[i]
        array_sum += numbers[i]

    max_score = initial_rotation_score
    previous_rotation_score = initial_rotation_score

    #Iterate to calculate subsequent rotation scores.
    for i in range(1, array_length):
        #Calculate next rotation using the trick.
        next_rotation_score = previous_rotation_score + array_sum - array_length * numbers[array_length - i]

        max_score = max(max_score, next_rotation_score)
        previous_rotation_score = next_rotation_score

    # Return the maximum rotation score observed.
    return max_score

Big(O) Analysis

Time Complexity

O(n) – The solution calculates the initial rotation score in O(n) time. Then, it iterates n-1 times, updating the rotation score in O(1) time during each iteration by using the clever trick involving the total sum and the last element of the previous rotation. The final maximum score calculation also takes O(1) time. Thus, the time complexity is dominated by the initial calculation and the n-1 iterations of O(1) operations, resulting in a total of O(n).

Space Complexity

O(1) – The algorithm calculates the initial score and the total sum of the input array. It then iteratively updates the score using only the previous score, the total sum, and the last element of the array. It also stores a variable to keep track of the maximum score found so far. Because the amount of extra space used (for the score, total sum, and max_score variable) does not depend on the size of the input array (N), the space complexity is constant.

Edge Cases

Case	How to Handle
Empty array	Return 0, as there is no rotation possible and the sum should be zero.
Array with one element	Return 0, as no rotation changes the value and F(0) will be 0.
Large array size (close to the maximum allowed)	The solution should use a memory-efficient approach (e.g., O(1) space) to avoid memory errors and consider possible integer overflows during the sum calculations.
Array containing large positive and negative numbers	Use long data types to store intermediate calculations like the sum and F(i) values to avoid integer overflow.
Array with all elements being zero	The solution should return 0 since all F(i) will be zero regardless of the rotation.
Array with elements having the same absolute value but different signs (+/-)	The algorithm should correctly compute the function values with both positive and negative numbers.
Array with very skewed distribution (one very large number and many small numbers)	Use appropriate data types to prevent overflow during sum and rotation calculations.
Integer overflow during summation of array elements and calculation of F(i)	Use long data type to store sum and F(i) to prevent overflow during calculations.

Empty array

How to Handle:

Return 0, as there is no rotation possible and the sum should be zero.

Array with one element

How to Handle:

Return 0, as no rotation changes the value and F(0) will be 0.

Large array size (close to the maximum allowed)

How to Handle:

The solution should use a memory-efficient approach (e.g., O(1) space) to avoid memory errors and consider possible integer overflows during the sum calculations.

Array containing large positive and negative numbers

How to Handle:

Use long data types to store intermediate calculations like the sum and F(i) values to avoid integer overflow.

Array with all elements being zero

How to Handle:

The solution should return 0 since all F(i) will be zero regardless of the rotation.

Array with elements having the same absolute value but different signs (+/-)

How to Handle:

The algorithm should correctly compute the function values with both positive and negative numbers.

Array with very skewed distribution (one very large number and many small numbers)

How to Handle:

Use appropriate data types to prevent overflow during sum and rotation calculations.

Integer overflow during summation of array elements and calculation of F(i)

How to Handle:

Use long data type to store sum and F(i) to prevent overflow during calculations.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the range of values for the integers in the input array?
Can the input array be empty or null? If so, what should the function return?
Are there any constraints on the size of the input array?
If there are multiple rotations that produce the same maximum value, is any one of them acceptable?
Can you provide a small example input and the corresponding expected output?

Brute Force Solution

Approach

Here's how the algorithm would work step-by-step:

Imagine the set of numbers sitting in a circle.
First, don't rotate the numbers at all, calculate the score based on their current positions.
Next, rotate the circle one position. Calculate the score again.
Keep rotating the circle by one position each time, calculating the score for each rotation.
Do this until you've tried every possible rotation, going all the way around the circle.
Compare the scores from all the different rotations.
The highest score is the answer.

Code Implementation

def rotate_function_brute_force(numbers):
    array_length = len(numbers)
    max_function_value = float('-inf')

    # Iterate through all possible rotations
    for rotation_index in range(array_length):
        function_value = 0

        # Calculate the function value for the current rotation
        for element_index in range(array_length):
            rotated_index = (element_index + rotation_index) % array_length
            function_value += element_index * numbers[rotated_index]

        # Update max_function_value if current rotation results in a higher value
        if function_value > max_function_value:
            max_function_value = function_value

    #Handle the case of empty array
    if array_length == 0:
        return 0

    return max_function_value

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

Here's how the algorithm would work step-by-step:

First, calculate the score of the original (unrotated) input.
Next, find the total sum of all the numbers in the input. This will be used repeatedly.
Now, use the following clever trick: The score of the next rotation can be easily calculated from the current rotation's score, using the total sum and the last element of the input. It is not required to recalculate it from scratch.
Repeat the process, rotating the input and calculating the score from the previous score until all rotations have been considered.
Keep track of the highest score found during this process.
Finally, return the highest score.

Code Implementation

def max_rotate_function(numbers):
    array_length = len(numbers)
    if array_length == 0:
        return 0

    # Calculate initial rotation score.
    initial_rotation_score = 0
    array_sum = 0
    for i in range(array_length):
        initial_rotation_score += i * numbers[i]
        array_sum += numbers[i]

    max_score = initial_rotation_score
    previous_rotation_score = initial_rotation_score

    #Iterate to calculate subsequent rotation scores.
    for i in range(1, array_length):
        #Calculate next rotation using the trick.
        next_rotation_score = previous_rotation_score + array_sum - array_length * numbers[array_length - i]

        max_score = max(max_score, next_rotation_score)
        previous_rotation_score = next_rotation_score

    # Return the maximum rotation score observed.
    return max_score

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Empty array	Return 0, as there is no rotation possible and the sum should be zero.
Array with one element	Return 0, as no rotation changes the value and F(0) will be 0.
Large array size (close to the maximum allowed)	The solution should use a memory-efficient approach (e.g., O(1) space) to avoid memory errors and consider possible integer overflows during the sum calculations.
Array containing large positive and negative numbers	Use long data types to store intermediate calculations like the sum and F(i) values to avoid integer overflow.
Array with all elements being zero	The solution should return 0 since all F(i) will be zero regardless of the rotation.
Array with elements having the same absolute value but different signs (+/-)	The algorithm should correctly compute the function values with both positive and negative numbers.
Array with very skewed distribution (one very large number and many small numbers)	Use appropriate data types to prevent overflow during sum and rotation calculations.
Integer overflow during summation of array elements and calculation of F(i)	Use long data type to store sum and F(i) to prevent overflow during calculations.

Empty array

How to Handle:

Return 0, as there is no rotation possible and the sum should be zero.

Array with one element

How to Handle:

Return 0, as no rotation changes the value and F(0) will be 0.

Large array size (close to the maximum allowed)

How to Handle:

The solution should use a memory-efficient approach (e.g., O(1) space) to avoid memory errors and consider possible integer overflows during the sum calculations.

Array containing large positive and negative numbers

How to Handle:

Use long data types to store intermediate calculations like the sum and F(i) values to avoid integer overflow.

Array with all elements being zero

How to Handle:

The solution should return 0 since all F(i) will be zero regardless of the rotation.

Array with elements having the same absolute value but different signs (+/-)

How to Handle:

The algorithm should correctly compute the function values with both positive and negative numbers.

Array with very skewed distribution (one very large number and many small numbers)

How to Handle:

Use appropriate data types to prevent overflow during sum and rotation calculations.

Integer overflow during summation of array elements and calculation of F(i)

How to Handle:

Use long data type to store sum and F(i) to prevent overflow during calculations.