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Minimum Number of Operations to Reinitialize a Permutation

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Arrays

You are given an even integer n. You initially have a permutation perm of size n where perm[i] == i (0-indexed).

In one operation, you will create a new array arr, and for each i:

If i % 2 == 0, then arr[i] = perm[i / 2].
If i % 2 == 1, then arr[i] = perm[n / 2 + (i - 1) / 2].

You will then assign arr to perm.

Return the minimum non-zero number of operations you need to perform on perm to return the permutation to its initial value.

Example 1:

Input: n = 2
Output: 1
Explanation: perm = [0,1] initially.
After the 1^st operation, perm = [0,1]
So it takes only 1 operation.

Example 2:

Input: n = 4
Output: 2
Explanation: perm = [0,1,2,3] initially.
After the 1^st operation, perm = [0,2,1,3]
After the 2^nd operation, perm = [0,1,2,3]
So it takes only 2 operations.

Example 3:

Input: n = 6
Output: 4

Constraints:

2 <= n <= 1000
n is even.

Try coding it in LeetCode

Minimum Number of Operations to Reinitialize a Permutation

Medium

Asked by:

16 views

Topics:

Arrays

You are given an even integer n. You initially have a permutation perm of size n where perm[i] == i (0-indexed).

In one operation, you will create a new array arr, and for each i:

If i % 2 == 0, then arr[i] = perm[i / 2].
If i % 2 == 1, then arr[i] = perm[n / 2 + (i - 1) / 2].

You will then assign arr to perm.

Return the minimum non-zero number of operations you need to perform on perm to return the permutation to its initial value.

Example 1:

Input: n = 2
Output: 1
Explanation: perm = [0,1] initially.
After the 1^st operation, perm = [0,1]
So it takes only 1 operation.

Example 2:

Input: n = 4
Output: 2
Explanation: perm = [0,1,2,3] initially.
After the 1^st operation, perm = [0,2,1,3]
After the 2^nd operation, perm = [0,1,2,3]
So it takes only 2 operations.

Example 3:

Input: n = 6
Output: 4

Constraints:

2 <= n <= 1000
n is even.

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 maximum size of the input permutation 'n'?
Does the input permutation always contain all numbers from 0 to n-1, and are they all unique?
If the initial permutation is already reinitialized (i.e., equals the initial state), should I return 0?
Are we looking for the minimum number of operations or any number of operations to reinitialize?
Is the value of 'n' always a power of 2?

Brute Force Solution

Approach

The problem asks us to find out how many steps it takes to return a shuffled list of numbers back to its original order. A brute force approach is to repeatedly perform the shuffling operation and check if the list is back to its original state after each shuffle.

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

Start with the initial list of numbers.
Perform the specified shuffling operation on the list.
Check if the shuffled list is now in the same order as the original list.
If it is, we're done and can say how many shuffles it took.
If it's not, repeat the shuffling operation on the current shuffled list.
Keep doing this until the list returns to its original order, counting each shuffle as we go.

Code Implementation

def reinitialize_permutation_brute_force(list_length):
    original_permutation = list(range(list_length))
    current_permutation = list(range(list_length))
    operations_count = 0

    while True:
        new_permutation = [0] * list_length

        # Perform the permutation operation as described in the problem.
        for index in range(list_length):
            if index % 2 == 0:
                new_permutation[index] = current_permutation[index // 2]
            else:
                new_permutation[index] = current_permutation[list_length // 2 + (index - 1) // 2]

        current_permutation = new_permutation
        operations_count += 1

        # Check if the current permutation matches the original.

        if current_permutation == original_permutation:
            
            return operations_count

Big(O) Analysis

Time Complexity

O(n^2) – The provided solution uses a brute force approach where we repeatedly shuffle the permutation until it returns to its original order. Each shuffle operation involves iterating through all 'n' elements of the array. We also compare each shuffled array with the original array, which again takes O(n) time. The number of shuffles required is at most 'n' in the worst-case scenario. Therefore, we have at most n shuffle operations each taking O(n) time, thus leading to a time complexity of O(n * n) = O(n^2).

Space Complexity

O(N) – The brute-force approach described involves repeatedly shuffling the list. The shuffled list must be stored, which takes O(N) space where N is the number of elements in the original list. Although the original list can be overwritten to save space, the explanation doesn't specify an in-place shuffle. Therefore, the shuffled list is the dominant auxiliary data structure. Hence the overall auxiliary space complexity is O(N).

Optimal Solution

Approach

The key idea is to simulate the permutation operation and track when the array returns to its original state. Instead of performing the operation until we find the original state, we can exploit the cyclical nature of the permutation and look for patterns to speed up the process.

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

Start with an initial state where the array is in its original sorted order.
Perform a single permutation operation, following the specific rules of the problem (even positions get values from the first half, odd positions get values from the second half).
Check if the array is back to its original sorted order. If it is, we're done; the number of operations performed is the answer.
If the array is not back to its original state, perform the permutation operation again, keeping track of the total number of operations.
Repeat the permutation operation and the check until the array returns to its original state.
The total number of operations needed to return to the original state is the answer.

Code Implementation

def reinitialize_permutation(number):    original_array = list(range(number))
    current_array = list(range(number))
    operations_count = 0
    
    while True:
        new_array = [0] * number
        for index in range(number):
            if index % 2 == 0:
                new_index = index // 2
            else:
                new_index = number // 2 + (index - 1) // 2
            new_array[index] = current_array[new_index]
            
        current_array = new_array
        operations_count += 1

        # Check if the array has returned to its original state.
        if current_array == original_array:
            return operations_count

Big(O) Analysis

Time Complexity

O(n^2) – The algorithm simulates the permutation operation until the array returns to its initial state. In each operation, we iterate through all n elements of the array to create the new permutation. The outer loop represents the number of permutation operations needed to return to the original state, which in the worst case, could require O(n) operations. Consequently, since each permutation operation requires O(n) work and it may take O(n) such operations, the overall time complexity is O(n * n) = O(n^2).

Space Complexity

O(N) – The provided solution requires creating a new array of size N to store the permuted array after each operation. The original array is also of size N. The solution also uses a few integer variables for loop counters and operation counts, which take constant space. Therefore, the dominant space usage comes from the creation of the temporary array for each permutation operation, which scales linearly with the input size N.

Edge Cases

Case	How to Handle
n = 1: Permutation of size 1	Should return 0 because the array is already initialized.
n = 2: Smallest non-trivial permutation	Requires one operation to revert to the initial state [0, 1].
n is a large power of 2 (e.g., 1024): Worst case for naive simulation	A naive simulation might be slow; an optimized approach that identifies repeating cycles is needed.
Permutation returns to original state early	The solution needs to detect when the permutation returns to its original state and avoid unnecessary iterations.
Integer overflow with large n during index calculations	Use appropriate data types (e.g., long in Java/C++) to prevent integer overflow when calculating indices.
n is odd	The formula `i -> (2i) % n if i < n/2 else (2i + 1 - n) % n` handles odd n correctly.
Very large n nearing the maximum allowed by memory constraints	Need to consider memory usage and possibly explore more memory-efficient algorithms if a large n leads to memory issues.
Negative n or n=0	Handle invalid input by either throwing an exception or returning a predefined error value like -1, depending on requirements.

n = 1: Permutation of size 1

How to Handle:

Should return 0 because the array is already initialized.

n = 2: Smallest non-trivial permutation

How to Handle:

Requires one operation to revert to the initial state [0, 1].

n is a large power of 2 (e.g., 1024): Worst case for naive simulation

How to Handle:

A naive simulation might be slow; an optimized approach that identifies repeating cycles is needed.

Permutation returns to original state early

How to Handle:

The solution needs to detect when the permutation returns to its original state and avoid unnecessary iterations.

Integer overflow with large n during index calculations

How to Handle:

Use appropriate data types (e.g., long in Java/C++) to prevent integer overflow when calculating indices.

n is odd

How to Handle:

The formula `i -> (2*i) % n if i < n/2 else (2*i + 1 - n) % n` handles odd n correctly.

Very large n nearing the maximum allowed by memory constraints

How to Handle:

Need to consider memory usage and possibly explore more memory-efficient algorithms if a large n leads to memory issues.

Negative n or n=0

How to Handle:

Handle invalid input by either throwing an exception or returning a predefined error value like -1, depending on requirements.

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 maximum size of the input permutation 'n'?
Does the input permutation always contain all numbers from 0 to n-1, and are they all unique?
If the initial permutation is already reinitialized (i.e., equals the initial state), should I return 0?
Are we looking for the minimum number of operations or any number of operations to reinitialize?
Is the value of 'n' always a power of 2?

Brute Force Solution

Approach

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

Start with the initial list of numbers.
Perform the specified shuffling operation on the list.
Check if the shuffled list is now in the same order as the original list.
If it is, we're done and can say how many shuffles it took.
If it's not, repeat the shuffling operation on the current shuffled list.
Keep doing this until the list returns to its original order, counting each shuffle as we go.

Code Implementation

def reinitialize_permutation_brute_force(list_length):
    original_permutation = list(range(list_length))
    current_permutation = list(range(list_length))
    operations_count = 0

    while True:
        new_permutation = [0] * list_length

        # Perform the permutation operation as described in the problem.
        for index in range(list_length):
            if index % 2 == 0:
                new_permutation[index] = current_permutation[index // 2]
            else:
                new_permutation[index] = current_permutation[list_length // 2 + (index - 1) // 2]

        current_permutation = new_permutation
        operations_count += 1

        # Check if the current permutation matches the original.

        if current_permutation == original_permutation:
            
            return operations_count

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

Start with an initial state where the array is in its original sorted order.
Perform a single permutation operation, following the specific rules of the problem (even positions get values from the first half, odd positions get values from the second half).
Check if the array is back to its original sorted order. If it is, we're done; the number of operations performed is the answer.
If the array is not back to its original state, perform the permutation operation again, keeping track of the total number of operations.
Repeat the permutation operation and the check until the array returns to its original state.
The total number of operations needed to return to the original state is the answer.

Code Implementation

def reinitialize_permutation(number):    original_array = list(range(number))
    current_array = list(range(number))
    operations_count = 0
    
    while True:
        new_array = [0] * number
        for index in range(number):
            if index % 2 == 0:
                new_index = index // 2
            else:
                new_index = number // 2 + (index - 1) // 2
            new_array[index] = current_array[new_index]
            
        current_array = new_array
        operations_count += 1

        # Check if the array has returned to its original state.
        if current_array == original_array:
            return operations_count

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
n = 1: Permutation of size 1	Should return 0 because the array is already initialized.
n = 2: Smallest non-trivial permutation	Requires one operation to revert to the initial state [0, 1].
n is a large power of 2 (e.g., 1024): Worst case for naive simulation	A naive simulation might be slow; an optimized approach that identifies repeating cycles is needed.
Permutation returns to original state early	The solution needs to detect when the permutation returns to its original state and avoid unnecessary iterations.
Integer overflow with large n during index calculations	Use appropriate data types (e.g., long in Java/C++) to prevent integer overflow when calculating indices.
n is odd	The formula `i -> (2i) % n if i < n/2 else (2i + 1 - n) % n` handles odd n correctly.
Very large n nearing the maximum allowed by memory constraints	Need to consider memory usage and possibly explore more memory-efficient algorithms if a large n leads to memory issues.
Negative n or n=0	Handle invalid input by either throwing an exception or returning a predefined error value like -1, depending on requirements.

n = 1: Permutation of size 1

How to Handle:

Should return 0 because the array is already initialized.

n = 2: Smallest non-trivial permutation

How to Handle:

Requires one operation to revert to the initial state [0, 1].

n is a large power of 2 (e.g., 1024): Worst case for naive simulation

How to Handle:

A naive simulation might be slow; an optimized approach that identifies repeating cycles is needed.

Permutation returns to original state early

How to Handle:

The solution needs to detect when the permutation returns to its original state and avoid unnecessary iterations.

Integer overflow with large n during index calculations

How to Handle:

Use appropriate data types (e.g., long in Java/C++) to prevent integer overflow when calculating indices.

n is odd

How to Handle:

The formula `i -> (2*i) % n if i < n/2 else (2*i + 1 - n) % n` handles odd n correctly.

Very large n nearing the maximum allowed by memory constraints

How to Handle:

Need to consider memory usage and possibly explore more memory-efficient algorithms if a large n leads to memory issues.

Negative n or n=0

How to Handle:

Handle invalid input by either throwing an exception or returning a predefined error value like -1, depending on requirements.