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Elimination Game

Medium

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Topics:

RecursionDynamic Programming

You have a list arr of all integers in the range [1, n] sorted in a strictly increasing order. Apply the following algorithm on arr:

Starting from left to right, remove the first number and every other number afterward until you reach the end of the list.
Repeat the previous step again, but this time from right to left, remove the rightmost number and every other number from the remaining numbers.
Keep repeating the steps again, alternating left to right and right to left, until a single number remains.

Given the integer n, return the last number that remains in arr.

Example 1:

Input: n = 9
Output: 6
Explanation:
arr = [1, 2, 3, 4, 5, 6, 7, 8, 9]
arr = [2, 4, 6, 8]
arr = [2, 6]
arr = [6]

Example 2:

Input: n = 1
Output: 1

Constraints:

1 <= n <= 10⁹

Try coding it in LeetCode

Elimination Game

Medium

Asked by:

35 views

Topics:

RecursionDynamic Programming

You have a list arr of all integers in the range [1, n] sorted in a strictly increasing order. Apply the following algorithm on arr:

Starting from left to right, remove the first number and every other number afterward until you reach the end of the list.
Repeat the previous step again, but this time from right to left, remove the rightmost number and every other number from the remaining numbers.
Keep repeating the steps again, alternating left to right and right to left, until a single number remains.

Given the integer n, return the last number that remains in arr.

Example 1:

Input: n = 9
Output: 6
Explanation:
arr = [1, 2, 3, 4, 5, 6, 7, 8, 9]
arr = [2, 4, 6, 8]
arr = [2, 6]
arr = [6]

Example 2:

Input: n = 1
Output: 1

Constraints:

1 <= n <= 10⁹

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 value of 'n'? Is 'n' guaranteed to be a positive integer?
For the elimination rounds, if there's an odd number of elements, how do we decide which element to eliminate in the final step?
Is the goal to return the *value* of the last remaining number, and should it be returned as an integer?
Could 'n' ever be zero or one? What should the function return in those cases?
Are there any memory constraints I should consider given the potential value of 'n'?

Brute Force Solution

Approach

The brute force strategy for this game involves simulating the entire elimination process as described. We directly follow the rules of the game and eliminate numbers step by step until only one remains. This method guarantees finding the answer, but it might take a long time, especially for large starting numbers.

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

Start with a list of all the numbers from 1 to the starting number.
Eliminate the first number, then every other number from left to right.
Then, eliminate from right to left, again eliminating the last remaining number and then every other number.
Keep alternating between eliminating from left to right and from right to left, following the same pattern.
Continue these eliminations until only one number is left in the list.
The last remaining number is the final answer.

Code Implementation

def elimination_game_brute_force(starting_number):
    numbers_list = list(range(1, starting_number + 1))
    left_to_right = True

    while len(numbers_list) > 1:
        if left_to_right:
            # Eliminate every other number from left to right
            new_numbers_list = []
            for index in range(len(numbers_list)):
                if (index + 1) % 2 == 0:
                    new_numbers_list.append(numbers_list[index])
            numbers_list = new_numbers_list

        else:
            # Eliminate every other number from right to left
            new_numbers_list = []
            for index in range(len(numbers_list) - 1, -1, -1):
                if (len(numbers_list) - index) % 2 == 0:
                    new_numbers_list.insert(0, numbers_list[index])
            numbers_list = new_numbers_list

        # Alternate the direction of elimination
        left_to_right = not left_to_right

    # Return the last remaining number
    return numbers_list[0]

Big(O) Analysis

Time Complexity

O(n^2) – The brute force approach simulates the elimination process. Initially, we have 'n' numbers. In each round, we iterate through the list and eliminate elements. In the worst case, when alternating directions, we might iterate almost the entire remaining list in each elimination round. Since we repeat this elimination process until only one number is left, and each round potentially involves examining a significant fraction of the remaining numbers, the total number of operations can approach n * n/2. This results in a time complexity of O(n^2).

Space Complexity

O(N) – The provided solution simulates the elimination process by maintaining a list of numbers from 1 to N. This list is modified at each step as numbers are eliminated. Therefore, the auxiliary space used is directly proportional to the input number N, as we need to store all numbers from 1 to N initially. The space needed for the list grows linearly with the input size N. Thus, the space complexity is O(N).

Optimal Solution

Approach

The problem can be efficiently solved by simulating the elimination process. Instead of physically removing elements, we can track the start, direction, and step size of the eliminations until only one number remains.

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

Imagine the numbers are in a row.
Figure out which number will be the first to be eliminated in the initial round based on whether the elimination starts from left or right.
Determine the step size of the elimination. The step size doubles in each round because the interval between remaining numbers doubles.
Update the start of remaining numbers if the elimination starts from left. If it starts from right, and the number of remaining numbers is odd, the start also needs updating.
Reverse the elimination direction (from left to right or right to left) for the next round.
Halve the number of remaining numbers.
Repeat steps 2-6 until only one number remains. That number is the answer.

Code Implementation

def elimination_game(number):
    start_of_remaining = 1
    step_size = 1
    remaining_numbers = number
    left_to_right = True

    while remaining_numbers > 1:
        # If left to right, update start always.
        # If right to left and odd, update start.
        if left_to_right or remaining_numbers % 2 == 1:
            start_of_remaining += step_size

        remaining_numbers //= 2
        step_size *= 2
        left_to_right = not left_to_right

    return start_of_remaining

Big(O) Analysis

Time Complexity

O(log n) – The described solution simulates the elimination process by repeatedly halving the number of remaining elements (n). Each iteration calculates the next starting point and doubles the step size. The loop continues as long as the number of remaining elements is greater than 1. Since n is repeatedly divided by 2, the number of iterations is logarithmic with respect to n. Thus, the time complexity is O(log n).

Space Complexity

O(1) – The algorithm uses a fixed number of variables to store the start of the remaining numbers, the step size, the number of remaining numbers, and the direction of elimination. No auxiliary data structures that scale with the input size N are used. Therefore, the auxiliary space complexity is constant, or O(1).

Edge Cases

Case	How to Handle
n is zero or negative	Return 0 immediately as the elimination game is not defined for non-positive n.
n is 1	Return 1 immediately, as it is the only remaining number.
Large n leading to potential integer overflow in calculations	Use long or appropriate data type to prevent integer overflow during intermediate calculations.
n is a power of 2	The result will always be 1 for powers of 2, which should be handled correctly by iterative elimination.
n is close to the maximum possible integer value	Ensure the loop variables and intermediate calculations don't exceed integer limits by using a wider data type.
The last number standing is not 1	Iterate the elimination process until only one number remains.
Alternating elimination direction leading to inconsistent state	Ensure elimination direction is flipped correctly after each round.
Number of remaining elements is even or odd after each elimination round	The elimination logic should handle both even and odd cases correctly by updating the remaining element count.

n is zero or negative

How to Handle:

Return 0 immediately as the elimination game is not defined for non-positive n.

n is 1

How to Handle:

Return 1 immediately, as it is the only remaining number.

Large n leading to potential integer overflow in calculations

How to Handle:

Use long or appropriate data type to prevent integer overflow during intermediate calculations.

n is a power of 2

How to Handle:

The result will always be 1 for powers of 2, which should be handled correctly by iterative elimination.

n is close to the maximum possible integer value

How to Handle:

Ensure the loop variables and intermediate calculations don't exceed integer limits by using a wider data type.

The last number standing is not 1

How to Handle:

Iterate the elimination process until only one number remains.

Alternating elimination direction leading to inconsistent state

How to Handle:

Ensure elimination direction is flipped correctly after each round.

Number of remaining elements is even or odd after each elimination round

How to Handle:

The elimination logic should handle both even and odd cases correctly by updating the remaining element count.

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 value of 'n'? Is 'n' guaranteed to be a positive integer?
For the elimination rounds, if there's an odd number of elements, how do we decide which element to eliminate in the final step?
Is the goal to return the *value* of the last remaining number, and should it be returned as an integer?
Could 'n' ever be zero or one? What should the function return in those cases?
Are there any memory constraints I should consider given the potential value of 'n'?

Brute Force Solution

Approach

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

Start with a list of all the numbers from 1 to the starting number.
Eliminate the first number, then every other number from left to right.
Then, eliminate from right to left, again eliminating the last remaining number and then every other number.
Keep alternating between eliminating from left to right and from right to left, following the same pattern.
Continue these eliminations until only one number is left in the list.
The last remaining number is the final answer.

Code Implementation

def elimination_game_brute_force(starting_number):
    numbers_list = list(range(1, starting_number + 1))
    left_to_right = True

    while len(numbers_list) > 1:
        if left_to_right:
            # Eliminate every other number from left to right
            new_numbers_list = []
            for index in range(len(numbers_list)):
                if (index + 1) % 2 == 0:
                    new_numbers_list.append(numbers_list[index])
            numbers_list = new_numbers_list

        else:
            # Eliminate every other number from right to left
            new_numbers_list = []
            for index in range(len(numbers_list) - 1, -1, -1):
                if (len(numbers_list) - index) % 2 == 0:
                    new_numbers_list.insert(0, numbers_list[index])
            numbers_list = new_numbers_list

        # Alternate the direction of elimination
        left_to_right = not left_to_right

    # Return the last remaining number
    return numbers_list[0]

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

Imagine the numbers are in a row.
Figure out which number will be the first to be eliminated in the initial round based on whether the elimination starts from left or right.
Determine the step size of the elimination. The step size doubles in each round because the interval between remaining numbers doubles.
Update the start of remaining numbers if the elimination starts from left. If it starts from right, and the number of remaining numbers is odd, the start also needs updating.
Reverse the elimination direction (from left to right or right to left) for the next round.
Halve the number of remaining numbers.
Repeat steps 2-6 until only one number remains. That number is the answer.

Code Implementation

def elimination_game(number):
    start_of_remaining = 1
    step_size = 1
    remaining_numbers = number
    left_to_right = True

    while remaining_numbers > 1:
        # If left to right, update start always.
        # If right to left and odd, update start.
        if left_to_right or remaining_numbers % 2 == 1:
            start_of_remaining += step_size

        remaining_numbers //= 2
        step_size *= 2
        left_to_right = not left_to_right

    return start_of_remaining

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
n is zero or negative	Return 0 immediately as the elimination game is not defined for non-positive n.
n is 1	Return 1 immediately, as it is the only remaining number.
Large n leading to potential integer overflow in calculations	Use long or appropriate data type to prevent integer overflow during intermediate calculations.
n is a power of 2	The result will always be 1 for powers of 2, which should be handled correctly by iterative elimination.
n is close to the maximum possible integer value	Ensure the loop variables and intermediate calculations don't exceed integer limits by using a wider data type.
The last number standing is not 1	Iterate the elimination process until only one number remains.
Alternating elimination direction leading to inconsistent state	Ensure elimination direction is flipped correctly after each round.
Number of remaining elements is even or odd after each elimination round	The elimination logic should handle both even and odd cases correctly by updating the remaining element count.

n is zero or negative

How to Handle:

Return 0 immediately as the elimination game is not defined for non-positive n.

n is 1

How to Handle:

Return 1 immediately, as it is the only remaining number.

Large n leading to potential integer overflow in calculations

How to Handle:

Use long or appropriate data type to prevent integer overflow during intermediate calculations.

n is a power of 2

How to Handle:

The result will always be 1 for powers of 2, which should be handled correctly by iterative elimination.

n is close to the maximum possible integer value

How to Handle:

Ensure the loop variables and intermediate calculations don't exceed integer limits by using a wider data type.

The last number standing is not 1

How to Handle:

Iterate the elimination process until only one number remains.

Alternating elimination direction leading to inconsistent state

How to Handle:

Ensure elimination direction is flipped correctly after each round.

Number of remaining elements is even or odd after each elimination round

How to Handle:

The elimination logic should handle both even and odd cases correctly by updating the remaining element count.