algoadvance

In this problem, you are given a 0-indexed integer array nums whose length is a power of 2.

Apply the following algorithm on nums repeatedly until the length of nums becomes 1:

  1. Create a new 0-indexed integer array newNums of length nums.length / 2.
  2. For every even index i where 0 <= i < nums.length / 2, assign the value of newNums[i] as:
    • newNums[i] = min(nums[2 * i], nums[2 * i + 1]) if i is even.
    • newNums[i] = max(nums[2 * i], nums[2 * i + 1]) if i is odd.
  3. Replace the array nums with newNums.
  4. Repeat the entire process starting from step 1.

Return the last number that remains in nums.

Clarifying Questions

  1. Can nums contain negative numbers or is it only non-negative integers?
  2. Is there an upper limit to the size of the input array?
  3. Are there any constraints on the values within the nums array?

Clarifications:

  1. Yes, nums can contain negative numbers.
  2. The size of nums is guaranteed to be a power of 2, but no specific upper limit is provided.
  3. There are no constraints on the values of the numbers within nums.

Code

Let’s proceed to implement this in Python.

def minMaxGame(nums):
    while len(nums) > 1:
        newNums = []
        for i in range(len(nums) // 2):
            if i % 2 == 0:
                newNums.append(min(nums[2 * i], nums[2 * i + 1]))
            else:
                newNums.append(max(nums[2 * i], nums[2 * i + 1]))
        nums = newNums
    return nums[0]

# Test the function with an example input
print(minMaxGame([1, 3, 5, 2, 4, 8, 2, 2]))  # Expected output depends on the input sequence.

Strategy

  1. Initialization: Start with the array nums.
  2. Iteration until single element:
    • Create a new array called newNums at each iteration.
    • For each pair of elements in nums:
      1. If the index is even, take the minimum of the pair.
      2. If the index is odd, take the maximum of the pair.
    • Replace nums with newNums.
  3. Termination: When the length of nums becomes 1, return the single element in nums.

Time Complexity

Thus, the overall time complexity is O(n log n).

This approach effectively reduces the problem size at each step, ensuring efficient execution even for larger inputs.

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