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Leetcode 2293. Min Max Game

Problem Statement

You are given a 0-indexed integer array nums whose length is a power of 2.

Apply the following algorithm repeatedly until there is only one element remaining:

  1. Let n be the length of nums. If n == 1, end the process. Otherwise, create a new 0-indexed integer array newNums of length n / 2.
  2. For every even index i where 0 <= i < n / 2, assign the value of newNums[i] as min(nums[2 * i], nums[2 * i + 1]).
  3. For every odd index i where 0 <= i < n / 2, assign the value of newNums[i] as max(nums[2 * i], nums[2 * i + 1]).
  4. Replace the array nums with newNums.
  5. Repeat the entire process starting from step 1.

Return the last number that remains in nums.

Clarifying Questions

  1. Q: What is the length range of the input array nums? A: The length of the array will always be a power of 2, and within a reasonable range for typical array operations.
  2. Q: Can the elements of nums be negative? A: Yes, the elements of nums can be any integers, both negative and positive.
  3. Q: When constructing newNums, how should we handle an element if i is neither even nor odd? A: Each index i will always be either even or odd by definition of integers.

Strategy

  1. Base Case: If the length of nums is 1, return nums[0].
  2. Iterate through the array in steps of 2 to form the newNums array.
    • For every even index in newNums, take the minimum of the corresponding two elements from nums.
    • For every odd index in newNums, take the maximum of the corresponding two elements from nums.
  3. Replace nums with newNums and repeat until nums has only one element.

Code

class Solution {
    public int minMaxGame(int[] nums) {
        while (nums.length > 1) {
            int n = nums.length;
            int[] newNums = new int[n / 2];
            for (int i = 0; i < n / 2; i++) {
                if (i % 2 == 0) {
                    newNums[i] = Math.min(nums[2 * i], nums[2 * i + 1]);
                } else {
                    newNums[i] = Math.max(nums[2 * i], nums[2 * i + 1]);
                }
            }
            nums = newNums;
        }
        return nums[0];
    }
}

Time Complexity

The time complexity of this algorithm is O(n), where n is the length of the initial nums array:

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