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Leetcode 2293. Min Max Game

Problem Statement

Leetcode problem 2293: Min Max Game

You are given an integer array nums. Our goal is to repeatedly perform the following operation until we have only one number left in the array:

1. Create a new integer array newNums.
2. For i in the range of 0 to [n/2 - 1], where n is the current length of nums, replace newNums[i] with:
   • min(nums[2 * i], nums[2 * i + 1]) if i is even.
   • max(nums[2 * i], nums[2 * i + 1]) if i is odd.

Return the only number left in the array after performing the operations repeatedly.

Clarifying Questions

1. Length of array: Can the length of the array be any positive integer, or is it constrained to be even?
   • Answer: Assume the length of the array is always a power of 2.
2. Range of values: Are there any constraints on the values within the array?
   • Answer: Assume elements of the array are integers within a typical range for array values.

Strategy

1. Initialization: Start with the given array nums.
2. Iterative Reduction: Repeat the process of constructing newNums from nums until the array length becomes 1.
3. Array Definition: For each step, iterate over half the length of the current nums array to build the newNums array.
4. Conditional Assignment: For each index, check if it’s even or odd and place the minimum or maximum of pairs in newNums.
5. Update State: Replace nums with newNums and repeat until a single element remains.

Code

#include <vector>
#include <algorithm>

int minMaxGame(std::vector<int>& nums) {
    while (nums.size() > 1) {
        std::vector<int> newNums(nums.size() / 2);
        for (int i = 0; i < newNums.size(); ++i) {
            if (i % 2 == 0) {
                newNums[i] = std::min(nums[2 * i], nums[2 * i + 1]);
            } else {
                newNums[i] = std::max(nums[2 * i], nums[2 * i + 1]);
            }
        }
        nums = newNums;
    }
    return nums[0];
}

Time Complexity

• Iteration Step: In each iteration, the size of nums is halved, resulting in ((n/2, n/4, …, 1)).
• Work Done per Iteration: Within each iteration, a complete pass over the elements is required ((O(n))).
• Total Iterations: Since the array size is reduced exponentially, there are (log_2(n)) iterations.

Combining these, the overall time complexity is:

[ O(n) \text{ work per iteration} \times O(\log n) \text{ iterations} = O(n \log n) ]

This approach ensures efficient handling of the problem within the constraints.

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