2293. Min Max Game

Problem Description

You have an array of integers, nums, whose length is guaranteed to be a power of 2. A specific algorithm needs to be applied to this array until only one number is left. The steps of the algorithm are as follows:

1. If the length of nums is 1, stop.
2. Create a new array named newNums of length n / 2, where n is the current length of nums.
3. Fill newNums by iterating through nums in pairs. For every even index i, set newNums[i] to the smaller number in the pair from nums. For every odd index i, set newNums[i] to the larger number in the pair.
4. Replace nums with newNums.
5. Repeat the process starting from step 1.

The goal is to return the single remaining number after repeating this process.

Intuition

Understanding the algorithm is critical. You repeatedly shrink the nums array by half until it consists of a single element. Each iteration consists of pairwise comparisons. For even-indexed pairs, you select the smaller number, and for odd-indexed pairs, you select the larger number.

The approach involves:

1. Using a loop to reduce the array size until one element is left.
2. Iterating over the current nums to create the newNums based on the even and odd index rule.
3. Overwriting nums with newNums after each iteration.

The transformation defined by the algorithm ensures that after each iteration, the size of nums is halved. The bitwise operations (n >>= 1, i << 1, i << 1 | 1) in the solution are efficient ways to perform calculations needed for indexing and updating the array. The expression n >>= 1 halves n, i << 1 doubles i (used to find the base index of a pair), and i << 1 | 1 finds the immediate next index (the second element of the pair).

By representing pairs of elements as a and b, the code is more readable and easier to work with. The condition if i % 2 == 0 checks if the index i is even, using the smaller value of the pair, otherwise using the larger value for odd indices.

This algorithmic process is inherently divide-and-conquer in nature, breaking down the problem into smaller pieces and then combining the results. Since the length of the array is a power of 2, we are assured that we will end with exactly one element.

Solution Approach

The solution uses a simple while loop to repeatedly apply the steps of the algorithm as follows:

• The condition for the while loop while n > 1 checks whether there is more than one element left in nums. If only one element is left, we have reached the end of the process.
• Inside the loop, the operation n >>= 1 efficiently halves n. This operation is equivalent to n = n // 2 but is quicker since it is a bitwise operation.
• A for loop for i in range(n) is then used to iterate through the first n elements of the array and perform the pairwise comparison and assignment to newNums. Instead of explicitly creating a newNums array, the solution saves space by reusing the first half of the original nums array to store the new values.
• The variables a and b represent the elements of the current pair being considered, where a is at the index i << 1 (double the index i) and b is at the index i << 1 | 1 (one more than double the index i). The use of bitwise operations makes the index calculation efficient.
• The condition if i % 2 == 0 checks if the index i is even, and if so, the code uses min(a, b) to select the smaller of the two elements. If i is odd, the code uses max(a, b) to select the larger of the two elements.
• After iterating through all pairs, the new values are already stored in the first half of nums, and we can repeat the process until only one element remains.

There are no additional data structures used in the implementation aside from basic variables and the input array, which highlights the in-place nature of this solution. This implementation benefits from space efficiency by not needing extra arrays. The solution emphasizes an understanding of bitwise manipulation (>>, <<, |) to perform operations on indices, a common pattern for algorithms with a binary nature or when dealing with arrays where the size is a power of 2.

The idea of the solution is quite straightforward once you understand how the array evolves after each round of the algorithm, and no complex data structures or algorithms are needed aside from a couple of loops and simple conditions.

Example Walkthrough

Let's walk through a hypothetical example to better understand the solution approach.

Suppose the input array is [3, 1, 4, 2, 8, 6, 5, 7]. The length of this array is 8, which is a power of 2 (2^3).

Here's how we apply the algorithm step by step:

Iteration 1:

• The current length of nums is 8. We need to create a newNums array of length 8 / 2 = 4.
• We iterate through nums in pairs: (3, 1), (4, 2), (8, 6), (5, 7).
• We fill newNums by selecting the smaller number for the first pair and the larger number for the second pair: newNums = [1, 4, 8, 7].

Iteration 2:

• The reduced nums is now [1, 4, 8, 7].
• We halve the length again for newNums, which will now have a length of 2.
• Iterating in pairs: (1, 4), (8, 7).
• Following the even-odd index rule, newNums becomes: [1, 8].

Iteration 3:

• Now nums is [1, 8].
• We halve the length again, and since it will be 1, this will be the final iteration.
• There's only one pair: (1, 8).
• Following the even-odd index rule, newNums ends up as just [1].

After the third iteration, nums is reduced to a single element array [1]. The algorithm stops here as the length of nums is 1. The remaining single number, 1, is the answer and will be returned by the algorithm.

In summary, the sequence of nums arrays through the iterations would look like this:

1. [3, 1, 4, 2, 8, 6, 5, 7]
2. [1, 4, 8, 7]
3. [1, 8]
4. [1]

The result, which is the last remaining number after all reductions, is 1.

Solution Implementation

Time and Space Complexity

The given code performs a series of actions to transform an initial list of numbers into just one number according to the rules of the "MinMax Game". For each round, the number of elements in the list is halved, and the new elements are either the minimum or maximum of pairs of the original elements, depending on their position.

Time Complexity:

The time complexity of this code is analyzed based on the number of operations needed to reach the final number. Initially, there are n elements, and in each subsequent round, the number of elements is halved. This halving continues until there is just one element left. The number of rounds is equal to log2(n) since we are halving the element count successively.

In each round, the operations performed are constant-time comparisons and assignments. For the first round, there are n/2 operations, for the second round n/4, then n/8, and so on, until we perform just one operation. When you sum these operations, it forms a geometric series: n/2 + n/4 + n/8 + ... + 1. The sum of this geometric series converges to n, which means the total number of operations will be O(n).

Space Complexity:

The space complexity is determined by the amount of extra space used. The given code modifies the input list nums in place, and does not use any additional space that grows with the input size. The space used for variables is constant, and does not depend on the size of the input.

Hence, the space complexity is O(1), indicating constant space complexity.

Learn more about how to find time and space complexity quickly using problem constraints.