2974. Minimum Number Game

Problem Description

In this problem, we have a simple game involving two players, Alice and Bob, and an integer array nums that has an even number of elements. The game follows a specific order of operations in each round:

  1. Alice starts by removing the smallest element from the nums array.
  2. Bob follows by also removing the now smallest element from the nums array.
  3. Bob then adds the element he removed to the array, arr.
  4. Alice adds the element she removed to the array, arr.

This process repeats until the nums array is empty. The task is to simulate this game and return the final state of the arr array.

To solve this problem, you need to think about how to efficiently track and remove the smallest elements and in which order they should be added to the arr array.


The core of the solution lies in consistently selecting the two smallest elements from the nums array with each round. To do this efficiently, we use a data structure called a min-heap, which allows us to access and remove the smallest element in an optimally fast manner.

The intuition behind using a min-heap is that it maintains the elements of nums in a way that we can always quickly extract the minimum element. After initializing the heap with the elements of nums, we repeatedly perform the following actions:

  1. Remove the smallest element a (Alice's move).
  2. Remove the next smallest element b (Bob's move).
  3. Add element b to the arr array first (Bob's addition to arr).
  4. Add element a to the arr array second (Alice's addition to arr).

This effectively simulates the game's rules, and we continue this process until the nums array (min-heap) is empty. The resulting arr array will be the final output of the algorithm, which is the order in which elements were appended to arr by Bob and Alice according to the game's rules.

Learn more about Sorting and Heap (Priority Queue) patterns.

Solution Approach

The solution to this problem is implemented using a priority queue, which is a type of data structure that always gives priority to the element with the least value.

In Python, the priority queue can easily be implemented using the heapq module, which turns a regular list into a min-heap. Below is a step-by-step breakdown of the solution approach:

  1. Heapify the Array: We start by turning the original array nums into a min-heap using the heapify function from the heapq module. The heapify process reorders the array into a heap, so that heappop() can be used in the subsequent steps to always get the smallest item efficiently.

  2. Initialize the Result Array: We define an empty list called ans, which will serve as the arr array where Alice and Bob will append their removed elements based on the game's rules.

  3. Simulate the Game Rounds: a. In a loop that runs as long as there are elements in nums, we simulate a single round by performing two heappop operations to get the two smallest elements from nums, a and b. Here, a will be the smallest element and b will be the second smallest. b. We append b to the ans list first (since Bob appends his choice to arr first), and then append a.

  4. Final Output: After the loop ends (when nums is empty), we return the ans list. This list is now the final state of the arr array after all rounds of the game have been played according to the rules specified in the problem description.

Here are the core lines of code that correspond to each step enclosed in "`":

  • Convert nums into a min-heap: heapify(nums)
  • Loop until nums is empty, simulating the game rounds:
1while nums:
2    a, b = heappop(nums), heappop(nums)
3    ans.append(b)
4    ans.append(a)
  • Return the result: return ans

In conclusion, the use of a min-heap is crucial because it allows us to simulate the rules of the game in an efficient way since we are repeatedly needing to access and eliminate the minimum elements from the array.

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Example Walkthrough

Let's illustrate how this solution approach works with a small example. Assume we have the following integer array, nums:

1[3, 1, 4, 2]

First, we convert nums into a min-heap using heapify(nums). For our small-sized array, the heapify process might not change the arrangement much, but for a larger dataset, the heap would significantly help in extracting the minimal elements efficiently.

Now our array, represented as a min-heap, looks like this:

1[1, 2, 4, 3]

Here's how we simulate the game rounds:

  1. heappop(nums) removes 1 (Alice's move) and the heap now looks like [2, 3, 4].
  2. heappop(nums) again removes 2 (Bob's move), resulting in a heap of [3, 4].
  3. We append 2 to ans first (Bob's addition to arr), so ans now looks like [2].
  4. We append 1 to ans (Alice's addition to arr), making ans look like [2, 1].

The nums heap after the first round is now [3, 4]. We repeat the process:

  1. heappop(nums) removes 3 (Alice's move), and the heap is [4].
  2. heappop(nums) removes the last element 4 (Bob's move), and the heap is now empty.
  3. We append 4 to ans (Bob's addition to arr), so ans now looks like [2, 1, 4].
  4. Finally, we append 3 to ans (Alice's addition to arr), resulting in ans being [2, 1, 4, 3].

The nums array is now empty, so our process is complete. The final state of the arr array returned by our function is [2, 1, 4, 3].

In this example, we followed the steps of the solution approach exactly, prioritizing the removal of the smallest elements and appending them to the result array ans in the order specified by the game's rules. The use of a min-heap made the process of selecting the smallest elements straightforward and efficient.

Solution Implementation

1from heapq import heapify, heappop # Import necessary functions from the heapq module
3class Solution:
4    def numberGame(self, numbers):
5        """
6        Reorder numbers from a min-heap such that each element from the heap
7        is appended in alternating order to a new list.
9        :param numbers: List[int]
10        :return: List[int]
11        """
12        # Convert the list into a heap in place
13        heapify(numbers)
15        # Initialize an empty list to hold the reordered elements
16        reordered_numbers = []
18        # Continue until the heap is empty
19        while numbers:
20            # Pop the two smallest elements from the heap.
21            # Note that the list must have an even number of elements,
22            # otherwise, the second pop operation would fail when the list becomes empty.
23            first_num = heappop(numbers)
24            second_num = heappop(numbers)
26            # Append the numbers to the reordered list in alternating order
27            reordered_numbers.append(second_num)
28            reordered_numbers.append(first_num)
30        # Return the list with elements reordered
31        return reordered_numbers
1import java.util.PriorityQueue;
3class Solution {
5    // Method to reorder the numbers in a specific pattern.
6    public int[] numberGame(int[] numbers) {
7        // Create a min-heap priority queue to order numbers in increasing order.
8        PriorityQueue<Integer> minHeap = new PriorityQueue<>();
10        // Add all numbers in the input array into the priority queue.
11        for (int num : numbers) {
12            minHeap.offer(num);
13        }
15        // Create an array to store the answer.
16        int[] answer = new int[numbers.length];
17        // Initialize an index counter to zero.
18        int index = 0;
20        // Continue this process until the priority queue is empty.
21        while (!minHeap.isEmpty()) {
22            // Poll (remove) the smallest element from the queue.
23            int first = minHeap.poll();
24            // Check if there's another number to pair with the polled number.
25            if (!minHeap.isEmpty()) {
26                // Poll the next smallest element to pair with the first.
27                int second = minHeap.poll();
28                // Place the second item first in the answer array.
29                answer[index++] = second;
30            }
31            // Place the first item after the second item in the answer array.
32            answer[index++] = first;
33        }
35        // Return the re-ordered array as the answer.
36        return answer;
37    }
1#include <vector>
2#include <queue> // Required for priority_queue
3using namespace std;
5class Solution {
7    // Renamed the method to reflect standard naming practices and added comments
8    vector<int> numberGame(vector<int>& nums) {
9        // Use a min-heap to store the numbers in increasing order
10        priority_queue<int, vector<int>, greater<int>> minHeap;
12        // Add all numbers from the input vector to the min-heap
13        for (int num : nums) {
14            minHeap.push(num);
15        }
17        // Prepare an answer vector to store the results
18        vector<int> answer;
20        // Iterate until the min-heap is empty
21        while (!minHeap.empty()) {
22            int smaller = minHeap.top(); // Get the smallest number
23            minHeap.pop(); // Remove the smallest number from the min-heap
25            // Since the problem statement is not clear about what should be done if there is
26            // an odd number of elements in the heap, it's important to check if the heap is
27            // not empty before trying to access the top element again.
28            if (!minHeap.empty()) {
29                int larger = minHeap.top(); // Get the next smallest number
30                minHeap.pop(); // Remove it from the min-heap
32                answer.push_back(larger); // Add the larger of the two to the answer
33                answer.push_back(smaller); // Then add the smaller
34            } else {
35                // If there is an odd number of elements, the last element won't have a pair.
36                // Assuming we're to add it directly to the answer, as per the original code,
37                // which would otherwise throw an error if not handled.
38                answer.push_back(smaller);
39            }
40        }
42        // Return the final populated answer vector
43        return answer;
44    }
1// Import the MinPriorityQueue class from the 'typescript-collections' library.
2// Note: you must have this library installed for this code to work.
3import { MinPriorityQueue } from 'typescript-collections';
5// Takes in an array of numbers and arranges them in a specific order.
6// The smallest two numbers are dequeued from the priority queue; 
7// the second-dequeued (larger) number is placed before the first-dequeued (smaller) number in the output array.
8// This process repeats until the priority queue is empty.
9// @param nums - The array of numbers to be processed.
10// @returns An array of numbers arranged according to the specific game rule.
11function numberGame(nums: number[]): number[] {
12    // Create a new minimum priority queue to store the numbers.
13    const priorityQueue = new MinPriorityQueue<number>();
15    // Add each number to the priority queue.
16    for (const num of nums) {
17        priorityQueue.enqueue(num);
18    }
20    // Initialize the array that will hold the re-arranged numbers.
21    const result: number[] = [];
23    // Continue processing until the priority queue is empty.
24    while (!priorityQueue.isEmpty()) {
25        // Dequeue the smallest number.
26        const smallerNumber = priorityQueue.dequeue().element;
28        // Check if there is another number to pair with the one dequeued.
29        // If the queue is empty, push the last number into the result and break the loop.
30        if (priorityQueue.isEmpty()) {
31            result.push(smallerNumber);
32            break;
33        }
35        // Dequeue the next smallest number.
36        const largerNumber = priorityQueue.dequeue().element;
38        // Add the second (larger) number before the first (smaller) one into the result array.
39        result.push(largerNumber, smallerNumber);
40    }
42    // Return the re-arranged array of numbers.
43    return result;

Time and Space Complexity

The time complexity of the code provided operates in two primary steps, heapification of the nums list and the subsequent pop operations. Heapification is completed in O(n) time. After heapification, the while loop pops two elements for every iteration until the heap is empty. Since there are n/2 iterations and each pop operation from a heap is O(log n), the cumulative time complexity for the pop operations is O(n/2 * 2 * log n) = O(n * log n). Therefore, the overall time complexity of the code is O(n + n * log n), which simplifies to O(n * log n) as n * log n dominates n.

The space complexity of the code is largely governed by the ans array that at most will contain n elements (if all elements from the input list nums are added to it). The in-place heapify operation does not consume additional space proportional to the input. Thus, the space complexity is O(n).

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

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Question 1 out of 10

What's the output of running the following function using the following tree as input?

1def serialize(root):
2    res = []
3    def dfs(root):
4        if not root:
5            res.append('x')
6            return
7        res.append(root.val)
8        dfs(root.left)
9        dfs(root.right)
10    dfs(root)
11    return ' '.join(res)
1import java.util.StringJoiner;
3public static String serialize(Node root) {
4    StringJoiner res = new StringJoiner(" ");
5    serializeDFS(root, res);
6    return res.toString();
9private static void serializeDFS(Node root, StringJoiner result) {
10    if (root == null) {
11        result.add("x");
12        return;
13    }
14    result.add(Integer.toString(root.val));
15    serializeDFS(root.left, result);
16    serializeDFS(root.right, result);
1function serialize(root) {
2    let res = [];
3    serialize_dfs(root, res);
4    return res.join(" ");
7function serialize_dfs(root, res) {
8    if (!root) {
9        res.push("x");
10        return;
11    }
12    res.push(root.val);
13    serialize_dfs(root.left, res);
14    serialize_dfs(root.right, res);

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