# 2974. Minimum Number Game

## Description

You are given a 0-indexed integer array nums of even length and there is also an empty array arr. Alice and Bob decided to play a game where in every round Alice and Bob will do one move. The rules of the game are as follows:

• Every round, first Alice will remove the minimum element from nums, and then Bob does the same.
• Now, first Bob will append the removed element in the array arr, and then Alice does the same.
• The game continues until nums becomes empty.

Return the resulting array arr.

Example 1:

Input: nums = [5,4,2,3]
Output: [3,2,5,4]
Explanation: In round one, first Alice removes 2 and then Bob removes 3. Then in arr firstly Bob appends 3 and then Alice appends 2. So arr = [3,2].
At the begining of round two, nums = [5,4]. Now, first Alice removes 4 and then Bob removes 5. Then both append in arr which becomes [3,2,5,4].


Example 2:

Input: nums = [2,5]
Output: [5,2]
Explanation: In round one, first Alice removes 2 and then Bob removes 5. Then in arr firstly Bob appends and then Alice appends. So arr = [5,2].


Constraints:

• 1 <= nums.length <= 100
• 1 <= nums[i] <= 100
• nums.length % 2 == 0

## Solutions

Solution 1: Simulation + Priority Queue (Min Heap)

We can put the elements in the array $nums$ into a min heap one by one, and each time take out two elements $a$ and $b$ from the min heap, then put $b$ and $a$ into the answer array in turn, until the min heap is empty.

Time complexity is $O(n \times \log n)$, and space complexity is $O(n)$. Where $n$ is the length of the array $nums$.

Solution 2: Sorting + Swapping

We can sort the array $nums$, and then swap the positions of every two adjacent elements in sequence to get the answer array.

The time complexity is $O(n \times \log n)$, and the space complexity is $O(\log n)$. Where $n$ is the length of the array $nums$.

• class Solution {
public int[] numberGame(int[] nums) {
PriorityQueue<Integer> pq = new PriorityQueue<>();
for (int x : nums) {
pq.offer(x);
}
int[] ans = new int[nums.length];
int i = 0;
while (!pq.isEmpty()) {
int a = pq.poll();
ans[i++] = pq.poll();
ans[i++] = a;
}
return ans;
}
}

• class Solution {
public:
vector<int> numberGame(vector<int>& nums) {
priority_queue<int, vector<int>, greater<int>> pq;
for (int x : nums) {
pq.push(x);
}
vector<int> ans;
while (pq.size()) {
int a = pq.top();
pq.pop();
int b = pq.top();
pq.pop();
ans.push_back(b);
ans.push_back(a);
}
return ans;
}
};

• class Solution:
def numberGame(self, nums: List[int]) -> List[int]:
heapify(nums)
ans = []
while nums:
a, b = heappop(nums), heappop(nums)
ans.append(b)
ans.append(a)
return ans


• func numberGame(nums []int) (ans []int) {
pq := &hp{nums}
heap.Init(pq)
for pq.Len() > 0 {
a := heap.Pop(pq).(int)
b := heap.Pop(pq).(int)
ans = append(ans, b)
ans = append(ans, a)
}
return
}

type hp struct{ sort.IntSlice }

func (h *hp) Less(i, j int) bool { return h.IntSlice[i] < h.IntSlice[j] }
func (h *hp) Pop() interface{} {
old := h.IntSlice
n := len(old)
x := old[n-1]
h.IntSlice = old[0 : n-1]
return x
}
func (h *hp) Push(x interface{}) {
h.IntSlice = append(h.IntSlice, x.(int))
}

• function numberGame(nums: number[]): number[] {
const pq = new MinPriorityQueue();
for (const x of nums) {
pq.enqueue(x);
}
const ans: number[] = [];
while (pq.size()) {
const a = pq.dequeue().element;
const b = pq.dequeue().element;
ans.push(b, a);
}
return ans;
}


• use std::collections::BinaryHeap;
use std::cmp::Reverse;

impl Solution {
pub fn number_game(nums: Vec<i32>) -> Vec<i32> {
let mut pq = BinaryHeap::new();

for &x in &nums {
pq.push(Reverse(x));
}

let mut ans = Vec::new();

while let Some(Reverse(a)) = pq.pop() {
if let Some(Reverse(b)) = pq.pop() {
ans.push(b);
ans.push(a);
}
}

ans
}
}