Welcome to Subscribe On Youtube

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 $nums$.

• Java
• C++
• Python
• Go
• TypeScript
• RenderScript

• 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
      }
  }
  
  

All Problems

All Solutions