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Min Max Game LeetCode Solution

Problem – Min Max Game LeetCode Solution

You are given a 0-indexed integer array nums whose length is a power of 2.

Apply the following algorithm on nums:

  1. Let n be the length of nums. If n == 1end the process. Otherwise, create a new 0-indexed integer array newNums of length n / 2.
  2. For every even index i where 0 <= i < n / 2assign the value of newNums[i] as min(nums[2 * i], nums[2 * i + 1]).
  3. For every odd index i where 0 <= i < n / 2assign the value of newNums[i] as max(nums[2 * i], nums[2 * i + 1]).
  4. Replace the array nums with newNums.
  5. Repeat the entire process starting from step 1.

Return the last number that remains in nums after applying the algorithm.

Example 1:

Input: nums = [1,3,5,2,4,8,2,2]
Output: 1
Explanation: The following arrays are the results of applying the algorithm repeatedly.
First: nums = [1,5,4,2]
Second: nums = [1,4]
Third: nums = [1]
1 is the last remaining number, so we return 1.

Example 2:

Input: nums = [3]
Output: 3
Explanation: 3 is already the last remaining number, so we return 3.


  • 1 <= nums.length <= 1024
  • 1 <= nums[i] <= 109
  • nums.length is a power of 2.

Min Max Game LeetCode Solution in Python

#base case
if len(nums)==1:
            return nums[0]

#as long as my length doesn't become 1, I repeat the process
while len(nums)!=1:
	#we take a newnums everytime of size n//2
	newnums = [-1]*(len(nums)//2)
	for i in range(0,len(nums)//2):
		if i%2==0:
			newnums[i] = min(nums[2 * i], nums[2 * i + 1])
			newnums[i] = max(nums[2 * i], nums[2 * i + 1])
	#add the newarray to the older one.
	nums = newnums

return nums[0]

Min Max Game LeetCode Solution in Java

public int minMaxGame(int[] nums) {
        List<Integer> list = new ArrayList<>();
        for(int i:nums) list.add(i);
            List<Integer> dummy = new ArrayList<>();
            boolean f = true;
            for(int i=1;i<list.size();i+=2){
                if(f)  dummy.add(Math.min(list.get(i-1),list.get(i)));
                else  dummy.add(Math.max(list.get(i-1),list.get(i)));
                f = f ^ true;
            list = dummy;
        return list.get(0);

Min Max Game LeetCode Solution in C++

class Solution {
    vector<int> helper(vector<int>& nums,int n){
        vector<int> ans;
        for(int i=0;i<n/2;i++){
            if(i%2 == 0) ans.emplace_back(min(nums[2*i],nums[2*i+1]));
            else ans.emplace_back(max(nums[2 * i], nums[2 * i + 1]));
        return ans;
    int minMaxGame(vector<int>& nums) {
        int n=size(nums);
        if(n == 1) return nums[0];
           vector<int> arr = helper(nums,n);
            n = arr.size();
            nums = arr;
            if(n == 1) return  arr[0];
        return 0;
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