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Formatted question description: https://leetcode.ca/all/2293.html

# 2293. Min Max Game

• Difficulty: Easy.
• Related Topics: Array, Simulation.
• Similar Questions: Elimination Game, Find Triangular Sum of an Array.

## Problem

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

Apply the following algorithm on nums:

• Let n be the length of nums. If n == 1, end the process. Otherwise, create a new 0-indexed integer array newNums of length n / 2.

• For every even index i where 0 <= i < n / 2, assign the value of newNums[i] as min(nums[2 * i], nums[2 * i + 1]).

• For every odd index i where 0 <= i < n / 2, assign the value of newNums[i] as max(nums[2 * i], nums[2 * i + 1]).

• Replace the array nums with newNums.

• 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.


Constraints:

• 1 <= nums.length <= 1024

• 1 <= nums[i] <= 109

• nums.length is a power of 2.

## Solution (Java, C++, Python)

• class Solution {
public int minMaxGame(int[] nums) {
int n = nums.length;
if (n == 1) {
return nums[0];
}
int[] newNums = new int[n / 2];
for (int i = 0; i < n / 2; i++) {
if (i % 2 == 0) {
newNums[i] = Math.min(nums[2 * i], nums[2 * i + 1]);
} else {
newNums[i] = Math.max(nums[2 * i], nums[2 * i + 1]);
}
}
return minMaxGame(newNums);
}
}

############

class Solution {
public int minMaxGame(int[] nums) {
for (int n = nums.length; n > 1;) {
n >>= 1;
for (int i = 0; i < n; ++i) {
int a = nums[i << 1], b = nums[i << 1 | 1];
nums[i] = i % 2 == 0 ? Math.min(a, b) : Math.max(a, b);
}
}
return nums[0];
}
}

• class Solution:
def minMaxGame(self, nums: List[int]) -> int:
n = len(nums)
while n > 1:
n >>= 1
for i in range(n):
a, b = nums[i << 1], nums[i << 1 | 1]
nums[i] = min(a, b) if i % 2 == 0 else max(a, b)
return nums[0]

############

# 2293. Min Max Game
# https://leetcode.com/problems/min-max-game/

class Solution:
def minMaxGame(self, nums: List[int]) -> int:

while len(nums) != 1:
temp = []

for i in range(len(nums) // 2):
if i % 2 == 0:
temp.append(min(nums[i * 2], nums[i * 2 + 1]))
else:
temp.append(max(nums[i * 2], nums[i * 2 + 1]))

nums = temp

return nums[0]


• class Solution {
public:
int minMaxGame(vector<int>& nums) {
for (int n = nums.size(); n > 1;) {
n >>= 1;
for (int i = 0; i < n; ++i) {
int a = nums[i << 1], b = nums[i << 1 | 1];
nums[i] = i % 2 == 0 ? min(a, b) : max(a, b);
}
}
return nums[0];
}
};

• func minMaxGame(nums []int) int {
for n := len(nums); n > 1; {
n >>= 1
for i := 0; i < n; i++ {
a, b := nums[i<<1], nums[i<<1|1]
if i%2 == 0 {
nums[i] = min(a, b)
} else {
nums[i] = max(a, b)
}
}
}
return nums[0]
}

func min(a, b int) int {
if a < b {
return a
}
return b
}

func max(a, b int) int {
if a > b {
return a
}
return b
}

• function minMaxGame(nums: number[]): number {
for (let n = nums.length; n > 1; ) {
n >>= 1;
for (let i = 0; i < n; ++i) {
const a = nums[i << 1];
const b = nums[(i << 1) | 1];
nums[i] = i % 2 == 0 ? Math.min(a, b) : Math.max(a, b);
}
}
return nums[0];
}


• impl Solution {
pub fn min_max_game(mut nums: Vec<i32>) -> i32 {
let mut n = nums.len();
while n != 1 {
n >>= 1;
for i in 0..n {
nums[i] = (if i & 1 == 1 {
i32::max
} else {
i32::min
})(nums[i << 1], nums[i << 1 | 1])
}
}
nums[0]
}
}



Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).