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#2293 - Min Max Game
Problem Description
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.
Solution
/**
* @param {number[]} nums
* @return {number}
*/
var minMaxGame = function(nums) {
let array = nums.slice();
while (array.length > 1) {
const newArray = new Array(array.length / 2);
for (let i = 0; i < newArray.length; i++) {
if (i % 2 === 0) {
newArray[i] = Math.min(array[2 * i], array[2 * i + 1]);
} else {
newArray[i] = Math.max(array[2 * i], array[2 * i + 1]);
}
}
array = newArray;
}
return array[0];
};