Leetcode 2293: Min Max Game
You are given an integer array nums where the length of the array is a power of 2. The task is to apply a repeated algorithm to reduce the array size. In each iteration, if the index of an element is even, the new value at that index will be the minimum of two adjacent elements; if the index is odd, the new value will be the maximum. The process continues until only one element remains, which is the result.
Problem
Approach
Steps
Complexity
Input: You are given an array nums of integers where the length is a power of 2.
Example: Input: nums = [2, 4, 6, 8, 10, 12, 14, 16]
Constraints:
• 1 <= nums.length <= 1024
• nums.length is a power of 2
• 1 <= nums[i] <= 10^9
Output: Return the last number that remains in the array after applying the reduction algorithm repeatedly.
Example: Output: 8
Constraints:
Goal: The goal is to reduce the array by repeatedly applying the described transformation. In each iteration, alternate positions are filled with the minimum or maximum values of adjacent elements.
Steps:
• Step 1: Begin with the full array.
• Step 2: Create a new array by replacing elements at even indices with the minimum of two adjacent elements and elements at odd indices with the maximum.
• Step 3: Replace the original array with the new array and repeat the process.
• Step 4: Continue until only one element remains in the array, which is returned as the result.
Goal: The array length will always be a power of 2. The integer values in the array are between 1 and 10^9.
Steps:
Assumptions:
• The length of the array is guaranteed to be a power of 2.
• The integer values in the array are valid and within the specified range.
• Input: Input: nums = [1, 3, 5, 7]
• Explanation: In the first iteration: newNums = [min(1, 3), max(5, 7)] = [1, 7]. Then, the next iteration gives newNums = [1]. The last remaining element is 1.
• Input: Input: nums = [2, 4, 6, 8, 10, 12, 14, 16]
• Explanation: After the first iteration, nums = [2, 6, 10, 14], and in subsequent iterations, nums = [2, 10], and finally, nums = [8].
Approach: The approach involves recursively applying the reduction algorithm to the array until only one element remains. In each iteration, you need to build a new array based on the minimum and maximum values of adjacent pairs.
Observations:
• The array's size reduces by half in each iteration.
• The process continues recursively until only one element is left.
• The problem can be solved efficiently by using a recursive approach to apply the transformation on smaller arrays.
Steps:
• Step 1: Start with the full array and check if the length is 1 (base case).
• Step 2: If the array has more than one element, create a new array where even indices are filled with the minimum of adjacent elements and odd indices with the maximum.
• Step 3: Recursively apply this process until only one element is left.
Empty Inputs:
• The input will never be empty, as the length of the array is guaranteed to be at least 1.
Large Inputs:
• For large arrays (up to size 1024), ensure that the algorithm efficiently reduces the array size in each iteration.
Special Values:
• Arrays with identical values will always return that value as the result.
Constraints:
• The array will always be a power of 2, and the number of iterations will be logarithmic with respect to the array size.
int minMaxGame(vector<int>& nums) {
int n = nums.size();
if(n==1) return nums[0]; // Base case
vector<int> newNum(n/2);
for(int i=0; i<n/2; i++) {
if(i%2==0) newNum[i] = min(nums[2 * i], nums[2 * i + 1]);
else newNum[i] = max(nums[2 * i], nums[2 * i + 1]);
}
int res = minMaxGame(newNum); // Recursive call
return res;
}
1 : Function Declaration
int minMaxGame(vector<int>& nums) {
The function `minMaxGame` is declared, which takes a reference to a vector of integers `nums` and returns an integer. This function aims to reduce the input vector based on alternating min and max operations.
2 : Get Array Size
int n = nums.size();
The size of the input vector `nums` is calculated and stored in the variable `n`.
3 : Base Case Check
if(n==1) return nums[0]; // Base case
A base case is checked: if the vector has only one element, that element is returned as the result.
4 : Create New Vector
vector<int> newNum(n/2);
A new vector `newNum` of size `n/2` is created to store the results of the alternating min and max operations.
5 : Iterate Over Vector
for(int i=0; i<n/2; i++) {
A for-loop is initiated to iterate through the input vector `nums` in steps of two, processing pairs of elements.
6 : Min Operation
if(i%2==0) newNum[i] = min(nums[2 * i], nums[2 * i + 1]);
For even indexed iterations, the minimum of the two adjacent elements in `nums` is stored in `newNum[i]`.
7 : Max Operation
else newNum[i] = max(nums[2 * i], nums[2 * i + 1]);
For odd indexed iterations, the maximum of the two adjacent elements in `nums` is stored in `newNum[i]`.
8 : Recursive Call
int res = minMaxGame(newNum); // Recursive call
The function `minMaxGame` is called recursively with the newly created vector `newNum` as the input.
9 : Return Result
return res;
The result of the recursive call is returned.
Best Case: O(n)
Average Case: O(n)
Worst Case: O(n)
Description: The time complexity is O(n) for each iteration, and there are log(n) iterations in total. Hence, the overall complexity is O(n log n).
Best Case: O(n)
Worst Case: O(n)
Description: The space complexity is O(n) since we create a new array of half the size of the original array in each iteration.
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