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Leetcode 2974. Minimum Number Game

Problem Statement

You are given an integer array nums of even length n. The array is a permutation of the integers in the range [1, n]. Using a function minMaxVal(nums) defined as follows:

  1. Given a number x, nums[2*i] = x for some 0 <= i <= n/2
  2. Given a number y, nums[2*i + 1] = y for some 0 <= i <= n/2
  3. The function minMaxVal(nums) returns the minimum of all maximum values of the pairs (nums[2*i], nums[2*i + 1]) for 0 <= i < n/2.

Your task is to write a function findMinMaxVal that returns the maximum of minMaxVal results over all permutations of nums.

Clarifying Questions

  1. Input Size: What is the expected size range of the input list?
    • This determines how efficient our solution needs to be.
  2. Constraints: Are there any constraints on the integers, other than being a permutation of [1, n]?
  3. Output: Should we return the maximum minMaxVal?
    • Yes, we want to find the maximum value of minMaxVal over all possible permutations of the input list.

Since the array is guaranteed to be even-length and a permutation, the only constraints are its size and the range of integers. Let’s assume n is not excessively large to keep the problem tractable within normal constraints.

Strategy

Given the constraints, a brute-force approach that evaluates all permutations might not be practical, especially considering the factorial growth of permutations. Instead, we can devise a strategy that arranges pairs to minimize their maximum values in a controlled way.

Optimized Approach:

  1. Sort nums: Start by sorting the array, since having pairs formed from the sorted array will help control the maxima.
  2. Pair Elements: Pair the smallest available element with the largest element from the unsorted part that hasn’t been used. This minimizes the impact of large differences when calculating the max.

The key insight is:

Code

import java.util.Arrays;

public class MinimumNumberGame {
    public int findMinMaxVal(int[] nums) {
        Arrays.sort(nums);
        int maxMinVal = Integer.MIN_VALUE;
        int n = nums.length;
        
        for (int i = 0; i < n / 2; i++) {
            int maxValOfPair = Math.max(nums[i], nums[n - 1 - i]);
            maxMinVal = Math.max(maxMinVal, maxValOfPair);
        }
        
        return maxMinVal;
    }
    
    // Example test
    public static void main(String[] args) {
        MinimumNumberGame game = new MinimumNumberGame();
        int[] nums = {3, 5, 2, 7, 1, 6, 4, 8};
        System.out.println(game.findMinMaxVal(nums));  // Expected output should reflect the optimal arrangement.
    }
}

Time Complexity

The time complexity of the provided solution is primarily determined by the sorting step:

Hence, the overall time complexity is O(n log n).

This approach should be efficient for reasonably sized inputs and leverages sorting to strategically manage the problem constraints.

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