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Leetcode 3194. Minimum Average of Smallest and Largest Elements

Problem Statement

Given an array of integers, find the minimum possible average of the smallest and largest elements of any subarray of length at least two. Your task is to return that minimum average.

Clarifying Questions

1. What is the constraint on the length of the array?
   • The length of the array will be at least 2.
2. Can the array contain negative numbers?
   • Yes, the array can contain both negative and positive integers.
3. Is there a specific range for the values within the array?
   • The problem statement does not specify, so we will assume standard integer limits.
4. What should be done if the array contains duplicate elements?
   • Duplicates do not affect the problem; we’ll handle them as usual.
5. What should the solution return if it finds the minimum average?
   • The solution should return a floating-point number as the average.

Strategy

To solve this problem efficiently, we can follow these steps:

1. Initialize Minimum Average:
   • Initialize a variable to store the minimum average value found, starting with a very large value.
2. Two-Pointer Technique:
   • Use two nested loops to examine all subarrays with lengths greater than or equal to 2.
   • For each subarray, find the smallest and largest element.
3. Calculate Average:
   • For each subarray, calculate the average of the smallest and largest elements. Update the minimum average variable if the current average is smaller.
4. Return Result:
   • After examining all possible subarrays, return the smallest average found.

Code

public class Solution {
    public double findMinAverage(int[] nums) {
        double minAverage = Double.MAX_VALUE;

        for (int i = 0; i < nums.length; i++) {
            int minElement = nums[i];
            int maxElement = nums[i];

            for (int j = i + 1; j < nums.length; j++) {
                minElement = Math.min(minElement, nums[j]);
                maxElement = Math.max(maxElement, nums[j]);

                double currentAverage = (minElement + maxElement) / 2.0;
                minAverage = Math.min(minAverage, currentAverage);
            }
        }

        return minAverage;
    }
    
    public static void main(String[] args) {
        Solution sol = new Solution();
        int[] nums = {3, 1, 4, 1, 5, 9, 2};
        System.out.println(sol.findMinAverage(nums)); // Should output the minimum average.
    }
}

Time Complexity

• Initialization and final return: O(1)
• Two-pointer nested loop:
  • Outermost loop runs n times.
  • Inner loop runs n-1 times (on average).
• Finding minimum and maximum elements in each subarray:
  • In the worst case, we will check (n*(n-1))/2 subarrays.

Thus, the overall time complexity is O(n^2), where n is the length of the array. This is the best we can do for evaluating all subarrays longer than one element.

Space Complexity

The space complexity is O(1) since we are using only a few extra variables and not any additional space that scales with input size.

This solution efficiently calculates the minimum possible average of the smallest and largest elements in all possible subarrays of length at least two, adhering to the time and space constraints expected for the problem’s requirements.

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