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Leetcode 2465. Number of Distinct Averages

Problem Statement

Given an integer array nums, move the elements of nums around to form a new array. Return the number of distinct averages you can get.

• The average of an array is the sum of the elements of the array divided by the number of elements in the array.
• When you form a new array nums[i..j] from nums, then the average of nums[i..j] is the integer quotient of the sum of nums[i..j] divided by j-i+1.

You cannot simply get the averages of all possible subarrays, as this would be too time-consuming.

Clarifying Questions

1. Q: Are the elements of the array nums guaranteed to be integers?
   • A: Yes, the elements of the array are integers.
2. Q: Can the elements of nums be negative?
   • A: Yes, elements in nums can be negative.
3. Q: What should be the result if the array nums is empty?
   • A: If the array nums is empty, we should return 0, as there are no elements to form any averages.

Strategy

1. Sort the Array: To simplify the problem, we first sort the array nums.
2. Two-pointer Technique: Use two pointers, one starting at the beginning (left) and one starting at the end (right) of the array.
3. Pair Averages: Pair the smallest element with the largest element, calculate their average, and insert it into a set to ensure distinct values.
4. Continue Pairing: Move the left pointer to the right and the right pointer to the left, and repeat the pairing process until the pointers meet.
5. Distinct Averages: The number of distinct averages is simply the size of the set.

Code

import java.util.Arrays;
import java.util.HashSet;
import java.util.Set;

public class Solution {
    public int distinctAverages(int[] nums) {
        Arrays.sort(nums);
        Set<Double> averages = new HashSet<>();
        int left = 0, right = nums.length - 1;
        
        while (left < right) {
            double avg = ((double) nums[left] + nums[right]) / 2;
            averages.add(avg);
            left++;
            right--;
        }

        // If the array has an odd number of elements, handle the middle element
        if (left == right) {
            averages.add((double) nums[left]);
        }
        
        return averages.size();
    }
}

Time Complexity

• Sorting: Sorting the array takes O(n log n).
• Two-pointer traversal: Traversing the array with two pointers takes O(n) since each element is processed exactly once.

Overall, the time complexity of this solution is O(n log n) due to the sorting step.

Space Complexity

• Auxiliary Space: The primary extra space used is for the Set which holds at most O(n/2) average values in the worst case.
• Hence, the space complexity is O(n).

This approach ensures that we efficiently compute the number of distinct averages with a time complexity that is optimal for the problem constraints.

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