algoadvance

You are given a 0-indexed integer array nums of even length. As long as nums is not empty, you must repetitively:

  1. Find the minimum and maximum values in nums.
  2. Compute the average of these two values.
  3. Remove both the minimum and maximum values from nums.

Your task is to return the number of distinct averages computed.

Clarifying Questions

  1. Input Constraints?
    • The array length is even.
    • All elements are integers.
  2. Range of Values?
    • The values in the array can range from -10^6 to 10^6.
  3. Output?
    • Output should be an integer representing the number of distinct averages.

Now, we’ll proceed with implementing the solution.

Strategy

  1. Sort the Array: First, sort the given array to easily find the minimum and maximum values.

  2. Two Pointer Technique: Use two pointers, one starting from the beginning (left) and one from the end (right) of the sorted array.

  3. Compute Averages: Iteratively compute the average of the values at the two pointers, store the average in a set (to keep track of distinct averages), and move the pointers towards each other.

  4. Remove Values: After computing the average, increment the left pointer and decrement the right pointer to remove both the minimum and maximum from the array implicitly.

  5. Continue Until the Pointers Meet: Repeat steps 3-4 until the pointers meet.

  6. Return the Size of the Set: Finally, return the size of the set containing distinct averages.

Code Implementation

def distinct_averages(nums):
    # Sort the array to facilitate the two-pointer technique
    nums.sort()
    distinct_avg_set = set()
    
    left, right = 0, len(nums) - 1
    
    # Loop until pointers meet
    while left < right:
        # Compute the average of nums[left] and nums[right]
        avg = (nums[left] + nums[right]) / 2
        # Add it to the set of distinct averages
        distinct_avg_set.add(avg)
        # Move pointers
        left += 1
        right -= 1
    
    # The number of distinct averages is the size of the set
    return len(distinct_avg_set)

# Example usage:
nums = [4,1,4,0,3,5]
print(distinct_averages(nums))  # Output should be the number of distinct averages

Time Complexity

  1. Sorting: The sorting step will take (O(n \log n)) where (n) is the number of elements in the array.
  2. Two-Pointer Traversal: The traversal with two pointers takes (O(n/2)), which simplifies to (O(n)).

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

The space complexity is O(1) for the two-pointer technique plus O(D) for the set storing distinct averages where D is the number of distinct averages (in the worst case D could be up to (n/2)).

Try our interview co-pilot at AlgoAdvance.com