Introduction

This blog post covers a solution to a Leetcode problem that requires calculating the minimum number of operations needed so that all elements of an array are greater than or equal to a given threshold value. The task is both interesting and practical, showcasing the application of algorithm design for efficiency.

In this article, I’ll explain the problem, walk you through my thought process, and share the solution I came up with.

The Problem

Link to question

You are given a 0-indexed integer array nums, and an integer k.

In one operation, you will:

• Take the two smallest integers x and y in nums.
• Remove x and y from nums.
• Add min(x, y) * 2 + max(x, y) anywhere in the array.

Note that you can only apply the described operation if nums contains at least two elements.

Return the minimum number of operations needed so that all elements of the array are greater than or equal to k.

Examples

Example 1:

Input: nums = [2,11,10,1,3], k = 10
Output: 2
Explanation:
- In the first operation, we remove elements 1 and 2, then add 1 * 2 + 2 to nums. nums becomes equal to [4, 11, 10, 3].
- In the second operation, we remove elements 3 and 4, then add 3 * 2 + 4 to nums. nums becomes equal to [10, 11, 10].
At this stage, all the elements of nums are greater than or equal to 10 so we can stop.
It can be shown that 2 is the minimum number of operations needed so that all elements of the array are greater than or equal to 10.

Example 2:

Input: nums = [1,1,2,4,9], k = 20
Output: 4
Explanation:
- After one operation, nums becomes equal to [2, 4, 9, 3].
- After two operations, nums becomes equal to [7, 4, 9].
- After three operations, nums becomes equal to [15, 9].
- After four operations, nums becomes equal to [33].
At this stage, all the elements of nums are greater than 20 so we can stop.
It can be shown that 4 is the minimum number of operations needed so that all elements of the array are greater than or equal to 20.

My Solution

The goal is to calculate the minimum number of operations needed so that all elements of the array are greater than or equal to k. To achieve this, we can use a min-heap to efficiently get the two smallest elements in each operation.

Here is the implementation of the solution:

class Solution:
    def minOperations(self, nums: List[int], k: int) -> int:
        if not nums or not k:
            return 0

        heapq.heapify(nums)
        res = 0

        while nums[0] < k:
            if len(nums) < 2:
                return -1  # Not possible to reach the threshold

            x = heapq.heappop(nums)
            y = heapq.heappop(nums)
            new_element = min(x, y) * 2 + max(x, y)
            heapq.heappush(nums, new_element)
            res += 1

        return res

Complexity Analysis

• Time Complexity:

  • The solution involves heap operations which take O(log n) time.
  • In the worst case, we might need to perform O(n) operations.
  • Thus, the overall time complexity is O(n log n).

• Space Complexity:

  • We use a heap to store the elements, which takes O(n) space.
  • Hence, the space complexity is O(n).

This approach ensures that we handle large inputs efficiently by leveraging the properties of the heap data structure to perform the necessary operations.

Reflections and Takeaways

This problem is a great example of how to efficiently calculate operations using a heap data structure. By leveraging the properties of the heap, we can efficiently get the two smallest elements and perform the required operations. This approach ensures that we handle large inputs efficiently.

Let me know if you came up with a different approach—I’d love to hear it!