How do you solve Minimum Operations to Exceed Threshold Value II on LeetCode?

Minimum Operations to Exceed Threshold Value II (LeetCode #3066) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Using a priority queue allows for efficient retrieval of the smallest elements, reducing the time complexity.

What is the brute force solution for Minimum Operations to Exceed Threshold Value II?

The brute force approach for Minimum Operations to Exceed Threshold Value II is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves repeatedly finding the two smallest elements, performing the operation, and checking if all elements meet the threshold. This is straightforward but inefficient as it may require many iterations. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Minimum Operations to Exceed Threshold Value II?

Minimum Operations to Exceed Threshold Value II uses the following concepts: Array, Heap (Priority Queue), Simulation. The recommended patterns to study are: Heap (Priority Queue), Simulation.

What are the interview tips for Minimum Operations to Exceed Threshold Value II?

Always consider the data structure that best fits the problem — a priority queue is ideal here. Explain your thought process clearly; interviewers appreciate understanding your reasoning. Practice similar problems to become familiar with patterns in operations on arrays.

What are common mistakes when solving Minimum Operations to Exceed Threshold Value II?

Not using a priority queue, which leads to inefficient solutions. Failing to check if there are enough elements to perform the operation.

#3066

Minimum Operations to Exceed Threshold Value II

Medium

ArrayHeap (Priority Queue)SimulationHeap (Priority Queue)Simulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(n)

💡

Intuition

Time O(n log n)Space O(n)

Using a priority queue (min-heap) allows us to efficiently retrieve and remove the two smallest elements, significantly reducing the time complexity of the operations.

⚙️

Algorithm

4 steps

1Step 1: Initialize a min-heap with all elements from nums.
2Step 2: While the smallest element in the heap is less than k and there are at least two elements, extract the two smallest elements.
3Step 3: Compute the new element using the formula min(x, y) * 2 + max(x, y) and insert it back into the heap.
4Step 4: Count the number of operations performed and return it.

solution.py15 lines

1# Full working Python code
2import heapq
3
4def min_operations(nums: List[int], k: int) -> int:
5    heapq.heapify(nums)
6    operations = 0
7    while nums and nums[0] < k:
8        if len(nums) < 2:
9            return -1
10        x = heapq.heappop(nums)
11        y = heapq.heappop(nums)
12        new_elem = min(x, y) * 2 + max(x, y)
13        heapq.heappush(nums, new_elem)
14        operations += 1
15    return operations

ℹ

Complexity note: Using a priority queue allows us to efficiently manage the smallest elements, leading to O(log n) time for each insertion and removal, resulting in O(n log n) overall.

1Using a priority queue allows for efficient retrieval of the smallest elements, reducing the time complexity.
2The operation effectively increases the minimum values in the array, which is crucial for reaching the threshold.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.