When faced with the problem, my initial thought was, "min-heap." However, my mind quickly introduced a note of skepticism: "Will this actually work?" This was a challenge that seemed straightforward at first glance but revealed its complexity through the peculiar nature of the required operation. Once solved, though, it provided a gratifying sense of accomplishment. The Problem You are given a list of scores, each representing a player's performance. Your goal is to ensure that every score is at least as high as a specified target score, using the minimum number of operations. The catch? You can't simply increment values directly. Instead, you have access to a unique operation: One Operation: - Select the smallest score in the list and double it. - Repeat this process until all scores are at least the target score, or return -1 if it is impossible to achieve. Example: - Scores: [4, 1, 7, 3, 6] - Target Score: 5 To tackle this problem effectively, it's important to break it down into manageable steps. Initial Instincts A min-heap, a data structure that allows efficient retrieval of the smallest element, seems like a natural choice. Here’s why: - Efficiency: Doubling the smallest score ensures you make the most significant improvement with each operation, minimizing the total number needed. - Dynamic Adjustment: As scores change, a min-heap can quickly update which score is the new smallest, keeping the solution optimal. However, the critical question remains: can we guarantee that this approach will always find the minimum number of operations? Analyzing the Problem Let's consider the properties of the operation and how they affect the scores: - Doubling the Smallest Score: Each time you double the smallest score, it becomes larger, potentially changing the next smallest score in subsequent iterations. - Target Constraint: If any score is below the target, it must be doubled until it meets or exceeds the target. - Impossibility Check: If doubling the smallest score repeatedly cannot bring all scores to the target level, the task is impossible, and you should return -1. Step-by-Step Solution To implement the solution using a min-heap: 1. Initialize the heap with all the scores. 2. Count the number of operations needed. 3. While the smallest score in the heap is less than the target score: - Extract the smallest score. - Double it. - Reinsert the doubled score into the heap. - Increment the operation count. 4. If all scores are now at or above the target, return the operation count. 5. If it's determined that the target cannot be reached, return -1. Example Walkthrough Let's walk through the example step by step: 1. Initial State: - Scores: [4, 1, 7, 3, 6] - Target Score: 5 - Operations: 0 First Operation: Smallest score: 1 Double it: 1 * 2 = 2 Update scores: [4, 2, 7, 3, 6] Operations: 1 Second Operation: Smallest score: 2 Double it: 2 * 2 = 4 Update scores: [4, 4, 7, 3, 6] Operations: 2 Third Operation: Smallest score: 3 Double it: 3 * 2 = 6 Update scores: [4, 4, 7, 6, 6] Operations: 3 Final Check: All scores are now greater than or equal to 5. Return the operation count: 3 Practical Implementation Here’s a Python implementation of the solution: ```python import heapq def min_operations(scores, target_score): # Initialize the min-heap heapq.heapify(scores) # Count the number of operations operations = 0 # Continue until all scores are at or above the target while scores[0] < target_score: # Extract the smallest score smallest_score = heapq.heappop(scores) # Double the smallest score doubled_score = smallest_score * 2 # Reinsert the doubled score heapq.heappush(scores, doubled_score) # Increment the operation count operations += 1 # Check if it's impossible if smallest_score >= target_score: return -1 return operations ``` Conclusion While the min-heap approach might initially seem too straightforward, it is indeed the optimal solution. By focusing on the smallest score and doubling it iteratively, you ensure that each operation brings you closer to the target in the most efficient way possible. This problem highlights the importance of breaking down complex tasks and leveraging the right data structures to achieve optimal results. For those not familiar with min-heaps, they offer an elegant way to manage and adjust priorities dynamically, making them invaluable tools in algorithm design.