How do you solve Remove Stones to Minimize the Total on LeetCode?

Remove Stones to Minimize the Total (LeetCode #1962) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(k log n) time complexity and O(n) space complexity. Key insight: Always target the pile with the maximum stones to minimize the total effectively.

What is the brute force solution for Remove Stones to Minimize the Total?

The brute force approach for Remove Stones to Minimize the Total is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute force approach, we simply iterate through the piles and apply the operation on each pile, choosing the maximum pile each time. This is straightforward but inefficient as it may require multiple passes through the piles. The optimal Optimal Solution approach improves this to O(k log n).

What algorithm or data structure is used to solve Remove Stones to Minimize the Total?

Remove Stones to Minimize the Total uses the following concepts: Array, Greedy, Heap (Priority Queue). The recommended patterns to study are: Greedy Algorithms, Heap (Priority Queue).

What are the interview tips for Remove Stones to Minimize the Total?

Clearly explain your thought process and why you choose a particular approach. Consider edge cases, such as when k is larger than the total number of stones. Practice implementing both brute force and optimal solutions to understand trade-offs.

What are common mistakes when solving Remove Stones to Minimize the Total?

Not using a priority queue, leading to inefficient maximum retrieval. Failing to account for the floor division when updating pile sizes.

#1962

Remove Stones to Minimize the Total

Medium

ArrayGreedyHeap (Priority Queue)Greedy AlgorithmsHeap (Priority Queue)

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

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

In the optimal solution, we use a max-heap (priority queue) to efficiently retrieve and update the pile with the maximum stones. This allows us to perform the k operations in a more efficient manner.

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Algorithm

4 steps

1Step 1: Create a max-heap from the piles array.
2Step 2: For k times, extract the maximum pile from the heap.
3Step 3: Remove floor(max / 2) stones from this pile and push the updated pile back into the heap.
4Step 4: After k operations, sum the remaining stones in the heap.

solution.py9 lines

1import heapq
2
3def minStoneSum(piles, k):
4    max_heap = [-x for x in piles]
5    heapq.heapify(max_heap)
6    for _ in range(k):
7        max_pile = -heapq.heappop(max_heap)
8        heapq.heappush(max_heap, -(max_pile - max_pile // 2))
9    return -sum(max_heap)

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Complexity note: The time complexity is O(k log n) because each of the k operations involves extracting the maximum from the heap and inserting the updated pile back, both of which take logarithmic time relative to the number of piles.

1Always target the pile with the maximum stones to minimize the total effectively.
2Using a max-heap allows for efficient retrieval and updating of the maximum pile.

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