Last Stone Weight II Leetcode Problem 1049 [Python Solution]
In this blog post, we'll delve into the LeetCode problem Last Stone Weight II (Problem #1049).
We'll explore the problem statement, understand the constraints, and then take a deep dive into the efficient Python solution to solve it.
Problem Overview
Question: You are given an array of integers stones, where stones[i] represents the weight of the ith stone.
You are playing a game with these stones.
On each turn, you can choose any two stones and smash them together.
The results of this smash are as follows:
- If both stones have the same weight, both stones are destroyed.
- If the stones have different weights (x and y, with x <= y), the stone with weight x is destroyed, and the stone with weight y has its weight reduced to y – x.
At the end of the game, there is at most one stone left.
Your task is to return the smallest possible weight of the remaining stone.
If there are no stones left, return 0.
Example 1:
Input: stones = [2,7,4,1,8,1]
Output: 1
In this example, you can combine the stones as follows:
1. Combine 2 and 4 to get 2, resulting in [2,7,1,8,1].
2. Combine 7 and 8 to get 1, resulting in [2,1,1,1].
3. Combine 2 and 1 to get 1, resulting in [1,1,1].
4. Finally, combine 1 and 1 to get 0, resulting in [1].
The optimal value is 1.
Example 2:
Input: stones = [31,26,33,21,40]
Output: 5
Understanding Constraints
Before we dive into the solution, let's break down the problem's constraints.
- 1 <=
stones.length<= 30: The array can contain a maximum of 30 stones. - 1 <=
stones[i]<= 100: The weight of each stone falls within the range of 1 to 100.
Last Stone Weight II LeetCode Problem Solution
Bruteforce Approach
To solve this problem efficiently, we can take advantage of a clever observation.
The goal is to minimize the remaining stone's weight.
To do that, we need to split the stones into two groups as evenly as possible.
By achieving this balance, we can minimize the remaining weight.
This problem can be reduced to the bounded knapsack problem, a classic dynamic programming problem.
In bounded knapsack, you have a set of items, each with a weight and a value.
The goal is to find the maximum value you can obtain by selecting a subset of the items while respecting a weight limit.
In our case, we treat each stone's weight as an item, and the weight limit is half of the total weight of all stones.
The value of each item is its weight, and the goal is to find a subset of stones that maximize the total weight without exceeding the weight limit.
Let's implement this approach in Python:
def lastStoneWeightII(stones):
# Calculate the sum of all stone weights
stoneSum = sum(stones)
# Calculate the target weight for each pile
target = (stoneSum + 1) // 2
# Helper function for dynamic programming
def dfs(i, total):
# Base cases
if total >= target or i == len(stones):
return abs(total - (stoneSum - total))
if (i, total) in dp:
return dp[(i, total)]
# Choose not to include the current stone
not_included = dfs(i + 1, total)
# Choose to include the current stone
included = dfs(i + 1, total + stones[i])
# Update the cache with the minimum of the two choices
dp[(i, total)] = min(not_included, included)
return dp[(i, total)]
# Initialize the cache
dp = {}
# Start the dynamic programming process
return dfs(0, 0)
In this code, we first calculate the total sum of all stone weights and determine our target weight, which is half of the total weight.
The dfs function is a recursive function that explores all possible combinations of stones to find the optimal solution.
We use memoization (caching) to avoid recomputation of overlapping subproblems.
Efficient Approach
The efficient approach is implemented using dynamic programming and memoization.
It reduces the time complexity by avoiding redundant calculations.
The code is optimized for the given problem, and it provides a quick and efficient solution.
Time and Space Complexity
Time Complexity: The time complexity of the solution is O(n * total), where n is the number of stones and total is the total weight of all stones.
This is because we have a cache (dp) with at most n * total entries, and for each entry, we perform constant time operations.
Space Complexity: The space complexity is O(n * total) as well, considering the space required for the cache and the recursion stack.
Reasoning Behind Our Approach
The key insight behind our approach is to recognize that the problem can be transformed into a bounded knapsack problem, where we aim to divide the stones into two piles as evenly as possible to minimize the remaining stone's weight.
By treating each stone's weight as an item and using dynamic programming with memoization, we efficiently explore all possible combinations and find the optimal solution.
Our approach handles the problem's constraints and offers a quick and effective solution for finding the smallest possible weight of the remaining stone.
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Conclusion
The Last Stone Weight II problem can be effectively solved by transforming it into a bounded knapsack problem and using dynamic programming with memoization.
This approach optimally divides the stones into two piles, minimizing the remaining stone's weight.
By understanding the problem constraints and implementing our efficient approach, you can solve this problem on LeetCode with confidence.
For further details and to try out the code yourself, you can visit the problem on LeetCode: Last Stone Weight II on LeetCode.
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Happy coding!
