Last Stone Weight II

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ArraysDynamic Programming

You are given an array of integers stones where stones[i] is the weight of the i^th stone.

We are playing a game with the stones. On each turn, we choose any two stones and smash them together. Suppose the stones have weights x and y with x <= y. The result of this smash is:

If x == y, both stones are destroyed, and
If x != y, the stone of weight x is destroyed, and the stone of weight y has new weight y - x.

At the end of the game, there is at most one stone left.

Return the smallest possible weight of the left stone. If there are no stones left, return 0.

Example 1:

Input: stones = [2,7,4,1,8,1]
Output: 1
Explanation:
We can combine 2 and 4 to get 2, so the array converts to [2,7,1,8,1] then,
we can combine 7 and 8 to get 1, so the array converts to [2,1,1,1] then,
we can combine 2 and 1 to get 1, so the array converts to [1,1,1] then,
we can combine 1 and 1 to get 0, so the array converts to [1], then that's the optimal value.

Example 2:

Input: stones = [31,26,33,21,40]
Output: 5

Constraints:

1 <= stones.length <= 30
1 <= stones[i] <= 100

Try coding it in LeetCode

Last Stone Weight II

Medium

Asked by:

10 views

Topics:

ArraysDynamic Programming

You are given an array of integers stones where stones[i] is the weight of the i^th stone.

We are playing a game with the stones. On each turn, we choose any two stones and smash them together. Suppose the stones have weights x and y with x <= y. The result of this smash is:

If x == y, both stones are destroyed, and
If x != y, the stone of weight x is destroyed, and the stone of weight y has new weight y - x.

At the end of the game, there is at most one stone left.

Return the smallest possible weight of the left stone. If there are no stones left, return 0.

Example 1:

Input: stones = [2,7,4,1,8,1]
Output: 1
Explanation:
We can combine 2 and 4 to get 2, so the array converts to [2,7,1,8,1] then,
we can combine 7 and 8 to get 1, so the array converts to [2,1,1,1] then,
we can combine 2 and 1 to get 1, so the array converts to [1,1,1] then,
we can combine 1 and 1 to get 0, so the array converts to [1], then that's the optimal value.

Example 2:

Input: stones = [31,26,33,21,40]
Output: 5

Constraints:

1 <= stones.length <= 30
1 <= stones[i] <= 100

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the constraints on the size of the `stones` array and the weight of each stone?
Can the weight of a stone be zero?
Are all stone weights guaranteed to be positive integers?
If after some moves, all stones are crushed into a single stone with weight zero, should I return 0?
Are there any memory constraints to be aware of, given potentially large input sizes?

Brute Force Solution

Approach

Imagine you have a pile of stones, and you want to find the smallest possible weight left after smashing them together. A brute force method tries every possible way to divide the stones into two groups, and then smashes each group against each other.

Here's how the algorithm would work step-by-step:

First, consider every possible combination of stones to put into one group. This means trying every single subset of stones.
For each of these possible groups, the remaining stones will form the other group.
Calculate the total weight of each group.
Smash the stones in each group by subtracting the smaller weight from the larger weight. The result is the weight remaining after smashing each group.
Subtract the weight remaining from the smash between the two groups from one another. This is the weight remaining after combining our groups and smashing stones between the groups.
Compare the weights remaining from all of these combinations and keep track of the smallest one.
The smallest weight remaining after trying all the combinations is your answer.

Code Implementation

def last_stone_weight_ii_brute_force(stones):
    number_of_stones = len(stones)
    minimum_weight = float('inf')

    # Iterate through all possible subsets of stones
    for i in range(1 << number_of_stones):
        first_group_weight = 0
        second_group_weight = 0

        # Divide the stones into two groups based on the subset
        for stone_index in range(number_of_stones):
            if (i >> stone_index) & 1:
                first_group_weight += stones[stone_index]
            else:
                second_group_weight += stones[stone_index]

        # Calculate the weight difference between the two groups.
        current_weight = abs(first_group_weight - second_group_weight)

        # Find the minimum weight among all combinations
        minimum_weight = min(minimum_weight, current_weight)

    return minimum_weight

Big(O) Analysis

Time Complexity

O(2^n) – The described approach iterates through every possible subset of the input stones. With n stones, there are 2^n possible subsets since each stone can either be in the subset or not. For each subset, we calculate the weight and simulate the smashing process which takes O(n) time in the worst case to sum the weights of both sets. Thus, the overall time complexity is dominated by generating all subsets, resulting in O(2^n * n). However, since 2^n grows much faster than n, we simplify the overall time complexity to O(2^n).

Space Complexity

O(2^N) – The algorithm considers every possible subset of stones, implying the creation of all possible combinations. Generating each subset implicitly requires storing its elements, leading to a maximum of 2^N subsets, where N is the number of stones. While the explanation doesn't explicitly state the creation of a data structure to hold all subsets simultaneously, the process of iterating through them and calculating weights suggests that intermediate subset representations contribute to auxiliary space. Therefore, the space complexity grows exponentially with the number of stones as the algorithm explores the power set. This space is dominated by the exponential number of combinations to consider, resulting in a space complexity of O(2^N).

Optimal Solution

Approach

The problem involves finding the minimum possible weight difference after smashing stones together. The key insight is to divide the stones into two groups such that the difference between their sums is minimized, effectively turning it into a subset sum problem.

Here's how the algorithm would work step-by-step:

Imagine you're trying to split the stones into two piles so their weights are as close as possible.
Calculate the total weight of all the stones. This total is important for finding the target weight for one of our piles.
Divide the total weight by two. This result is our target weight for one of the piles. We want to create a pile with a weight as close as possible to this target without exceeding it.
Figure out all the possible weights you can make by combining different stones (up to the target weight we just calculated). This is like exploring all possible subsets of stones.
Among all the possible subset weights you found, pick the weight that's closest to (but not more than) the target weight.
Now, to get the final answer, subtract the pile weight you found twice from the total weight. This gives you the minimum possible difference between the two piles.

Code Implementation

def lastStoneWeightII(stones):    total_weight = sum(stones)
    target_weight = total_weight // 2
    
    # dp[i] is True if subset with sum i is possible
    possible_subset_weights = [False] * (target_weight + 1)
    possible_subset_weights[0] = True
    
    for stone in stones:
        for current_weight in range(target_weight, stone - 1, -1):
            possible_subset_weights[current_weight] = possible_subset_weights[current_weight] or possible_subset_weights[current_weight - stone]

    # Find the largest possible sum that is <= target
    max_possible_weight = 0
    for current_weight in range(target_weight, -1, -1):
        if possible_subset_weights[current_weight]:
            max_possible_weight = current_weight
            break
    
    # The result is total - 2 * max, minimizing difference
    return total_weight - 2 * max_possible_weight

Big(O) Analysis

Time Complexity

O(n*sum) – The algorithm calculates the total sum of the stones, which we'll call 'sum'. It then uses dynamic programming to find the possible subset sums up to sum/2. The DP table has dimensions n x (sum/2 + 1), where n is the number of stones. We iterate through the stones array (outer loop of size n) and for each stone, we iterate through the possible subset sums from 0 to sum/2 (inner loop of size sum/2, approximated as sum). Thus, the time complexity is dominated by the nested loops, resulting in O(n*sum) where 'n' is the number of stones and 'sum' is the total weight of the stones.

Space Complexity

O(totalWeight/2) – The dominant space complexity stems from step 4, where we figure out all possible weights obtainable by combining stones up to a target weight. This is effectively storing intermediate results for subset sums. This process requires an auxiliary boolean array (or similar data structure) of size totalWeight/2 (rounded up), to keep track of which weights are achievable. Therefore, the auxiliary space used grows linearly with totalWeight/2. Thus the space complexity is O(totalWeight/2).

Edge Cases

Case	How to Handle
Empty input array	Return 0 since there are no stones to smash.
Single stone in the array	Return the single stone's weight as there's nothing to combine it with.
Array with two stones	Return the absolute difference between the two stone weights.
All stones have the same weight	The result will either be 0 (if the count is even) or the stone's weight (if the count is odd); the standard algorithm correctly calculates this.
Array contains very large stone weights that could lead to integer overflow during summation	Use a larger integer type (e.g., long) or modulo operation to prevent overflow during summation.
Maximum size input array (e.g., n=30), combined with stone weights nearing the constraint limit	Dynamic programming solutions should handle this efficiently, though careful memory management and potentially using bitsets can optimize further, ensuring compliance within time and memory limits.
Stone weights are heavily skewed (e.g., one very large stone, many small stones)	The dynamic programming approach will explore a wide range of possible subsets regardless of skew, so no special handling is needed.
Input array contains duplicate stone weights	The dynamic programming solution inherently handles duplicates by considering all possible combinations of stone weights.