How do you solve Put Marbles in Bags on LeetCode?

Put Marbles in Bags (LeetCode #2551) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(1) space complexity. Key insight: Only the endpoints of the bags matter for calculating scores.

What is the time complexity of Put Marbles in Bags?

The optimal solution for Put Marbles in Bags runs in O(n log n) time and uses O(1) space. The complexity is dominated by the sorting step, which is O(n log n), while the subsequent calculations are linear, O(n).

What is the brute force solution for Put Marbles in Bags?

The brute force approach for Put Marbles in Bags is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible ways to divide the marbles into k bags and calculating the score for each distribution. This method is straightforward but inefficient for larger inputs. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Put Marbles in Bags?

Put Marbles in Bags uses the following concepts: Array, Greedy, Sorting, Heap (Priority Queue). The recommended patterns to study are: Greedy, Sorting.

What are the interview tips for Put Marbles in Bags?

Clearly explain your thought process and how you arrived at your solution. Consider edge cases, such as when k equals the length of the weights array. Practice explaining your solution to someone else to solidify your understanding.

What are common mistakes when solving Put Marbles in Bags?

Failing to consider that only the endpoints contribute to the score. Not sorting the weights before calculating scores.

#2551

Put Marbles in Bags

Hard

ArrayGreedySortingHeap (Priority Queue)GreedySorting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n log n)Space O(1)

The optimal approach leverages the fact that only the endpoints of the bags matter for calculating scores. By focusing on the largest and smallest weights, we can maximize and minimize the score efficiently.

⚙️

Algorithm

4 steps

1Step 1: Sort the weights array.
2Step 2: The minimum score is calculated using the first k smallest weights for the left endpoints and the last k smallest weights for the right endpoints.
3Step 3: The maximum score is calculated using the first k largest weights for the left endpoints and the last k largest weights for the right endpoints.
4Step 4: Return the difference between the maximum and minimum scores.

solution.py7 lines

1# Full working Python code
2
3def maxMinDifference(weights, k):
4    weights.sort()
5    min_score = sum(weights[i] + weights[-(i + 1)] for i in range(k))
6    max_score = sum(weights[-(i + 1)] + weights[i] for i in range(k))
7    return max_score - min_score

ℹ

Complexity note: The complexity is dominated by the sorting step, which is O(n log n), while the subsequent calculations are linear, O(n).

1Only the endpoints of the bags matter for calculating scores.
2Sorting the weights allows us to easily access the smallest and largest weights.

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