How do you solve Removing Minimum Number of Magic Beans on LeetCode?

Removing Minimum Number of Magic Beans (LeetCode #2171) 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: Choosing the right target for equalization is crucial; it should be based on the bags with the least beans.

What is the time complexity of Removing Minimum Number of Magic Beans?

The optimal solution for Removing Minimum Number of Magic Beans runs in O(n log n) time and uses O(1) space. The time complexity is O(n log n) due to the sorting step, while the space complexity is O(1) since we are using a constant amount of extra space.

What is the brute force solution for Removing Minimum Number of Magic Beans?

The brute force approach for Removing Minimum Number of Magic Beans is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking every possible way to equalize the number of beans in the bags by removing beans from each bag. This method is simple but inefficient as it considers all combinations. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Removing Minimum Number of Magic Beans?

Removing Minimum Number of Magic Beans uses the following concepts: Array, Greedy, Sorting, Enumeration, Prefix Sum. The recommended patterns to study are: Greedy, Sorting, Prefix Sum.

What are the interview tips for Removing Minimum Number of Magic Beans?

Always clarify the problem requirements and constraints before diving into coding. Think about edge cases, such as when all bags have the same number of beans or when there are only one or two bags. Explain your thought process clearly while coding; it shows your understanding and helps the interviewer follow your logic.

What are common mistakes when solving Removing Minimum Number of Magic Beans?

Not considering the sorted order of bags, which can lead to incorrect calculations of removals. Failing to optimize the approach, leading to unnecessary computations.

#2171

Removing Minimum Number of Magic Beans

Medium

ArrayGreedySortingEnumerationPrefix SumGreedySortingPrefix Sum

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

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

The optimal solution leverages sorting and prefix sums to efficiently calculate the minimum beans to remove. By sorting the bags, we can easily determine how many beans to remove for each potential target based on the number of bags that would remain non-empty.

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Algorithm

4 steps

1Step 1: Sort the array of beans in ascending order.
2Step 2: Calculate the prefix sums of the sorted array to quickly compute the total number of beans in the remaining bags.
3Step 3: Iterate through the sorted array, and for each bag, calculate the total beans to remove if we decide to keep all bags from this point onward.
4Step 4: Keep track of the minimum removal count.

solution.py8 lines

1def minimumBeans(beans):
2    beans.sort()
3    total_beans = sum(beans)
4    min_removal = float('inf')
5    for i in range(len(beans)):
6        removal = total_beans - (len(beans) - i) * beans[i]
7        min_removal = min(min_removal, removal)
8    return min_removal

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Complexity note: The time complexity is O(n log n) due to the sorting step, while the space complexity is O(1) since we are using a constant amount of extra space.

1Choosing the right target for equalization is crucial; it should be based on the bags with the least beans.
2Sorting helps in efficiently determining how many beans to remove, as it allows us to use prefix sums.

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