Removing Minimum Number of Magic Beans

Medium

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ArraysGreedy AlgorithmsTwo Pointers

You are given an array of positive integers beans, where each integer represents the number of magic beans found in a particular magic bag.

Remove any number of beans (possibly none) from each bag such that the number of beans in each remaining non-empty bag (still containing at least one bean) is equal. Once a bean has been removed from a bag, you are not allowed to return it to any of the bags.

Return the minimum number of magic beans that you have to remove.

Example 1:

Input: beans = [4,1,6,5]
Output: 4
Explanation: 
- We remove 1 bean from the bag with only 1 bean.
  This results in the remaining bags: [4,0,6,5]
- Then we remove 2 beans from the bag with 6 beans.
  This results in the remaining bags: [4,0,4,5]
- Then we remove 1 bean from the bag with 5 beans.
  This results in the remaining bags: [4,0,4,4]
We removed a total of 1 + 2 + 1 = 4 beans to make the remaining non-empty bags have an equal number of beans.
There are no other solutions that remove 4 beans or fewer.

Example 2:

Input: beans = [2,10,3,2]
Output: 7
Explanation:
- We remove 2 beans from one of the bags with 2 beans.
  This results in the remaining bags: [0,10,3,2]
- Then we remove 2 beans from the other bag with 2 beans.
  This results in the remaining bags: [0,10,3,0]
- Then we remove 3 beans from the bag with 3 beans. 
  This results in the remaining bags: [0,10,0,0]
We removed a total of 2 + 2 + 3 = 7 beans to make the remaining non-empty bags have an equal number of beans.
There are no other solutions that removes 7 beans or fewer.

Constraints:

1 <= beans.length <= 10⁵
1 <= beans[i] <= 10⁵

Removing Minimum Number of Magic Beans

Medium

Asked by:

1 view

Topics:

ArraysGreedy AlgorithmsTwo Pointers

You are given an array of positive integers beans, where each integer represents the number of magic beans found in a particular magic bag.

Remove any number of beans (possibly none) from each bag such that the number of beans in each remaining non-empty bag (still containing at least one bean) is equal. Once a bean has been removed from a bag, you are not allowed to return it to any of the bags.

Return the minimum number of magic beans that you have to remove.

Example 1:

Input: beans = [4,1,6,5]
Output: 4
Explanation: 
- We remove 1 bean from the bag with only 1 bean.
  This results in the remaining bags: [4,0,6,5]
- Then we remove 2 beans from the bag with 6 beans.
  This results in the remaining bags: [4,0,4,5]
- Then we remove 1 bean from the bag with 5 beans.
  This results in the remaining bags: [4,0,4,4]
We removed a total of 1 + 2 + 1 = 4 beans to make the remaining non-empty bags have an equal number of beans.
There are no other solutions that remove 4 beans or fewer.

Example 2:

Input: beans = [2,10,3,2]
Output: 7
Explanation:
- We remove 2 beans from one of the bags with 2 beans.
  This results in the remaining bags: [0,10,3,2]
- Then we remove 2 beans from the other bag with 2 beans.
  This results in the remaining bags: [0,10,3,0]
- Then we remove 3 beans from the bag with 3 beans. 
  This results in the remaining bags: [0,10,0,0]
We removed a total of 2 + 2 + 3 = 7 beans to make the remaining non-empty bags have an equal number of beans.
There are no other solutions that removes 7 beans or fewer.

Constraints:

1 <= beans.length <= 10⁵
1 <= beans[i] <= 10⁵

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 `beans` array and the range of values within it?
Can the input array `beans` be empty or null?
Are all the elements in the `beans` array guaranteed to be non-negative positive integers?
If all bags already contain the same number of beans, should I return 0?
Are we optimizing for the most efficient solution, considering the constraints on input size?

Brute Force Solution

Approach

The brute force approach involves trying out every single possibility of how many beans to remove. It's like we're testing out all scenarios until we find the one that requires removing the fewest beans. We do this by checking each possible target value that all the bean piles can be reduced to.

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

Consider each possible number of beans in the input as a potential target for the final number of beans in each pile.
For each target value, go through each bean pile.
Calculate how many beans would need to be removed from that pile to reach the target value.
Sum up the number of beans removed from all piles to reach the target value.
Remember this total number of removed beans.
After considering all possible target values, find the target value that resulted in removing the fewest beans overall.
Report this minimum number of beans to be removed.

Code Implementation

def remove_minimum_magic_beans_brute_force(magic_beans):
    minimum_removed_beans = float('inf')

    # Consider each number of beans as a potential target
    for potential_target in magic_beans:

        total_removed_beans = 0

        # Calculate the number of beans to remove for current target
        for bean_count in magic_beans:
            beans_to_remove = bean_count - potential_target

            # Only count removals if the target is less than the current pile
            if beans_to_remove > 0:
                total_removed_beans += beans_to_remove

        # Update min if current target requires fewer removals
        minimum_removed_beans = min(minimum_removed_beans, total_removed_beans)

    # The target of zero must be considered
    total_removed_beans_zero = sum(magic_beans)
    minimum_removed_beans = min(minimum_removed_beans, total_removed_beans_zero)

    return minimum_removed_beans

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through each of the n bean counts in the input array to consider it as a potential target. For each target value, it iterates through all n bean counts again to calculate the number of beans to remove to reach that target. This nested loop structure results in approximately n * n operations. Therefore, the time complexity is O(n²).

Space Complexity

O(1) – The provided algorithm iterates through potential target values and bean piles, calculating the number of beans to remove. It keeps track of the minimum number of removed beans seen so far. No auxiliary data structures that scale with the input size (N, the number of bean piles) are created; only a few variables for storing intermediate sums and the minimum value are needed. Thus, the space complexity is constant.

Optimal Solution

Approach

To minimize the magic beans removed, we want to keep as many beans as possible on the same level. The key idea is to find the most frequent level and then remove beans from all the other bags to match that level.

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

First, consider all the bags of beans. We will choose a single height and bring all other bags to this height.
For each bag, calculate the total number of beans that would be removed if we made all other bags match its bean count.
Choose the bag whose bean count requires removing the least number of beans across all the other bags.
Return the minimum number of beans removed that was computed in the previous step.

Code Implementation

def removing_minimum_number_of_magic_beans(
    magic_beans: list[int],
) -> int:
    total_beans = sum(magic_beans)
    magic_beans.sort()
    minimum_removed = float('inf')

    # Iterate through each bag to find the optimal height
    for i, current_bean_count in enumerate(magic_beans):
        # Calculate beans removed if all bags matched current height
        removed_beans_count = total_beans - (
            len(magic_beans) - i
        ) * current_bean_count

        # Update minimum if current removal is less
        minimum_removed = min(minimum_removed, removed_beans_count)

    return minimum_removed

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(n log n) – The dominant factor in the time complexity is sorting the input array of size n. This sorting step typically takes O(n log n) time using efficient algorithms like merge sort or quicksort. The subsequent iteration through the sorted array to calculate the removal cost for each unique bean count takes O(n) time. Since O(n log n) dominates O(n), the overall time complexity is O(n log n).

Space Complexity

O(1) – The algorithm iterates through the input array of magic beans and calculates the number of beans to remove for each bag. It only keeps track of the minimum number of beans removed so far, which is stored in a single variable. Therefore, the algorithm uses a constant amount of extra space regardless of the input size N, where N is the number of bags of beans.

Edge Cases

Case	How to Handle
Null or empty input array	Return 0 immediately since there are no beans to remove.
Input array with only one element	Return 0 because all bags already have the same number of beans.
Input array with all elements having the same value	Return 0 because no beans need to be removed.
Large input array (performance consideration)	Sort the array to efficiently determine the optimal target value; time complexity is O(n log n).
Integer overflow when calculating the total number of beans for removal	Use a 64-bit integer type (long in Java/C++) to store the sum to prevent potential overflow.
Input array containing large integer values that could lead to overflow when summing	Use long data type to prevent overflow when calculating the total sum and bean removals.
Array with very skewed distribution (one very large value compared to others)	The sorting and iterative calculation will still correctly identify the optimal target value.
Array contains zeros	The algorithm functions correctly with zeros, treating them as a valid number of beans in a bag; if zero is the target value all other beans will be removed.