Leetcode 2171: Removing Minimum Number of Magic Beans

Input: The input consists of an array of integers, where each integer represents the number of beans in a particular magic bag.

Example: [8, 3, 7, 5]

Constraints:

• 1 <= beans.length <= 10^5

• 1 <= beans[i] <= 10^5

Output: The output should be a single integer representing the minimum number of beans that need to be removed.

Example: 7

Constraints:

• The output must be an integer value representing the minimum beans to be removed.

Goal: The goal is to remove the minimum number of beans while ensuring that all remaining non-empty bags have the same number of beans.

Steps:

• Calculate the sum of all beans in the array.

• Sort the array to facilitate efficient calculation of removals.

• For each possible number of beans to keep in each bag, compute the total removals required, and track the minimum removals.

Goal: The input has constraints on the length of the array and the size of the integers in the array.

Steps:

• 1 <= beans.length <= 10^5

• 1 <= beans[i] <= 10^5

Assumptions:

• The number of bags (beans.length) is at least 1.

• The integers representing the beans in each bag are positive.

• Input: [8, 3, 7, 5]

• Explanation: In this example, the optimal solution is to remove 7 beans in total by making the bags have 3 beans each.

Approach: The approach is to calculate the total sum of beans and explore the different possible configurations of bags with equal numbers of beans.

Observations:

• Sorting the array allows efficient calculation of the total number of beans to be removed.

• The problem involves iterating over the bags to find the most efficient way of removing beans.

Steps:

• Calculate the sum of all the beans in the array.

• Sort the array to identify the optimal target number of beans per bag.

• For each possible number of beans to keep, calculate the total number of removals needed.

Empty Inputs:

• The array will not be empty as per the constraints.

Large Inputs:

• Ensure the solution handles large arrays efficiently with lengths up to 10^5.

Special Values:

• Consider cases where all the bags contain the same number of beans.

Constraints:

• Ensure the solution works within the provided input constraints.

long long minimumRemoval(vector<int>& beans) {
    int n = beans.size();
    long long sum = accumulate(begin(beans), end(beans), 0L);
    sort(beans.begin(), beans.end());
    long long res = LLONG_MAX;
    for (int i = 0; i < n; i++)
        res = min(res, sum - (long long) (n - i) * beans[i]);
    
    return res;
}

1 : Function Definition

long long minimumRemoval(vector<int>& beans) {

Define the function `minimumRemoval`, which takes a vector of integers `beans` and returns a long long integer. This function aims to calculate the minimum beans that need to be removed.

2 : Variable Initialization

    int n = beans.size();

Initialize the variable `n` to the size of the `beans` vector, representing the total number of elements in the vector.

3 : Sum Calculation

    long long sum = accumulate(begin(beans), end(beans), 0L);

Use the `accumulate` function to calculate the total sum of the `beans` vector. The initial value of the sum is 0L, and the result is stored in `sum`.

4 : Sorting

    sort(beans.begin(), beans.end());

Sort the `beans` vector in ascending order. Sorting helps in calculating the minimum removals efficiently.

5 : Variable Initialization

    long long res = LLONG_MAX;

Initialize the variable `res` to `LLONG_MAX`, representing the minimum removal result, which will be updated during the iteration.

6 : Loop Start

    for (int i = 0; i < n; i++)

Start a loop that iterates through each element in the `beans` vector. The variable `i` tracks the current index.

7 : Update Minimum Removal

        res = min(res, sum - (long long) (n - i) * beans[i]);

For each iteration, update `res` with the minimum value between the current `res` and the difference between `sum` and the calculated removal cost for the current index `i`.

8 : Return Result

    return res;

Return the final result, which is the minimum number of beans that need to be removed.

Best Case: O(n log n)

Average Case: O(n log n)

Worst Case: O(n log n)

Description: The time complexity is dominated by the sorting step, which takes O(n log n).

Best Case: O(1)

Worst Case: O(n)

Description: The space complexity is O(n) due to the space required for sorting the array.

Leetcode 2171: Removing Minimum Number of Magic Beans

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