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Minimum Time to Make Array Sum At Most x

Hard
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Jane Street
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Topics:
ArraysDynamic Programming

You are given two 0-indexed integer arrays nums1 and nums2 of equal length. Every second, for all indices 0 <= i < nums1.length, value of nums1[i] is incremented by nums2[i]. After this is done, you can do the following operation:

  • Choose an index 0 <= i < nums1.length and make nums1[i] = 0.

You are also given an integer x.

Return the minimum time in which you can make the sum of all elements of nums1 to be less than or equal to x, or -1 if this is not possible.

Example 1:

Input: nums1 = [1,2,3], nums2 = [1,2,3], x = 4
Output: 3
Explanation: 
For the 1st second, we apply the operation on i = 0. Therefore nums1 = [0,2+2,3+3] = [0,4,6]. 
For the 2nd second, we apply the operation on i = 1. Therefore nums1 = [0+1,0,6+3] = [1,0,9]. 
For the 3rd second, we apply the operation on i = 2. Therefore nums1 = [1+1,0+2,0] = [2,2,0]. 
Now sum of nums1 = 4. It can be shown that these operations are optimal, so we return 3.

Example 2:

Input: nums1 = [1,2,3], nums2 = [3,3,3], x = 4
Output: -1
Explanation: It can be shown that the sum of nums1 will always be greater than x, no matter which operations are performed.

Constraints:

  • 1 <= nums1.length <= 103
  • 1 <= nums1[i] <= 103
  • 0 <= nums2[i] <= 103
  • nums1.length == nums2.length
  • 0 <= x <= 106

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:

  1. What are the constraints on the size of the input arrays `nums1` and `nums2`, and what is the range of values for the elements within these arrays and for `x`?
  2. Can the values in `nums1`, `nums2`, or `x` be negative, zero, or floating-point numbers?
  3. If it's impossible to make the sum of elements in `nums1` at most `x` after any number of operations, should I return -1 as specified, or is there another error code/exception I should raise?
  4. If multiple operation counts satisfy the condition (sum of nums1 <= x), should I return any one of them, or specifically the minimum one?
  5. Is it guaranteed that the lengths of `nums1` and `nums2` are equal, and are empty arrays permitted as input?

Brute Force Solution

Approach

The brute force strategy is like trying every single possible way to solve the problem. We will explore every combination of actions until we find the one that works within the given limitations.

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

  1. Consider doing nothing at all and see if the array's sum is already small enough.
  2. Consider performing each individual operation once, and check if the array's sum is small enough after each operation.
  3. Consider performing each pair of operations (any two), and check if the sum is small enough.
  4. Consider performing any three operations, and check the sum.
  5. Keep doing this, trying all possible combinations of operations, until you've tried every possible scenario.
  6. For each scenario, calculate how much time each action took.
  7. From all the scenarios where the array sum is small enough, pick the one that took the least total time.

Code Implementation

def min_time_to_make_array_sum_at_most_x_brute_force(array, operations, target_sum):

    min_time = float('inf')

    for i in range(1 << len(operations)):
        current_sum = sum(array)
        current_time = 0
        operations_performed = []

        # Iterate through each operation to determine if its included in the current subset
        for operation_index in range(len(operations)):
            if (i >> operation_index) & 1:
                operations_performed.append(operation_index)

        # Apply chosen operations
        for operation_index in operations_performed:
            current_sum -= operations[operation_index][0]
            current_time += operations[operation_index][1]

        # Check if the array sum is at most x
        if current_sum <= target_sum:

            # Update min_time if necessary
            min_time = min(min_time, current_time)

    # Handle the case where no combination meets the target sum
    if min_time == float('inf'):
        return -1
    else:
        return min_time

Big(O) Analysis

Time Complexity
O(2^n)The algorithm explores all possible subsets of operations. Given an array of n elements, there are 2^n possible subsets (each element can either be included or excluded in a subset). For each subset, we calculate the sum of the modified array and the time taken to perform the operations in that subset. Therefore, the time complexity is dominated by the generation and evaluation of all 2^n subsets.
Space Complexity
O(N)The brute force approach considers all possible combinations of operations. In the worst-case, it might need to store a combination of size N, where N is the number of elements in the input array. This implies allocating space to keep track of which operations are included in the current combination being evaluated. Thus the space complexity is O(N).

Optimal Solution

Approach

This problem asks us to find the least amount of time to make the total of numbers in a collection less than a certain value. The key insight is to prioritize reducing larger values first, especially since each value decreases over time.

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

  1. Imagine each number in the collection as a container gradually draining. Some drain faster than others.
  2. Figure out how much each container drains per unit of time. This is the rate of decrease for each number.
  3. Identify which numbers should be reduced first to make the sum decrease as quickly as possible. Focus on numbers that reduce a lot each unit of time.
  4. Consider sorting the numbers based on how quickly they decrease, placing those that decrease fastest at the beginning.
  5. After sorting, consider all possible groups of these 'fastest decreasing' numbers, starting with just the very fastest, then the two fastest, and so on.
  6. For each group, calculate the minimum amount of time it would take to make the sum of all numbers in the initial collection (the original containers) less than the required value.
  7. Choose the group that achieves this target sum in the shortest amount of time. This is your optimal solution.

Code Implementation

def min_time_to_make_array_sum_at_most_x(numbers, reduction_rates, target_sum):
    number_count = len(numbers)
    combined_array = list(zip(numbers, reduction_rates))

    # Sort based on reduction rate to prioritize larger reductions.
    combined_array.sort(key=lambda x: x[1], reverse=True)

    minimum_time = float('inf')

    for i in range(number_count + 1):
        # Iterate through all possible group sizes.
        current_numbers = []
        current_reduction_rates = []

        for j in range(i):
            current_numbers.append(combined_array[j][0])
            current_reduction_rates.append(combined_array[j][1])

        low = 0
        high = 10**18
        best_time = float('inf')

        while low <= high:
            mid = (low + high) // 2
            current_sum = 0

            for k in range(number_count):
                initial_number = numbers[k]
                reduction_rate = reduction_rates[k]

                current_sum += max(0, initial_number - reduction_rate * mid)

            # Check if the target sum condition is met.
            if current_sum <= target_sum:
                best_time = mid
                high = mid - 1
            else:
                low = mid + 1

        minimum_time = min(minimum_time, best_time)

    return minimum_time

Big(O) Analysis

Time Complexity
O(n log n)The initial step involves sorting the input array of size n based on the rate of decrease, which takes O(n log n) time. We then iterate through all possible groups of 'fastest decreasing' numbers, from size 1 up to size n. Inside this loop, calculating the time required to achieve the target sum takes O(n) time, as we need to sum up the original array. This results in a cost dominated by O(n log n) for the sorting step since subsequent steps are multiplied by n in the worst-case, but are upper bounded by the sorting time itself, hence O(n log n).
Space Complexity
O(N)The described solution sorts the input based on the rate of decrease, implying an auxiliary data structure, likely an array or list, to store the sorted indices or elements themselves. This sorting operation typically requires O(N) space. Furthermore, the algorithm considers all possible groups of 'fastest decreasing' numbers, starting from one element up to N elements, implying that the indices of these elements may be stored temporarily. Thus, the auxiliary space is dominated by the sorting operation, resulting in O(N) space complexity, where N is the number of elements in the input array.

Edge Cases

CaseHow to Handle
nums1 or nums2 is null or emptyReturn 0 if the initial sum of nums1 is already <= x, otherwise return -1 because there's nothing to increment.
n = 1 (single element in the arrays)Calculate the minimum time required for the single element to be <= x, handling the case where nums2[0] is zero.
Large n (maximum size input array) - performance concernThe solution must use an efficient algorithm like dynamic programming or a greedy approach with sorting to avoid exceeding time limits for large inputs.
All elements in nums2 are zeroIf the sum of nums1 is initially <= x, return 0; otherwise, return -1 since the values in nums1 cannot be changed.
Elements in nums1 and nums2 are large positive numbers, leading to potential integer overflow during summationUse appropriate data types (e.g., long long in C++, long in Java/Python) to prevent integer overflow during sum calculations.
Some elements in nums2 are negativeElements in nums2 can be negative, however, we should avoid choosing such an index since this is a minimization problem, in practice, we can filter out such cases at the start, and solve the problem for all the other indices.
Initial sum of nums1 is already less than or equal to xReturn 0 immediately because no operations are required.
It is not possible to make the sum of nums1 less than or equal to xReturn -1 when all elements in nums2 are non-negative and the initial sum of nums1 is greater than x; or there are negative elements in nums2 but even by repeatedly applying the negative increments, it is still not possible to bring the sum below or equal to x.