Minimum Operations to Make Numbers Non-positive

Hard

Citadel

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Topics:

ArraysGreedy Algorithms

You are given two positive integer arrays nums1 and nums2.

In one operation, you can choose an index i and subtract 1 from both nums1[i] and nums2[i].

Return the minimum number of operations required to make all elements in at least one of nums1 or nums2 non-positive.

Example 1:

Input: nums1 = [2,3,4,2], nums2 = [1,5,2,4]
Output: 5
Explanation: One of the optimal solutions is to perform the following operations:
For the first index, we do the operation once, nums1[0] = 1 and nums2[0] = 0.
For the second index, we do the operation twice, nums1[1] = 1 and nums2[1] = 3.
For the third index, we do the operation once, nums1[2] = 3 and nums2[2] = 1.
For the fourth index, we do the operation once, nums1[3] = 1 and nums2[3] = 3.
In total we need 1 + 2 + 1 + 1 = 5 operations to make at least one of the arrays non-positive.

Example 2:

Input: nums1 = [2,4,3], nums2 = [2,2,3]
Output: 4
Explanation: One of the optimal solutions is to perform the following operations:
For the first index, we do the operation once, nums1[0] = 1 and nums2[0] = 1.
For the second index, we do the operation once, nums1[1] = 3 and nums2[1] = 1.
For the third index, we do the operation twice, nums1[2] = 1 and nums2[2] = 1.
In total we need 1 + 1 + 2 = 4 operations to make at least one of the arrays non-positive.

Constraints:

1 <= nums1.length == nums2.length <= 10⁵
1 <= nums1[i], nums2[i] <= 10⁹

Try coding it in LeetCode

Minimum Operations to Make Numbers Non-positive

Hard

Citadel

1 view

Topics:

ArraysGreedy Algorithms

You are given two positive integer arrays nums1 and nums2.

In one operation, you can choose an index i and subtract 1 from both nums1[i] and nums2[i].

Return the minimum number of operations required to make all elements in at least one of nums1 or nums2 non-positive.

Example 1:

Input: nums1 = [2,3,4,2], nums2 = [1,5,2,4]
Output: 5
Explanation: One of the optimal solutions is to perform the following operations:
For the first index, we do the operation once, nums1[0] = 1 and nums2[0] = 0.
For the second index, we do the operation twice, nums1[1] = 1 and nums2[1] = 3.
For the third index, we do the operation once, nums1[2] = 3 and nums2[2] = 1.
For the fourth index, we do the operation once, nums1[3] = 1 and nums2[3] = 3.
In total we need 1 + 2 + 1 + 1 = 5 operations to make at least one of the arrays non-positive.

Example 2:

Input: nums1 = [2,4,3], nums2 = [2,2,3]
Output: 4
Explanation: One of the optimal solutions is to perform the following operations:
For the first index, we do the operation once, nums1[0] = 1 and nums2[0] = 1.
For the second index, we do the operation once, nums1[1] = 3 and nums2[1] = 1.
For the third index, we do the operation twice, nums1[2] = 1 and nums2[2] = 1.
In total we need 1 + 1 + 2 = 4 operations to make at least one of the arrays non-positive.

Constraints:

1 <= nums1.length == nums2.length <= 10⁵
1 <= nums1[i], nums2[i] <= 10⁹

Try coding it in LeetCode

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 values of nums1, nums2, k1, and k2? Specifically, what are the minimum and maximum possible values for each?
Can the values in nums1 and nums2 be negative, zero, or only positive?
If it's impossible to decrement elements to fully use k1 and k2 decrements while keeping the numbers non-negative, what should the function return? Can k1 and k2 be zero?
Are the input arrays nums1 and nums2 guaranteed to be of the same length? If not, what should I do?
Can you provide a concrete example to illustrate how k1 and k2 decrement operations are distributed across the array elements to minimize the sum of squares?

Brute Force Solution

Approach

We need to make a list of numbers non-positive by adding or subtracting a value from each number. The brute force approach involves trying every single possible combination of additions and subtractions to see what works.

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

Consider the first number in the list.
Try adding the value to it, and also try subtracting the value from it.
For each of those two possibilities, move on to the second number in the list.
Again, try both adding and subtracting the value from the second number, creating new possibilities.
Continue this process for all the numbers in the list, exploring every single possible combination of adding or subtracting the value for each number.
After trying every combination, check each one to see how many operations (additions or subtractions) it took to make all the numbers non-positive.
Finally, select the combination that required the fewest operations to achieve the goal.

Code Implementation

def minimum_operations_brute_force(numbers):
    min_operations_count = float('inf')

    def solve(current_numbers, operations_count):
        nonlocal min_operations_count

        all_non_positive = all(number <= 0 for number in current_numbers)

        if all_non_positive:
            min_operations_count = min(min_operations_count, operations_count)
            return

        # If the current number of operations is already greater than the current minimum, then prune
        if operations_count >= min_operations_count:
            return

        for index in range(len(numbers)):
            if current_numbers[index] > 0:

                # Explore the possibility of adding 1
                next_numbers_add = current_numbers[:]
                next_numbers_add[index] += 1
                solve(next_numbers_add, operations_count + 1)

                # Explore the possibility of subtracting 1
                next_numbers_subtract = current_numbers[:]
                next_numbers_subtract[index] -= 1
                solve(next_numbers_subtract, operations_count + 1)

    solve(numbers, 0)
    return min_operations_count

Big(O) Analysis

Time Complexity

O(2^n) – The algorithm explores all possible combinations of adding or subtracting a value for each of the n numbers in the input list. For each number, there are two choices (add or subtract). This leads to 2 * 2 * ... * 2 (n times) possibilities, which is 2^n. The algorithm then checks each of these 2^n combinations to see if they make all numbers non-positive and counts the operations. Therefore, the time complexity is O(2^n).

Space Complexity

O(2^N) – The provided brute-force approach explores every combination of adding or subtracting a value from each number in the list. This implies a branching factor of 2 for each of the N numbers, resulting in a recursion tree with a depth of N and potentially 2^N paths to explore. Each path represents a potential combination of operations that needs to be stored, likely through recursive call stacks, to determine the minimum number of operations. Therefore, the space complexity is proportional to the number of possible combinations, leading to an auxiliary space of O(2^N).

Optimal Solution

Approach

The problem asks for the fewest actions to make all numbers non-positive. The key idea is to realize each number can be made non-positive independently and then we just add up the minimum number of operations for each. For each number, the optimal strategy involves incrementing the special value (if it helps) until that number becomes non-positive.

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

For each number in the set, figure out how many times you need to increment the special value to make that number non-positive.
If the special value starts at zero or a positive number, you don't need to do anything special to it to help make the number non-positive. You can just directly determine the minimum increments needed.
If the special value is a negative number, incrementing it can potentially lower the needed increments. Keep incrementing it until it reaches zero or the point where additional increments would not lower the amount of total operations needed.
Add up the minimum increments you found for each number. This total is the minimum number of operations needed for the entire set.

Code Implementation

def min_operations_to_non_positive(numbers, reduction_factor):
    total_operations = 0

    for number in numbers:
        # Numbers already non-positive need no operations.
        if number <= 0:
            continue

        # Find the minimum operations to make number non-positive.
        operations_needed = (number + reduction_factor - 1) // reduction_factor
        total_operations += operations_needed

    return total_operations

Big(O) Analysis

Time Complexity

O(n*m) – The algorithm iterates through each of the n numbers in the input set. For each number, in the worst case, it might need to increment the special value multiple times until it reaches zero or the point where further increments no longer reduce the number of operations needed to make the number non-positive. If we denote the maximum number of increments needed for the special value as m, then the inner loop runs m times for each of the n numbers. Therefore, the overall time complexity is O(n*m).

Space Complexity

O(1) – The algorithm iterates through the input numbers and calculates increments for each number independently, without storing intermediate results in any data structure that scales with the input size. Only a few constant space variables (such as the current number, the special value, and the increment count) are used within the loop. The space used does not depend on the number of input values (N). Therefore, the auxiliary space complexity is constant.

Edge Cases

Case	How to Handle
nums1 or nums2 is null or empty	Return 0 since the sum of squares will be 0 for an empty array.
k1 or k2 is zero	The function should still correctly compute the sum of squares without any decrement operations performed, using the original arrays.
k1 or k2 is larger than the sum of the elements in nums1 or nums2 respectively	Clamp the decremented value to 0, preventing negative array elements and invalid calculations.
All elements in nums1 and nums2 are already zero	The function should return 0, as no operations are needed, and the sum of squares is already minimized at 0.
Large input arrays and large values of k1 and k2 causing integer overflow during calculation of sum of squares	Use long long data type to store intermediate and final results to prevent overflow.
Arrays containing negative numbers	Clamp the values to zero during decrement if they become negative, which ensures proper processing, since we can't decrement a number below 0.
k1 + k2 is larger than the sum of all elements in nums1 and nums2	Clamp the decremented values to 0 so the sum of squares calculation can handle this scenario correctly by forcing non-negative final array elements.
The optimal solution requires decrementing some elements in nums1 to zero and other elements in nums2 to zero.	The solution should use a priority queue to always decrement the largest elements until k1 and k2 are exhausted, correctly finding the minimum sum of squares.