Minimum Operations to Make Array Equal to Target

Hard

Amazon

3 views

Topics:

Arrays

You are given two positive integer arrays nums and target, of the same length.

In a single operation, you can select any subarray of nums and increment each element within that subarray by 1 or decrement each element within that subarray by 1.

Return the minimum number of operations required to make nums equal to the array target.

For example:

nums = [3,5,1,2], target = [4,6,2,4]

Output: 2

Explanation:

We will perform the following operations to make nums equal to target:
- Increment nums[0..3] by 1, nums = [4,6,2,3]. The first element is incremented to 4, the second to 6, the third to 2, and the fourth to 3.
- Increment nums[3..3] by 1, nums = [4,6,2,4]. The fourth element is incremented to 4.
nums = [1,3,2], target = [2,1,4]

Output: 5

Explanation:

We will perform the following operations to make nums equal to target:
- Increment nums[0..0] by 1, nums = [2,3,2]
- Decrement nums[1..1] by 1, nums = [2,2,2]
- Decrement nums[1..1] by 1, nums = [2,1,2]
- Increment nums[2..2] by 1, nums = [2,1,3]
- Increment nums[2..2] by 1, nums = [2,1,4]

What is the most efficient algorithm to solve this problem?

Minimum Operations to Make Array Equal to Target

Hard

Amazon

3 views

Topics:

Arrays

You are given two positive integer arrays nums and target, of the same length.

In a single operation, you can select any subarray of nums and increment each element within that subarray by 1 or decrement each element within that subarray by 1.

Return the minimum number of operations required to make nums equal to the array target.

For example:

nums = [3,5,1,2], target = [4,6,2,4]

Output: 2

Explanation:

We will perform the following operations to make nums equal to target:
- Increment nums[0..3] by 1, nums = [4,6,2,3]. The first element is incremented to 4, the second to 6, the third to 2, and the fourth to 3.
- Increment nums[3..3] by 1, nums = [4,6,2,4]. The fourth element is incremented to 4.
nums = [1,3,2], target = [2,1,4]

Output: 5

Explanation:

We will perform the following operations to make nums equal to target:
- Increment nums[0..0] by 1, nums = [2,3,2]
- Decrement nums[1..1] by 1, nums = [2,2,2]
- Decrement nums[1..1] by 1, nums = [2,1,2]
- Increment nums[2..2] by 1, nums = [2,1,3]
- Increment nums[2..2] by 1, nums = [2,1,4]

What is the most efficient algorithm to solve this problem?

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 possible value ranges for the numbers in the array and the target?
Can the array be empty or null?
If it's impossible to make the array equal to the target, what should the function return?
What data type should I use for the array elements and the target (e.g., integers, floats)?
Are there any constraints on the number of operations allowed, or is the goal to find the absolute minimum regardless of the count?

Brute Force Solution

Approach

The goal is to make a collection of numbers match a desired target number. The brute force method tries every single possible combination of changes to the numbers, one at a time. It's like trying every possibility until we find the right one.

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

Start by trying to change the first number to the target value.
Then, explore changing the first two numbers, then the first three, and so on until we've changed them all.
For each set of changes, check if the entire collection of numbers now matches the target.
If it does match, record the changes we made.
Repeat this process, exploring all other possible changes to the numbers.
After we've checked all the possibilities, find the combination of changes that requires the fewest steps or the smallest total amount of change.

Code Implementation

def min_operations_brute_force(numbers, target):
    minimum_operations = float('inf')
    number_of_elements = len(numbers)

    for i in range(2 ** number_of_elements):
        current_operations = 0
        modified_numbers = []
        
        # Create a modified array based on the current bitmask
        for index in range(number_of_elements):
            if (i >> index) & 1:
                modified_numbers.append(target)
                current_operations += abs(numbers[index] - target)

            else:
                modified_numbers.append(numbers[index])

        # If all elements are target, update the min operations
        if all(number == target for number in modified_numbers):

            # This is where we track the smallest number of operations
            minimum_operations = min(minimum_operations, current_operations)

    if minimum_operations == float('inf'):
        return -1
    else:
        return minimum_operations

Big(O) Analysis

Time Complexity

O(2^n) – The described brute force approach explores all possible subsets of the input array of size n. For each element, we either choose to modify it towards the target or not. This binary decision for each element results in 2^n possible combinations of modifications to check. Therefore, the algorithm’s runtime grows exponentially with the input size n. Checking if a particular combination achieves the target takes O(n) time, which is dominated by the 2^n factor.

Space Complexity

O(N) – The described brute force approach implicitly explores all possible subsets of changes to the array of N numbers. While the explanation doesn't explicitly state the use of a separate data structure to store these combinations, the process of 'recording the changes we made' suggests the need to store at least one such combination temporarily. In the worst-case, we may need to keep track of N changes, and the recursion stack depth could also reach N in the process of exploring different options. Therefore, the auxiliary space complexity scales linearly with the input size N, resulting in O(N) space.

Optimal Solution

Approach

This problem asks us to find the fewest steps to transform one set of numbers into another. The trick is to focus on the changes needed between neighboring numbers, rather than the entire set at once. By looking at differences, we can simplify the transformation process.

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

First, calculate the difference between each adjacent pair of numbers in the set of numbers.
Next, we want to count how many positive changes versus negative changes are required.
Add up all of the positive changes. This is how many additions will be required to change the numbers.
The total amount of positive changes represents the minimum number of operations to make the initial set equal to the target set.

Code Implementation

def min_operations(numbers): 
    operations_count = 0
    
    # Need at least two numbers to calculate difference
    if len(numbers) < 2:
        return 0

    differences = []
    for i in range(1, len(numbers)): 
        differences.append(numbers[i] - numbers[i-1])

    # Sum all positive differences
    for difference in differences:
        # Only count positive changes
        if difference > 0:
            operations_count += difference
            
    return operations_count

Big(O) Analysis

Time Complexity

O(n) – The solution iterates through the input array once to calculate the differences between adjacent elements. It then iterates through these differences to sum up the positive changes. Both iterations are linear with respect to the input size n, where n is the number of elements in the input array. Therefore, the overall time complexity is O(n).

Space Complexity

O(1) – The solution calculates differences between adjacent pairs and sums positive changes. It uses only a few constant space variables to store these differences and the cumulative sum. Therefore, the auxiliary space used does not depend on the input size N (the number of elements in the array). The space complexity is constant.

Edge Cases

Case	How to Handle
Null or empty input array	Return 0 (or appropriate default indicating no operations needed) as there's nothing to process; check for null/empty at start.
Array with only one element	If the single element equals the target, return 0; otherwise return -1 indicating not possible; check length at start.
Array with very large numbers (potential integer overflow)	Use long data type to avoid integer overflow during intermediate calculations and final return.
Array with negative numbers	The algorithm should correctly handle negative numbers since addition and subtraction work as expected; no special treatment needed.
Array with zeros	Zero values should be handled correctly during comparisons and arithmetical operations without causing divide by zero or logical errors.
No solution exists to reach the target	The algorithm should return -1 (or appropriate indicator) when the target cannot be reached, after exhausting all possible operations.
All elements in the array are the same value	The algorithm still needs to iterate and potentially apply decrement/increment operations if the common value doesn't match the target.
Maximum-sized input array (performance)	The algorithm should have a time complexity that scales reasonably well with array size to avoid exceeding time limits (e.g., O(n) or O(n log n) is acceptable).