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Minimum Operations to Reduce X to Zero

Medium

Minimum Operations to Reduce X to Zero

Medium

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 ranges for the numbers in the `nums` array and the value of `x`? Can they be negative or zero?
If it's impossible to reduce `x` to zero using the given operations, what should I return?
Are there any constraints on the size of the `nums` array? Should I be concerned about exceeding memory limits with certain approaches?
If there are multiple possible solutions (combinations of operations that minimize the number of operations), is any one acceptable, or is there a specific solution I should prioritize (e.g., leftmost elements)?
Are all the elements in `nums` integers?

Brute Force Solution

Approach

The brute force approach to this problem is all about trying every single possible combination. We'll explore every possible number of elements from the start and end of the given bunch of numbers and see if they add up to the target number. We'll keep track of the fewest elements needed to reach the target.

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

First, consider using no elements from the start or the end.
Next, consider using only the first element, then only the last element.
Then, try using the first two elements, the last two elements, the first element and the last element, and so on.
Keep going, trying every combination of elements from the start and the end.
For each combination, check if the sum of the selected elements equals the target number.
If it does equal the target number, remember how many elements you used to reach the target.
After trying all the combinations, find the smallest number of elements you needed to reach the target. That will be your answer.

Code Implementation

def min_operations_to_reduce_x_to_zero_brute_force(numbers, target):
    minimum_operations = float('inf')
    array_length = len(numbers)

    for left_elements in range(array_length + 1):
        for right_elements in range(array_length - left_elements + 1):
            #Ensures that left and right elements do not exceed total elements
            if left_elements + right_elements > array_length:
                continue

            current_sum = 0
            # Calculate the sum of elements from left and right sides
            for i in range(left_elements):
                current_sum += numbers[i]

            for i in range(right_elements):
                current_sum += numbers[array_length - 1 - i]

            # Check if the current sum equals the target
            if current_sum == target:
                minimum_operations = min(minimum_operations, left_elements + right_elements)

    #If no solution is found return -1 as per prompt
    if minimum_operations == float('inf'):
        return -1
    else:
        return minimum_operations

Big(O) Analysis

Time Complexity

O(2^n) – The provided brute force approach explores all possible combinations of elements from the start and end of the array. In the worst-case scenario, we essentially generate all possible subsets of the input array. Since an array of size n has 2^n subsets, the algorithm checks the sum of elements in each subset to see if it equals x. Therefore, the time complexity is directly proportional to the number of subsets, resulting in O(2^n).

Space Complexity

O(1) – The described brute force approach doesn't explicitly allocate any auxiliary data structures like lists, arrays, or hash maps. It primarily involves iterating through different combinations of elements from the input array. Therefore, the space complexity is determined by the storage of a few integer variables (e.g., indices, counters) to track the current combination's sum and length. These variables occupy a constant amount of space regardless of the input size N (the number of elements in the array). Hence, the auxiliary space complexity is O(1).

Optimal Solution

Approach

Instead of figuring out all the ways to remove numbers, we find the longest continuous section that adds up to a specific target. This target is calculated by understanding what's left when you subtract X from the total sum of all numbers. By finding the longest continuous section summing to this target, we know the fewest numbers were removed from the outside.

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

Calculate the total sum of all the numbers.
Determine the target sum by subtracting X from the total sum.
If the target sum is negative, it's impossible to reach zero; return -1.
Use a sliding window approach to find the longest continuous section of numbers that adds up to the target sum.
Start with an empty window and gradually expand it by including numbers from the beginning.
If the sum of numbers in the window is less than the target, keep expanding it.
If the sum of numbers in the window is greater than the target, shrink the window from the beginning.
If the sum of numbers in the window equals the target, record the length of this section.
Keep adjusting the window until you've gone through all the numbers.
The longest section found represents the most numbers we can keep, which means the fewest removed.
Subtract the length of the longest section from the total number of elements to get the minimum operations.

Code Implementation

def min_operations_to_reduce_x_to_zero(numbers, x):
    total_sum = sum(numbers)
    target_sum = total_sum - x

    if target_sum < 0:
        return -1

    maximum_length = -1
    current_sum = 0
    window_start = 0

    for window_end in range(len(numbers)):
        current_sum += numbers[window_end]

        # Shrink the window until current sum is no longer greater than target
        while current_sum > target_sum:
            current_sum -= numbers[window_start]
            window_start += 1

        # Update max length if the current window's sum equals the target
        if current_sum == target_sum:

            current_length = window_end - window_start + 1
            maximum_length = max(maximum_length, current_length)

    #If maximum_length is -1, that means target not found.
    if maximum_length == -1:
        return -1

    # Minimum operations is the original length minus the longest sub-array length.
    minimum_operations = len(numbers) - maximum_length

    return minimum_operations

Big(O) Analysis

Time Complexity

O(n) – The algorithm calculates the total sum in O(n) time. The sliding window approach iterates through the array using two pointers, left and right. Each element is visited at most twice (once by the right pointer expanding the window and once by the left pointer shrinking it). Therefore, the sliding window part of the algorithm also takes O(n) time. Thus the dominant time complexity is O(n).

Space Complexity

O(1) – The algorithm calculates the total sum which requires constant space. The sliding window approach uses a few integer variables (window start, window end, current sum, max length) to track the window's state. These variables consume a fixed amount of memory regardless of the input array's size, denoted as N. Therefore, the auxiliary space complexity is constant.

Edge Cases

Case	How to Handle
Empty array nums	Return -1 immediately as no operations are possible.
x is zero	Return 0 immediately, as no operations are needed.
nums contains only zeros and x > 0	Return -1, as no combination of zeros can reach a positive x.
Sum of all nums is less than x	Return -1, as no sequence of operations can reduce x to zero.
Sum of all nums equals x	Return the length of nums since taking all elements is a valid solution.
All elements in nums are identical and sum to greater than x, but no subset sums to x	Return -1, indicating no solution exists after checking all subsets of nums
Array contains negative numbers	This violates the problem constraint; return -1 or throw an exception indicating invalid input.
Integer overflow when calculating the sum of nums	Use long data type for intermediate sums to prevent potential overflow.

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 ranges for the numbers in the `nums` array and the value of `x`? Can they be negative or zero?
If it's impossible to reduce `x` to zero using the given operations, what should I return?
Are there any constraints on the size of the `nums` array? Should I be concerned about exceeding memory limits with certain approaches?
If there are multiple possible solutions (combinations of operations that minimize the number of operations), is any one acceptable, or is there a specific solution I should prioritize (e.g., leftmost elements)?
Are all the elements in `nums` integers?

Brute Force Solution

Approach

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

First, consider using no elements from the start or the end.
Next, consider using only the first element, then only the last element.
Then, try using the first two elements, the last two elements, the first element and the last element, and so on.
Keep going, trying every combination of elements from the start and the end.
For each combination, check if the sum of the selected elements equals the target number.
If it does equal the target number, remember how many elements you used to reach the target.
After trying all the combinations, find the smallest number of elements you needed to reach the target. That will be your answer.

Code Implementation

def min_operations_to_reduce_x_to_zero_brute_force(numbers, target):
    minimum_operations = float('inf')
    array_length = len(numbers)

    for left_elements in range(array_length + 1):
        for right_elements in range(array_length - left_elements + 1):
            #Ensures that left and right elements do not exceed total elements
            if left_elements + right_elements > array_length:
                continue

            current_sum = 0
            # Calculate the sum of elements from left and right sides
            for i in range(left_elements):
                current_sum += numbers[i]

            for i in range(right_elements):
                current_sum += numbers[array_length - 1 - i]

            # Check if the current sum equals the target
            if current_sum == target:
                minimum_operations = min(minimum_operations, left_elements + right_elements)

    #If no solution is found return -1 as per prompt
    if minimum_operations == float('inf'):
        return -1
    else:
        return minimum_operations

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

Calculate the total sum of all the numbers.
Determine the target sum by subtracting X from the total sum.
If the target sum is negative, it's impossible to reach zero; return -1.
Use a sliding window approach to find the longest continuous section of numbers that adds up to the target sum.
Start with an empty window and gradually expand it by including numbers from the beginning.
If the sum of numbers in the window is less than the target, keep expanding it.
If the sum of numbers in the window is greater than the target, shrink the window from the beginning.
If the sum of numbers in the window equals the target, record the length of this section.
Keep adjusting the window until you've gone through all the numbers.
The longest section found represents the most numbers we can keep, which means the fewest removed.
Subtract the length of the longest section from the total number of elements to get the minimum operations.

Code Implementation

def min_operations_to_reduce_x_to_zero(numbers, x):
    total_sum = sum(numbers)
    target_sum = total_sum - x

    if target_sum < 0:
        return -1

    maximum_length = -1
    current_sum = 0
    window_start = 0

    for window_end in range(len(numbers)):
        current_sum += numbers[window_end]

        # Shrink the window until current sum is no longer greater than target
        while current_sum > target_sum:
            current_sum -= numbers[window_start]
            window_start += 1

        # Update max length if the current window's sum equals the target
        if current_sum == target_sum:

            current_length = window_end - window_start + 1
            maximum_length = max(maximum_length, current_length)

    #If maximum_length is -1, that means target not found.
    if maximum_length == -1:
        return -1

    # Minimum operations is the original length minus the longest sub-array length.
    minimum_operations = len(numbers) - maximum_length

    return minimum_operations

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Empty array nums	Return -1 immediately as no operations are possible.
x is zero	Return 0 immediately, as no operations are needed.
nums contains only zeros and x > 0	Return -1, as no combination of zeros can reach a positive x.
Sum of all nums is less than x	Return -1, as no sequence of operations can reduce x to zero.
Sum of all nums equals x	Return the length of nums since taking all elements is a valid solution.
All elements in nums are identical and sum to greater than x, but no subset sums to x	Return -1, indicating no solution exists after checking all subsets of nums
Array contains negative numbers	This violates the problem constraint; return -1 or throw an exception indicating invalid input.
Integer overflow when calculating the sum of nums	Use long data type for intermediate sums to prevent potential overflow.