Taro Logo

Minimize Rounding Error to Meet Target

Medium
Asked by:
Profile picture
18 views
Topics:
Greedy Algorithms

You are given an array of strings prices where each string prices[i] is a price to a stock in the format $x.xx. The absolute error between a price to its nearest cent is no more than 0.005.

Return the sum of all prices after rounding to the nearest cent. Round each price separately. The answer should be represented as a string rounded to two decimal places.

Example 1:

Input: prices = ["$0.70", "$2.80", "$4.90"]
Output: "7.40"
Explanation: 
- $0.70 rounded to $0.70 is $0.70.
- $2.80 rounded to $2.80 is $2.80.
- $4.90 rounded to $4.90 is $4.90.
The sum is $0.70 + $2.80 + $4.90 = $8.40.

Example 2:

Input: prices = ["$1.50", "$2.50", "$3.50"]
Output: "8.00"
Explanation: 
- $1.50 rounded to $1.00 is $1.00.
- $2.50 rounded to $3.00 is $3.00.
- $3.50 rounded to $4.00 is $4.00.
The sum is $1.00 + $3.00 + $4.00 = $8.00.

Example 3:

Input: prices = ["$1.20", "$2.50", "$3.50"]
Output: "7.20"
Explanation: 
- $1.20 rounded to $1.00 is $1.00.
- $2.50 rounded to $2.00 is $2.00.
- $3.50 rounded to $4.00 is $4.00.
The sum is $1.00 + $2.00 + $4.00 = $7.00.

Constraints:

  • 1 <= prices.length <= 1000
  • prices[i] is in the format $x.xx where x is an integer and xx is between 00 and 99.
  • The absolute error between a price to its nearest cent is no more than 0.005.

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 range of values for the floating-point numbers in the `prices` array, and what is the data type used to represent them (e.g., float, double)?
  2. What are the constraints on the value of `target`? Is `target` guaranteed to be non-negative?
  3. Is the input array `prices` guaranteed to be non-empty, or should I handle the case of an empty input?
  4. If there are multiple ways to round the numbers that result in the same minimum absolute difference, is any one of them acceptable, or is there a specific rounding strategy I should prioritize?
  5. If it's impossible to reach the `target` value by rounding the numbers, what should I return? Should I return a specific error code, or perhaps `float('inf')`?

Brute Force Solution

Approach

The brute force approach attempts every possible combination of rounding up or rounding down each number. For each combination, we calculate the total rounding error and check if the sum of the rounded numbers matches the target. We then choose the combination with the smallest total rounding error.

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

  1. Consider the first number. We can either round it up or round it down.
  2. For each of those choices (rounding the first number up or down), consider the second number. Again, we can round it up or down. This creates two possibilities for each initial choice, so now we have four possibilities total.
  3. Continue this process for every number in the list. Each number doubles the number of possibilities we need to check.
  4. For each complete combination of rounding choices, calculate the sum of the rounded numbers.
  5. If the sum of the rounded numbers does not equal the target, then that entire rounding combination is invalid. Discard it and move on.
  6. If the sum of the rounded numbers equals the target, calculate the total rounding error for that combination. This is done by adding up the individual rounding errors for each number (the difference between the original number and the rounded number).
  7. Keep track of the smallest rounding error found so far. If a new combination has a smaller error, replace the current smallest error with the new one.
  8. After checking all possible combinations, the smallest rounding error that was found is the answer.

Code Implementation

def minimize_rounding_error_brute_force(prices, target):    minimum_error = float('inf')
    number_of_prices = len(prices)
    # Iterate through all possible rounding combinations.
    for i in range(2 ** number_of_prices):
        current_sum = 0.0
        current_error = 0.0
        rounded_numbers = []
        # Determine rounding for each price.
        for price_index in range(number_of_prices):
            if (i >> price_index) & 1:
                rounded_price = int(prices[price_index]) + 1
                current_error += abs(prices[price_index] - rounded_price)
                current_sum += rounded_price
                rounded_numbers.append(rounded_price)
            else:
                rounded_price = int(prices[price_index])
                current_error += abs(prices[price_index] - rounded_price)
                current_sum += rounded_price
                rounded_numbers.append(rounded_price)
        # Check if the rounded sum matches the target.
        if abs(current_sum - target) < 1e-6:
            # Update minimum error if necessary. 
            minimum_error = min(minimum_error, current_error)
    # Prevent returning infinity if no valid combination exists
    if minimum_error == float('inf'):
        return -1.0
    else:
        return minimum_error

Big(O) Analysis

Time Complexity
O(2^n)The algorithm explores all possible rounding combinations for n numbers, where each number can be rounded either up or down. This results in 2 possible choices for each of the n numbers, leading to 2 * 2 * ... * 2 (n times) which equals 2^n total combinations. For each combination, we iterate through all n numbers to calculate the sum of rounded values and the rounding error. Therefore, the time complexity is dominated by the exploration of all 2^n combinations, making it O(2^n).
Space Complexity
O(1)The brute force approach, as described, doesn't explicitly mention the creation of any auxiliary data structures like arrays, lists, or hash maps to store intermediate results. It iterates and compares values, keeping track of the smallest rounding error found so far using a single variable. Therefore, the extra space used remains constant, independent of the input size N (the number of numbers). So the space complexity is O(1).

Optimal Solution

Approach

The goal is to change some numbers to their closest whole numbers so that their total equals a target. Some numbers must be rounded up, and some must be rounded down, and we want to minimize the total error from rounding. The optimal approach prioritizes which numbers to round up or down based on how close they are to the nearest integers.

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

  1. Figure out how much you need to adjust the total by rounding all numbers down. This tells you how many numbers need to be rounded up.
  2. Calculate the fractional part of each number. This tells you how close each number is to the next higher integer.
  3. Sort the numbers based on their fractional parts, from largest to smallest. Numbers with larger fractional parts are closer to the next integer and should be rounded up first to minimize error.
  4. Round the top numbers (based on the sorted list) up and the rest down. This achieves the target with the smallest possible error.

Code Implementation

def minimize_error(numbers, target):
    total_down = 0
    fractional_parts = []

    for number in numbers:
        down = int(number)
        total_down += down
        fractional_part = number - down
        fractional_parts.append(fractional_part)

    # Calculate how many numbers need rounding up.
    up_count = target - total_down

    if up_count < 0 or up_count > len(numbers):
        return "-1"

    # Sort by fractional part to minimize error.
    sorted_indices = sorted(range(len(numbers)), key=lambda i: fractional_parts[i], reverse=True)

    total_error = 0.0

    # Round numbers based on sorted indices.
    for i in range(len(numbers)):
        if i < up_count:
            # Round up numbers with the largest fractional parts first.
            total_error += (1 - fractional_parts[sorted_indices[i]])

        else:
            # Round down the remaining numbers.
            total_error += fractional_parts[sorted_indices[i]]

    return "{:.3f}".format(total_error)

Big(O) Analysis

Time Complexity
O(n log n)The algorithm iterates through the n numbers to calculate the initial difference and the fractional parts, which takes O(n) time. Sorting the fractional parts dominates the time complexity. Sorting n numbers using an efficient algorithm like merge sort or quicksort takes O(n log n) time. Finally, another O(n) loop is used to perform the rounding based on the sorted indices. Therefore, the overall time complexity is dominated by the sorting step, resulting in O(n log n).
Space Complexity
O(N)The algorithm calculates the fractional part of each number, requiring an auxiliary array of size N to store these fractional parts, where N is the number of input numbers. Then, this array is sorted, either in place or requiring additional space proportional to N depending on the sorting algorithm implementation. Therefore the space complexity is O(N).

Edge Cases

prices is null or empty
How to Handle:
Return target if prices is null or empty, as no rounding can be performed.
target is negative
How to Handle:
The rounding error should be calculated using the absolute difference, so negative target is acceptable.
prices contains strings that cannot be parsed as floating-point numbers
How to Handle:
Throw an exception or return an error code indicating invalid input.
prices contains extremely large floating-point numbers that could lead to overflow issues
How to Handle:
Ensure that intermediate calculations do not exceed the maximum representable value of the used data type (e.g., double).
The sum of the floors/ceilings of all prices is already equal to the target
How to Handle:
Return 0 in this case, as no rounding error is necessary.
Precision issues in floating-point arithmetic might cause slight inaccuracies in the error calculation
How to Handle:
Use a small epsilon value (e.g., 1e-9) to compare floating-point numbers for equality to mitigate precision errors.
The problem scales poorly with a large number of prices if not optimized correctly
How to Handle:
Employ dynamic programming to efficiently store and reuse intermediate results, avoiding exponential time complexity.
The target is much larger or smaller than the sum of the rounded values no matter the rounding decisions
How to Handle:
The absolute difference calculation handles these extreme target values correctly.