Problem Description

In this problem, we are given an array of prices and a target value. We are asked to round each price either up (ceil) or down (floor) to an integer such that the sum of the rounded prices equals the target. Our objective is to minimize the rounding error, which is defined as the sum of the absolute differences between the original prices and their rounded values.

To clarify with an example, if we have prices prices = [0.70, 2.80, 4.90] and a target of 8, our goal is to round these prices to whole numbers such as 1, 3, and 4, respectively, so that their sum is 8. The rounding error in this case is |1-0.70| + |3-2.80| + |4-4.90| = 0.30 + 0.20 + 0.10 = 0.60.

Yet, there might be cases where it is impossible to round the prices to reach the exact target sum. For instance, if we had the target 10 in the previous example, there’s no way to round 0.70, 2.80, 4.90 to sum up to 10.

Intuition

To arrive at the solution, a good starting point is to consider the minimum and maximum sums we can get by rounding all prices down (using floor) or up (using ceil), respectively. We calculate the total sum of the floored values and count the items that have a fraction (i.e., numbers that are not already integers). If the target is outside the range of the minimum sum and the maximum sum (minimum sum <= target <= maximum sum), we return "-1" as it is impossible to achieve the target by rounding.

Given that the target is within reach, we want to minimize the error by carefully choosing which numbers to round up and which to round down. Since this rounding must add up to the target, we need to round up a specific number of prices, corresponding to the difference between the target and the minimum sum we calculated earlier. To minimize the error, we should prioritize rounding up numbers with the smallest fractional parts, as doing so will lead to the smallest possible error.

The solution approach is as follows:

1. Calculate the total sum of floor values of all the prices.
2. Calculate how much we need to add to the floored sum to reach the target.
3. Sort the fractional parts of each number in descending order. By doing this, we can ensure that when rounding up, we always add the smallest necessary amount to our total sum.
4. Compute the error contributed by the numbers we need to round up (first d numbers after sorting) and the error contributed by the numbers we leave as floor values.
5. Return the sum of these errors as a formatted string with three decimal places.

Learn more about Greedy and Math patterns.

Solution Approach

The solution uses simple data manipulation techniques and sorting algorithm to achieve the desired minimum error value. Following is the detailed explanation of how the Python code implements the solution:

1. Initial Summation and Difference Collection The first step in the code iterates through the given prices and calculates the sum of their floor values. This is done by iterating each price p in the prices array, converting it to a float, and then adding its floor value to mi. Simultaneously, the code checks if the given price has a non-zero fractional part (i.e., p is not an integer) and, if so, appends the fractional part to the arr list by using d := p - int(p).

   This approach employs basic looping and arithmetic operations, and utilizes a list to collect all fractional parts for later processing.

2. Checking the Feasibility After initial summation and difference collection, we verify whether it's possible to hit the target by rounding. This is checked by the condition if not mi <= target <= mi + len(arr): return "-1". Here, mi denotes the sum of the floor values, and mi + len(arr) denotes the sum if we were to round all prices up. If the target is not within this range, the code determines it's not feasible to reach the target and immediately returns "−1".

3. Sorting and Error Calculation If the target is achievable, then the arr list, which contains all the differences between each price and its floor value, is sorted in descending order using arr.sort(reverse=True). This allows us to select the smallest fractional parts first when we round up, which minimizes the error.

4. Computing the Total Rounding Error The total rounding error is computed by taking into account the numbers that we have to round up to reach the target. The difference between the target and initial sum mi gives us d, the count of prices we must round up. The sum of rounding errors for prices rounded up is sum(arr[:d]), and for those rounded down is sum(arr[d:]). We subtract the first from d and add the second to find the total error.

5. Formatting the Result Finally, the code computes the answer and formats it to 3 decimal places using the formatted string f'{ans:.3f}', which completes our rounding error calculation.

This solution is efficient as it makes use of the properties of numbers and their rounding behaviors, along with sorting, which runs in O(n log n) time complexity for Python's Timsort algorithm, where n is the number of prices given.

The primary data structure utilized here is the list, for collecting the fractional parts and later sorting them. The use of list slicing and aggregation functions like sum() makes the code concise and efficient.

The given solution elegantly combines simple mathematical observation with a well-known sorting algorithm to solve what initially appears to be a complex problem. This implementation underlines the effectiveness of combining fundamental programming constructs and algorithms to solve practical computational problems.

Example Walkthrough

Let's step through the solution approach using the example where prices = [1.50, 2.50, 3.40] and the target sum is 7.

1. Initial Summation and Difference Collection

   Considering the prices array:

   • For 1.50, its floor value is 1 and the fractional part is 1.50 - 1 = 0.50.
   • For 2.50, its floor value is 2 and the fractional part is 2.50 - 2 = 0.50.
   • For 3.40, its floor value is 3 and the fractional part is 3.40 - 3 = 0.40.

   The sum of floor values, mi, is 1 + 2 + 3 = 6, and the fractional parts arr are [0.50, 0.50, 0.40].

2. Checking the Feasibility

   The minimum sum mi is 6 and if we were to round all prices up, the maximum sum would be 6 + 3 = 9. The target 7 is within this range (6 <= 7 <= 9), so proceeding is possible.

3. Sorting and Error Calculation

   We sort arr in descending order, resulting in [0.50, 0.50, 0.40]. It remains intact as all elements are in the correct order.

4. Computing the Total Rounding Error

   We need to increase our sum from 6 to 7, hence d = 7 - 6 = 1. This means we need to round up one price. To minimize error, we round up the price with the smallest fractional part, which is 0.40.

   The rounding error for rounding up:

   • We have one value to round up, which is 0.40, so error from it is 1 - 0.40 = 0.60.

   The rounding error for numbers we don't round up:

   • The two values we keep as floor values are the two 0.50s, and their errors remain 0.50 + 0.50 = 1.00.

   Therefore, the total error is 0.60 + 1.00 = 1.60.

5. Formatting the Result

   We format the total error to three decimal places, resulting in the answer: "1.600".

This example illustrates that by sorting the fractional parts and rounding up only the required number of smallest fractional parts, we optimize the rounding error while achieving our target sum. Using this strategy allows us to determine the best numbers to round up to meet our target sum with the minimal rounding difference.

Solution Implementation

Time and Space Complexity

Time Complexity

The function minimizeError has several components, each contributing to the total time complexity:

1. Conversion of strings to floats: The loop converts each string in the prices list to a float. This process takes O(n) where n is the number of strings since each string is processed once.

2. Rounding down floats and collecting fractional parts: Within the same loop, the code rounds down the float to the nearest integer using int(p) and collects the fractional part if any. This is also O(n) time.

3. Sorting fractional parts: The fractional parts are sorted in reverse order which has a time complexity of O(n log n).

4. Summing specific portions of the array: The algorithm sums parts of the array, which is linear with respect to the number of elements summed. In the worst case, this is O(n).

Therefore, the overall time complexity of minimizeError is dominated by the sorting step, which makes the time complexity O(n log n).

Space Complexity

Regarding the space complexity:

1. Storing fractional parts: The array arr stores at most n fractional parts. Hence, its space complexity is O(n).

2. Return value formatting: String formatting for the return value does not use additional space that scales with the input size.

Therefore, the total space complexity of the function is O(n).

Learn more about how to find time and space complexity quickly using problem constraints.