Solution Explanation

This problem asks to find the minimum rounding error when rounding a list of prices to meet a target sum. The rounding can be either flooring or ceiling. The solution uses a greedy approach.

Approach:

1. Preprocessing:

   • Iterate through the prices array. Convert each price string to a floating-point number.
   • Calculate the integer part (int(price)) and the fractional part (price - int(price)).
   • Accumulate the sum of the integer parts into mi.
   • If the fractional part is non-zero, add it to the arr list. This list will contain only the fractional parts that need to be considered for rounding.

2. Feasibility Check:

   • If the target is less than mi (sum of integer parts) or greater than mi plus the number of fractional parts in arr, then it's impossible to achieve the target sum through rounding. Return "-1" in this case.

3. Greedy Rounding:

   • Calculate d, the difference between the target and mi. This represents how many fractional parts need to be rounded up.
   • Sort the arr list in descending order. This is crucial for the greedy strategy to work optimally. We want to round up the largest fractional parts first to minimize the overall error.
   • Iterate d times, subtracting each of the largest d fractional parts from ans.
   • Iterate through the remaining fractional parts (those not rounded up), adding them to ans.

4. Return Value:

   • Format ans to three decimal places and return it as a string.

Time Complexity Analysis:

• The preprocessing step iterates through the prices array once, which takes O(n) time, where n is the length of the prices array.
• Sorting the arr list takes O(n log n) time in the worst case.
• The remaining steps iterate through arr a constant number of times, so they take O(n) time.

Therefore, the overall time complexity is dominated by the sorting step, making it O(n log n).

Space Complexity Analysis:

• The space used is proportional to the size of the arr list, which is at most n in size (in the case where all prices have fractional parts).
• Therefore, the space complexity is O(n).

Code Explanation (Python):

The Python code efficiently implements this approach:

class Solution:
    def minimizeError(self, prices: List[str], target: int) -> str:
        mi = 0
        arr = []
        for p in prices:
            p = float(p)
            mi += int(p)
            if d := p - int(p):
                arr.append(d)
        if not mi <= target <= mi + len(arr):
            return "-1"
        d = target - mi
        arr.sort(reverse=True)
        ans = d - sum(arr[:d]) + sum(arr[d:])
        return f'{ans:.3f}'
 

The code is concise and readable due to Python's expressive features. The if d := p - int(p): line uses the walrus operator for a compact way to check and assign the fractional part. The sum(arr[:d]) and sum(arr[d:]) expressions efficiently calculate the sums of the relevant parts of the sorted array. The f-string formatting neatly handles the decimal place requirements. Similar logic and efficiency are present in the Java and C++ versions. The Go version uses math.Floor, strconv.ParseFloat, and fmt.Sprintf for similar functionality but with Go's specific libraries.