README_EN.md

1058. Minimize Rounding Error to Meet Target

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Description

Given an array of prices [p1,p2...,pn] and a target, round each price pi to Roundi(pi) so that the rounded array [Round1(p1),Round2(p2)...,Roundn(pn)] sums to the given target. Each operation Roundi(pi) could be either Floor(pi) or Ceil(pi).

Return the string "-1" if the rounded array is impossible to sum to target. Otherwise, return the smallest rounding error, which is defined as Σ |Roundi(pi) - (pi)| for i from 1 to n, as a string with three places after the decimal.

 

Example 1:

Input: prices = ["0.700","2.800","4.900"], target = 8
Output: "1.000"
Explanation:
Use Floor, Ceil and Ceil operations to get (0.7 - 0) + (3 - 2.8) + (5 - 4.9) = 0.7 + 0.2 + 0.1 = 1.0 .

Example 2:

Input: prices = ["1.500","2.500","3.500"], target = 10
Output: "-1"
Explanation: It is impossible to meet the target.

Example 3:

Input: prices = ["1.500","2.500","3.500"], target = 9
Output: "1.500"

 

Constraints:

• 1 <= prices.length <= 500
• Each string prices[i] represents a real number in the range [0.0, 1000.0] and has exactly 3 decimal places.
• 0 <= target <= 106

Solutions

Python3

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}'

Java

class Solution {
    public String minimizeError(String[] prices, int target) {
        int mi = 0;
        List<Double> arr = new ArrayList<>();
        for (String p : prices) {
            double price = Double.valueOf(p);
            mi += (int) price;
            double d = price - (int) price;
            if (d > 0) {
                arr.add(d);
            }
        }
        if (target < mi || target > mi + arr.size()) {
            return "-1";
        }
        int d = target - mi;
        arr.sort(Collections.reverseOrder());
        double ans = d;
        for (int i = 0; i < d; ++i) {
            ans -= arr.get(i);
        }
        for (int i = d; i < arr.size(); ++i) {
            ans += arr.get(i);
        }
        DecimalFormat df = new DecimalFormat("#0.000");
        return df.format(ans);
    }
}

C++

class Solution {
public:
    string minimizeError(vector<string>& prices, int target) {
        int mi = 0;
        vector<double> arr;
        for (auto& p : prices) {
            double price = stod(p);
            mi += (int) price;
            double d = price - (int) price;
            if (d > 0) {
                arr.push_back(d);
            }
        }
        if (target < mi || target > mi + arr.size()) {
            return "-1";
        }
        int d = target - mi;
        sort(arr.rbegin(), arr.rend());
        double ans = d;
        for (int i = 0; i < d; ++i) {
            ans -= arr[i];
        }
        for (int i = d; i < arr.size(); ++i) {
            ans += arr[i];
        }
        string s = to_string(ans);
        return s.substr(0, s.find('.') + 4);
    }
};

Go

func minimizeError(prices []string, target int) string {
	arr := []float64{}
	mi := 0
	for _, p := range prices {
		price, _ := strconv.ParseFloat(p, 64)
		mi += int(math.Floor(price))
		d := price - float64(math.Floor(price))
		if d > 0 {
			arr = append(arr, d)
		}
	}
	if target < mi || target > mi+len(arr) {
		return "-1"
	}
	d := target - mi
	sort.Float64s(arr)
	ans := float64(d)
	for i := 0; i < d; i++ {
		ans -= arr[len(arr)-i-1]
	}
	for i := d; i < len(arr); i++ {
		ans += arr[len(arr)-i-1]
	}
	return fmt.Sprintf("%.3f", ans)
}

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