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Minimize Rounding Error to Meet Target in C++
Suppose we have an array of prices P [p1,p2...,pn] and a target value, we have to round each price Pi to Roundi(Pi) so that the rounded array [Round1(P1),Round2(P2)...,Roundn(Pn)] sums to the given target value. Here each operation Roundi(pi) could be either Floor(Pi) or Ceil(Pi).
We have to return the string "-1" if the rounded array is impossible to sum to target. Otherwise, return the smallest rounding error, which will be (as a string with three places after the decimal) defined as −
$\displaystyle\sum\limits_{i-1}^n |Round_{i} (???? ) - ????$
So if the input is like [“0.700”, “2.800”, “4.900”], and the target is 8. Use floor or ceil operations to get (0.7 - 0) + (3 - 2.8) + (5 - 4.9) = 0.7 + 0.2 + 0.1 = 1.0
To solve this, we will follow these steps −
ret := 0
make one priority queue pq for (double and array) type complex data
for i in range 0 to size of prices
x := double value of prices[i]
low := floor of x
high := ceiling of x
if low is not high
diff := (high - x) – (x - low)
insert diff into pq
target := target – low
ret := ret + (x - low)
if target > size of pq or target < 0, then return “-1”
while target is not 0
ret := ret + top of pq, delete from pq
- d
decrease target by 1
s := ret as string
return substring s by taking number up to three decimal places
Let us see the following implementation to get better understanding −
Example
#include <bits/stdc++.h>
using namespace std;
struct Comparator{
bool operator()(double a, double b) {
return !(a < b);
}
};
class Solution {
public:
string minimizeError(vector<string>& prices, int target) {
double ret = 0;
priority_queue < double, vector < double >, Comparator > pq;
for(int i = 0; i < prices.size(); i++){
double x = stod(prices[i]);
double low = floor(x);
double high = ceil(x);
if(low != high){
double diff = ((high - x) - (x - low));
pq.push(diff);
}
target -= low;
ret += (x - low);
}
if(target > pq.size() || target < 0) return "-1";
while(target--){
ret += pq.top();
pq.pop();
}
string s = to_string (ret);
return s.substr (0, s.find_first_of ('.', 0) + 4);
}
};
main(){
vector<string> v = {"0.700","2.800","4.900"};
Solution ob;
cout << (ob.minimizeError(v, 8));
}Input
["0.700","2.800","4.900"] 8
Output
"1.000"
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