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Minimum number of points to be removed to get remaining points on one side of axis using C++.
Problem statement
We are given N points in a Cartesian plane. Our task is to find the minimum number of points that should be removed in order to get the remaining points on one side of any axis.
If given input is {(10, 5), (-2, -5), (13, 8), (-14, 7)} then if we remove (-2, -5) then all remaining points are above X-axis.
Hence answer is 1.
Algorithm
1. Finds the number of points on all sides of the X-axis and Y-axis 2. Return minimum from both of them
Example
#include <iostream>
#include <algorithm>
#define SIZE(arr) (sizeof(arr) / sizeof(arr[0]))
using namespace std;
struct point{
int x, y;
};
int minPointsToBeRemoved(point arr[], int n){
int a = 0, b = 0, c = 0, d = 0;
for (int i = 0; i < n; i++){
if (arr[i].x >= 0)
b++;
else if (arr[i].x <= 0)
a++;
if (arr[i].y <= 0)
d++;
else if (arr[i].y >= 0)
c++;
}
return min({a, d, c, b});
}
int main(){
point arr[] = {{10, 5}, {-2, -5}, {13, 8}, {-14, 7}};
cout << "Minimum points to be removed = " <<
minPointsToBeRemoved(arr, SIZE(arr)) << endl;
return 0;
}Output
When you compile and execute the above program. It generates the following output −
Minimum points to be removed = 1
Updated on: 31-Oct-2019
524 Views
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