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Count triplet pairs (A, B, C) of points in 2-D space that satisfy the given condition in C++
We are given an input of N points on a 2-D space. The goal is to find the count of triplets of points from the input such that one point is the mid-point on the line between the other two. i.e if triplet is (A,B,C) then B is midpoint of A and C ( or any other combination of A,B,C ).
We will do this by inserting all points as pairs <int,int> into a vector. Then add all pairs from this vector in set. By taking two points from the set check if the sum of (x,y) coordinates divided by 2 exist in the same set. If yes increment triplet count.
Let’s understand with examples.
Input
{ 1,2 }, { 4,2} , { 2,1 } , { 7,2 } N=4 pairsOutput
Count of triplet pairs that satisfy the given condition are: 1
Explanation
Here {4,2} is mid-point between {1,2} and {7,2}. Only 1 such tripletInput
{ 1,2 }, { 4,2} , { 2,1 } , { 5,2 }, { 8,1} , {1,1} N=6Output
Count of triplet pairs that satisfy the given condition are: 1
Explanation
No such triplet exist
Approach used in the below program is as follows
We are taking a vector of pairs of type <int,int>.
Each pair contains (x,y) coordinates.
Function mid_point(vector<pair<int, int>> vec, int size) takes a vector and it’s size as input and returns the number of triplets that satisfy the mid-point condition.
Take the initial variable count as 0 for such triplets.
Insert all pairs from the vector into a set<pair<int, int> > sets. It will have all unique points.
Traverse the vector using two for loops for each pair of points.
Store sum of x coordinates of both points in integer point_A and sum of y coordinates of both points in integer point_B.
If both these sums in point_A and point_B are even then check for mid-point condition.
If a pair(point_A/2,point_B/2) exists as a pair in the set means mid-point exists. Increment count of triplet.
In the end count will have number of triplets.
Return the count as a result at the end of the for loop.
Example
#include <bits/stdc++.h>
using namespace std;
int mid_point(vector<pair<int, int>> vec, int size){
int count = 0;
set<pair<int, int> > sets;
for (int i = 0; i < size; i++){
sets.insert(vec[i]);
}
for (int i = 0; i < size; i++){
for (int j = i + 1; j < size; j++){
int point_A = vec[i].first + vec[j].first;
int point_B = vec[i].second + vec[j].second;
if (point_A % 2 == 0 && point_B % 2 == 0){
if (sets.find(make_pair(point_A / 2, point_B / 2)) != sets.end()){
count++;
}
}
}
}
return count;
}
int main(){
vector<pair<int, int>> vec = { { 9, 2 }, { 5, 2 }, { 1, 2 } };
int size = vec.size();
cout<<"Count of triplet pairs (A, B, C) of points in 2-D space that satisfy the given condition are: "<<mid_point(vec, size);
}Output
If we run the above code it will generate the following output −
Count of triplet pairs (A, B, C) of points in 2-D space that satisfy the given condition are: 1
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