Count of Unique Direct Path Between N Points On a Plane

Given N factors on a aircraft, the place every level has a direct path connecting it to a distinct level, the duty is to rely the whole variety of distinctive direct paths between the factors.

Observe: The worth of N will at all times be larger than 2.

Examples:

Enter: N = 4
Output: 6
Rationalization: Consider 4 factors as a 4 sided polygon. There’ll 4 direct paths (sides of the polygon) in addition to 2 diagonals (diagonals of the polygon). Therefore the reply will likely be 6 direct paths.

Enter: N = 3
Output: 3
Rationalization: Consider 3 factors as a 3 sided polygon. There’ll 3 direct paths (sides of the polygon) in addition to 0 diagonals (diagonals of the polygon). Therefore the reply will likely be 3 direct paths.

Method: The given drawback will be solved utilizing an commentary that for any N-sided there are (variety of sides + variety of diagonals) direct paths. For any N-sided polygon, there are N sides and N*(N – 3)/2 diagonals. Due to this fact, the whole variety of direct paths is given by N + (N * (N – 3))/2.

Under is the implementation of the above strategy:

C++

#embody <bits/stdc++.h>

utilizing namespace std;

int countDirectPath(int N)

{

return N + (N * (N - 3)) / 2;

}

int predominant()

{

int N = 5;

cout << countDirectPath(N);

return 0;

}

Time Complexity: O(1)
Auxiliary Area: O(1)

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