# Number of Integral Points between Two Points in C++

In this tutorial, we are going to write a program the finds the number of integral points between the given two points.

The number of points between two given points will be gcd(abs(x2), abs(y1-y2)) - 1.

If the line joining is parallel to x-axis, then the number of integral points will be abs(y1 - y2) - 1.

If the line joining is parallel to y-axis, then the number of integral points will be abs(x1 - x2) - 1

If the x points of both points are equal, then they are parallel to the x-axis. If the y points of both points are equal, then they are parallel to the y-axis.

Let's see an example.

Input

pointOne = [1, 5]
pointTwo = [1, 3]

Output

1


## Algorithm

• Initialise two points.
• Check whether they are parallel to x-axis or not.
• If they are parallel to the x-axis, then use the formula abs(y1 - y2) - 1.
• Check whether they are parallel to y-axis or not.
• If they are parallel to the y-axis, then use the formula abs(x1 - x2) - 1.
• If they are not parallel to any of the axes, then use the formula gcd(abs(x1-x2), abs(y1- y2)) - 1.
• Compute the result and print it.

## Implementation

Following is the implementation of the above algorithm in C++

#include <bits/stdc++.h>
using namespace std;
int gcd(int a, int b) {
if (b == 0) {
return a;
}
return gcd(b, a % b);
}
int getCount(int pointOne[], int pointTwo[]) {
if (pointOne == pointTwo) {
return abs(pointOne - pointTwo) - 1;
}
if (pointOne == pointTwo) {
return abs(pointOne - pointTwo) - 1;
}
return gcd(abs(pointOne - pointTwo), abs(pointOne - pointTwo)) - 1;
}
int main() {
int pointOne[] = {1, 3}, pointTwo[] = {10, 12};
cout << getCount(pointOne, pointTwo) << endl;
return 0;
}

## Output

If you run the above code, then you will get the following result.

8