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# 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[0] == pointTwo[0]) { return abs(pointOne[1] - pointTwo[1]) - 1; } if (pointOne[1] == pointTwo[1]) { return abs(pointOne[0] - pointTwo[0]) - 1; } return gcd(abs(pointOne[0] - pointTwo[0]), abs(pointOne[1] - pointTwo[1])) - 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

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