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Aroup Goldar Dhruba

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# LeetCode: Check If It Is a Straight Line

### Problem Statement

You are given an array coordinates, coordinates[i] = [x, y], where [x, y] represents the coordinate of a point. Check if these points make a straight line in the XY plane.

### Example

Example 1:

Input: coordinates = [[1,2],[2,3],[3,4],[4,5],[5,6],[6,7]]
Output: true


Example 2:

Input: coordinates = [[1,1],[2,2],[3,4],[4,5],[5,6],[7,7]]
Output: false


Constraints:

• 2 <= coordinates.length <= 1000
• coordinates[i].length == 2
• -10^4 <= coordinates[i][0], coordinates[i][1] <= 10^4
• coordinates contains no duplicate point.

### Solution Thought Process

Let's solve a simple problem first. We are given 3 coordinates (x1, y1), (x2,y2) , (x3,y3) . We are told to find if those 3 points form a line. What we would do is, find the tangent of those points and compare them. If they are the same, then we can say that the lines are equal.

$tan1 = \frac{y1-y2}{x1-x2} \newline \newline tan2 = \frac{y2-y3}{x2-x3}$

Say the three points are (0,2), (10,14), (30,38)

$tan1 = \frac{2-14}{0-10} = \frac{-12}{-10} = \frac {6}{5} \newline \newline tan2 = \frac{14-38}{10-30} = \frac{-24}{-20} = \frac{6}{5}$

So we can say that those points form a line.

Following this mathematical formula, we can scan all the points in the array, considering only two at a time. For reducing the complexity of the floating division, we will separate the tangent into two parts - numerator and denominator. For example, in the above formula, numerator = 6 and denominator = 5.

• First, we find the numerator for those two points (the difference between y coordinates).
• We find the denominator for those two points (the difference between x coordinates).
• We find the gcd of numerator and denominator.
• We divide numerator and denominator by gcd to get the co-prime form for comparing. For comparing the tangents, we have to reduce them into their co-prime form.
• If it's the first element and the second element in the array, then we store the numerator and denominator in a separate value.
• For the further elements, we have to check the numerator and the denominator with the previously-stored numerator and denominator of the tangent. If they don't match, we return false.

• If every tangent matches with each other, then we return true. Those points do form a line, indeed!

### Solution

class Solution {
public:
bool checkStraightLine(vector<vector<int>>& coordinates) {
int tangent_numerator = -1, tangent_denominator = -1;
for(int i=0;i<coordinates.size()-1;i++)
{
int numerator = abs(coordinates[i][1] - coordinates[i+1][1]);
int denominator = abs(coordinates[i][0] - coordinates[i+1][0]);
int gcd = __gcd(numerator, denominator);
numerator /= gcd;
denominator /= gcd;
if(i==0)
{
tangent_numerator = numerator;
tangent_denominator = denominator;
}
else
{
if(numerator != tangent_numerator || denominator != tangent_denominator)
{
return false;
}
}
}
return true;
}
};


### Complexity

Time Complexity: O(n), we are just iterating over the points.

Space Complexity: O(1).