Count of integral coordinates that lies inside a Square
Given lower left and upper right coordinates (x1, y1) and (x2, y2) of a square, the task is to count the number of integral coordinates that lies strictly inside the square.
Input: x1 = 1, y1 = 1, x2 = 5, x3 = 5
Below is the square for the given coordinates:
Input: x1 = 1, y1 = 1, x2 = 4, x3 = 4
Approach: The difference between the x and y ordinates of the lower and upper right coordinates of the given squares gives the number integral points of x ordinates and y ordinates between opposite sides of square respectively. The total number of points that strictly lies inside the square is given by:
count = (x2 – x1 – 1) * (y2 – y1 – 1)
In the above figure:
1. The total number of integral points inside base of the square is (x2 – x1 – 1).
2. The total number of integral points inside height of the square is (y2 – y1 – 1).
These (x2 – x1 – 1) integrals points parallel to the base of the square repeats (y2 – y1 – 1) number of times. Therefore the total number of integral points is given by (x2 – x1 – 1)*(y2 – y1 – 1).
Below is the implementation of the above approach:
Time Complexity: O(1), as we are not using any loops.
Auxiliary Space: O(1), as we are not using any extra space.