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.
Examples:
Input: x1 = 1, y1 = 1, x2 = 5, x3 = 5
Output: 9
Explanation:
Below is the square for the given coordinates:
Input: x1 = 1, y1 = 1, x2 = 4, x3 = 4
Output: 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)
For Example:
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:
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function to calculate the integral // points inside a square void countIntgralPoints( int x1, int y1,
int x2, int y2)
{ cout << (y2 - y1 - 1) * (x2 - x1 - 1);
} // Driver Code int main()
{ int x1 = 1, y1 = 1;
int x2 = 4, y2 = 4;
countIntgralPoints(x1, y1, x2, y2);
return 0;
} |
// Java program for the above approach class GFG {
// Function to calculate the integral // points inside a square static void countIntgralPoints( int x1, int y1,
int x2, int y2)
{ System.out.println((y2 - y1 - 1 ) *
(x2 - x1 - 1 ));
} // Driver Code public static void main(String args[])
{ int x1 = 1 , y1 = 1 ;
int x2 = 4 , y2 = 4 ;
countIntgralPoints(x1, y1, x2, y2);
} } // This code is contributed by rutvik_56 |
# Python3 program for the above approach # Function to calculate the integral # points inside a square def countIntgralPoints(x1, y1, x2, y2):
print ((y2 - y1 - 1 ) * (x2 - x1 - 1 ))
# Driver Code if __name__ = = '__main__' :
x1 = 1
y1 = 1
x2 = 4
y2 = 4
countIntgralPoints(x1, y1, x2, y2)
# This code is contributed by Samarth |
// C# program for the above approach using System;
class GFG{
// Function to calculate the integral // points inside a square static void countIntgralPoints( int x1, int y1,
int x2, int y2)
{ Console.WriteLine((y2 - y1 - 1) *
(x2 - x1 - 1));
} // Driver code static void Main()
{ int x1 = 1, y1 = 1;
int x2 = 4, y2 = 4;
countIntgralPoints(x1, y1, x2, y2);
} } // This code is contributed by divyeshrabadiya07 |
<script> // Javascript program for the above approach // Function to calculate the integral // points inside a square function countIntgralPoints(x1, y1, x2, y2)
{ document.write( (y2 - y1 - 1) * (x2 - x1 - 1));
} // Driver Code var x1 = 1, y1 = 1;
var x2 = 4, y2 = 4;
countIntgralPoints(x1, y1, x2, y2); </script> |
4
Time Complexity: O(1), as we are not using any loops.
Auxiliary Space: O(1), as we are not using any extra space.