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.
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.