Count of integral coordinates that lies inside a Square

Namachila November 19, 2025

0 0 3 minutes read

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++

// 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

// 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

# 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#

// 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    

Javascript

<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>

Output:

Time Complexity: O(1), as we are not using any loops.

Auxiliary Space: O(1), as we are not using any extra space.

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