# 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 = 5Output:9Explanation:

Below is the square for the given coordinates:

Input:x1 = 1, y1 = 1, x2 = 4, x3 = 4Output: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:**

4

**Time Complexity:** O(1)

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