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# Count of integral points that lie at a distance D from origin

• Last Updated : 13 Jun, 2022

Given a positive integer D, the task is to find the number of integer coordinates (x, y) which lie at a distance D from origin.

Example:

Input: D = 1
Output: 4
Explanation: Total valid points are {1, 0}, {0, 1}, {-1, 0}, {0, -1}

Input: D = 5
Output: 12
Explanation: Total valid points are {0, 5}, {0, -5}, {5, 0}, {-5, 0}, {3, 4}, {3, -4}, {-3, 4}, {-3, -4}, {4, 3}, {4, -3}, {-4, 3}, {-4, -3}

Approach: This question can be simplified to count integer coordinates lying on the circumference of the circle centered at the origin, having a radius D and can be solved with the help of the Pythagoras theorem. As the points should be at a distance D from the origin, so they all must satisfy the equation x * x + y * y = D2 where (x, y) are the coordinates of the point.
Now, to solve the above problem, follow the below steps:

• Initialize a variable, say count that stores the total count of the possible pairs of coordinates.
• Iterate over all possible x coordinates and calculate the corresponding value of y as sqrt(D2 – y*y).
• Since every coordinate whose both x and y are positive integers can form a total of 4 possible valid pairs as {x, y}, {-x, y}, {-x, -y}, {x, -y} and increment the count each possible pair (x, y) by 4 in the variable count.
• Also, there is always an integer coordinate present at the circumference of the circle where it cuts the x-axis and y-axis because the radius of the circle is an integer. So add 4 in count, to compensate these points.
• After completing the above steps, print the value of count as the resultant count of pairs.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach` `#include ``using` `namespace` `std;` `// Function to find the total valid``// integer coordinates at a distance D``// from origin``int` `countPoints(``int` `D)``{``    ``// Stores the count of valid points``    ``int` `count = 0;` `    ``// Iterate over possible x coordinates``    ``for` `(``int` `x = 1; x * x < D * D; x++) {` `        ``// Find the respective y coordinate``        ``// with the pythagoras theorem``        ``int` `y = (``int``)``sqrt``(``double``(D * D - x * x));``        ``if` `(x * x + y * y == D * D) {``            ``count += 4;``        ``}``    ``}` `    ``// Adding 4 to compensate the coordinates``    ``// present on x and y axes.``    ``count += 4;` `    ``// Return the answer``    ``return` `count;``}` `// Driver Code``int` `main()``{``    ``int` `D = 5;``    ``cout << countPoints(D);` `    ``return` `0;``}`

## Java

 `// Java program for the above approach``import` `java.io.*;` `class` `GFG {` `// Function to find the total valid``// integer coordinates at a distance D``// from origin``static` `int` `countPoints(``int` `D)``{``  ` `    ``// Stores the count of valid points``    ``int` `count = ``0``;` `    ``// Iterate over possible x coordinates``    ``for` `(``int` `x = ``1``; x * x < D * D; x++) {` `        ``// Find the respective y coordinate``        ``// with the pythagoras theorem``        ``int` `y = (``int``)Math.sqrt((D * D - x * x));``        ``if` `(x * x + y * y == D * D) {``            ``count += ``4``;``        ``}``    ``}` `    ``// Adding 4 to compensate the coordinates``    ``// present on x and y axes.``    ``count += ``4``;` `    ``// Return the answer``    ``return` `count;``}` `// Driver Code``public` `static` `void` `main (String[] args)``{``    ``int` `D = ``5``;``    ``System.out.println(countPoints(D));``}``}` `// this code is contributed by shivanisinghss2110`

## Python3

 `# python 3 program for the above approach``from` `math ``import` `sqrt` `# Function to find the total valid``# integer coordinates at a distance D``# from origin``def` `countPoints(D):``  ` `    ``# Stores the count of valid points``    ``count ``=` `0` `    ``# Iterate over possible x coordinates``    ``for` `x ``in` `range``(``1``, ``int``(sqrt(D ``*` `D)), ``1``):` `        ``# Find the respective y coordinate``        ``# with the pythagoras theorem``        ``y ``=` `int``(sqrt((D ``*` `D ``-` `x ``*` `x)))``        ``if` `(x ``*` `x ``+` `y ``*` `y ``=``=` `D ``*` `D):``            ``count ``+``=` `4` `    ``# Adding 4 to compensate the coordinates``    ``# present on x and y axes.``    ``count ``+``=` `4` `    ``# Return the answer``    ``return` `count` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ``D ``=` `5``    ``print``(countPoints(D))``    ` `    ``# This code is contributed by SURENDRA_GANGWAR.`

## C#

 `// C# program for the above approach``using` `System;` `// Function to find the total valid``// integer coordinates at a distance D``// from origin``public` `class` `GFG{``    ``static` `int` `countPoints(``int` `D){``        ` `        ``// Stores the count of valid points``        ``int` `count = 0;``        ` `        ``// Iterate over possible x coordinates``        ``for``(``int` `x = 1; x*x < D*D; x++){``            ``int` `y = (``int``)Math.Sqrt((D * D - x * x));` `            ``// Find the respective y coordinate``            ``// with the pythagoras theorem``            ``if``(x * x + y * y == D * D){``              ``count += 4;  ``            ``}``       ``}``    ``// Adding 4 to compensate the coordinates``    ``// present on x and y axes.``    ` `    ``count += 4;` `    ``// Return the answer``    ``return` `count;``}``    ` `    ``// Driver Code` `    ``public` `static` `void` `Main(){``        ``int` `D = 5;``        ``Console.Write(countPoints(D));``    ``}``}` `// This code is contributed by gfgking`

## Javascript

 ``

Output

`12`

Time Complexity: O(R)
Auxiliary Space: O(1)

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