# Count Distinct Non-Negative Integer Pairs (x, y) that Satisfy the Inequality x*x + y*y < n

Given a positive number n, count all distinct Non-Negative Integer pairs (x, y) that satisfy the inequality x*x + y*y < n.

Examples:

```Input:  n = 5
Output: 6
The pairs are (0, 0), (0, 1), (1, 0), (1, 1), (2, 0), (0, 2)

Input: n = 6
Output: 8
The pairs are (0, 0), (0, 1), (1, 0), (1, 1), (2, 0), (0, 2),
(1, 2), (2, 1)```

A Simple Solution is to run two loops. The outer loop goes for all possible values of x (from 0 to ?n). The inner loops pick all possible values of y for the current value of x (picked by outer loop).

Following is the implementation of a simple solution.

## C++

 `#include ` `using` `namespace` `std;`   `// This function counts number of pairs (x, y) that satisfy` `// the inequality x*x + y*y < n.` `int` `countSolutions(``int` `n)` `{` `   ``int` `res = 0;` `   ``for` `(``int` `x = 0; x*x < n; x++)` `      ``for` `(``int` `y = 0; x*x + y*y < n; y++)` `         ``res++;` `   ``return` `res;` `}`   `// Driver program to test above function` `int` `main()` `{` `    ``cout << ``"Total Number of distinct Non-Negative pairs is "` `         ``<< countSolutions(6) << endl;` `    ``return` `0;` `}`

## Java

 `// Java code to Count Distinct ` `// Non-Negative Integer Pairs ` `// (x, y) that Satisfy the ` `// inequality x*x + y*y < n` `import` `java.io.*;`   `class` `GFG ` `{` `    ``// This function counts number` `    ``// of pairs (x, y) that satisfy` `    ``// the inequality x*x + y*y < n.` `    ``static` `int` `countSolutions(``int` `n)` `    ``{` `        ``int` `res = ``0``;` `        ``for` `(``int` `x = ``0``; x * x < n; x++)` `            ``for` `(``int` `y = ``0``; x * x + y * y < n; y++)` `                ``res++;` `                `  `        ``return` `res;` `    ``}`   `    ``// Driver program ` `    ``public` `static` `void` `main(String args[])` `    ``{` `        ``System.out.println ( ``"Total Number of distinct Non-Negative pairs is "` `                                                        ``+countSolutions(``6``));` `        `  `    ``}` `}`   `// `

## Python3

 `# Python3 implementation of above approach`   `# This function counts number of pairs ` `# (x, y) that satisfy` `# the inequality x*x + y*y < n.` `def` `countSolutions(n):`   `    ``res ``=` `0` `    ``x ``=` `0` `    ``while``(x ``*` `x < n):` `        ``y ``=` `0` `        ``while``(x ``*` `x ``+` `y ``*` `y < n):` `            ``res ``=` `res ``+` `1` `            ``y ``=` `y ``+` `1` `        ``x ``=` `x ``+` `1`   `    ``return` `res`   `# Driver program to test above function` `if` `__name__``=``=``'__main__'``:` `    ``print``(``"Total Number of distinct Non-Negative pairs is "``, ` `         ``countSolutions(``6``))`   `# This code is contributed by` `# Sanjit_Prasad`

## C#

 `// C# code to Count Distinct ` `// Non-Negative Integer Pairs ` `// (x, y) that Satisfy the ` `// inequality x*x + y*y < n` `using` `System;`   `class` `GFG {` `    `  `    ``// This function counts number` `    ``// of pairs (x, y) that satisfy` `    ``// the inequality x*x + y*y < n.` `    ``static` `int` `countSolutions(``int` `n)` `    ``{` `        ``int` `res = 0;` `        `  `        ``for` `(``int` `x = 0; x*x < n; x++)` `            ``for` `(``int` `y = 0; x*x + y*y < n; y++)` `                ``res++;` `                `  `        ``return` `res;` `    ``}`   `    ``// Driver program ` `    ``public` `static` `void` `Main()` `    ``{` `        ``Console.WriteLine( ``"Total Number of "` `          ``+ ``"distinct Non-Negative pairs is "` `                        ``+ countSolutions(6));` `    ``}` `}`   `// This code is contributed by Sam007.`

## PHP

 ``

## Javascript

 ``

Output:

`Total Number of distinct Non-Negative pairs is 8`

An upper bound for the time complexity of the above solution is O(n). The outer loop runs ?n times. The inner loop runs at most ?n times.

Auxiliary Space: O(1)

Using an Efficient Solution, we can find the count in O(?n) time. The idea is to first find the count of all y values corresponding to the 0 value of x. Let count of distinct y values be yCount. We can find yCount by running a loop and comparing yCount*yCount with n.
After we have the initial yCount, we can one by one increase the value of x and find the next value of yCount by reducing yCount.

## C++

 `// An efficient C program to find different (x, y) pairs that` `// satisfy x*x + y*y < n.` `#include ` `using` `namespace` `std;`   `// This function counts number of pairs (x, y) that satisfy` `// the inequality x*x + y*y < n.` `int` `countSolutions(``int` `n)` `{` `   ``int` `x = 0, yCount, res = 0;`   `   ``// Find the count of different y values for x = 0.` `   ``for` `(yCount = 0; yCount*yCount < n; yCount++) ;`   `   ``// One by one increase value of x, and find yCount for` `   ``// current x.  If yCount becomes 0, then we have reached` `   ``// maximum possible value of x.` `   ``while` `(yCount != 0)` `   ``{` `       ``// Add yCount (count of different possible values of y` `       ``// for current x) to result` `       ``res  +=  yCount;`   `       ``// Increment x` `       ``x++;`   `       ``// Update yCount for current x. Keep reducing yCount while` `       ``// the inequality is not satisfied.` `       ``while` `(yCount != 0 && (x*x + (yCount-1)*(yCount-1) >= n))` `         ``yCount--;` `   ``}`   `   ``return` `res;` `}`   `// Driver program to test above function` `int` `main()` `{` `    ``cout << ``"Total Number of distinct Non-Negative pairs is "` `         ``<< countSolutions(6) << endl;` `    ``return` `0;` `}`

## Java

 `// An efficient Java program to ` `// find different (x, y) pairs ` `// that satisfy x*x + y*y < n.` `import` `java.io.*;`   `class` `GFG ` `{` `    ``// This function counts number ` `    ``//of pairs (x, y) that satisfy` `    ``// the inequality x*x + y*y < n.` `    ``static` `int` `countSolutions(``int` `n)` `    ``{` `        ``int` `x = ``0``, yCount, res = ``0``;` `        `  `        ``// Find the count of different` `        ``// y values for x = 0.` `        ``for` `(yCount = ``0``; yCount * yCount < n; yCount++) ;` `        `  `        ``// One by one increase value of x,` `        ``// and find yCount forcurrent x. If ` `        ``// yCount becomes 0, then we have reached` `        ``// maximum possible value of x.` `        ``while` `(yCount != ``0``)` `        ``{` `            ``// Add yCount (count of different possible` `            ``// values of y for current x) to result` `            ``res += yCount;` `            `  `            ``// Increment x` `            ``x++;` `            `  `            ``// Update yCount for current x. Keep reducing ` `            ``// yCount while the inequality is not satisfied.` `            ``while` `(yCount != ``0` `&& (x * x + (yCount - ``1``)` `                                  ``* (yCount - ``1``) >= n))` `            ``yCount--;` `            `  `        ``}` `        ``return` `res;` `        `  `    ``}` `    `  `    ``// Driver program ` `    ``public` `static` `void` `main(String args[])` `    ``{` `        ``System.out.println ( ``"Total Number of distinct Non-Negative pairs is "` `                             ``+countSolutions(``6``)) ;` `        `  `    ``}` `}`   `// `

## Python3

 `# An efficient python program to` `# find different (x, y) pairs ` `# that satisfy x*x + y*y < n.`   `# This function counts number of` `# pairs (x, y) that satisfy the` `# inequality x*x + y*y < n.` `def` `countSolutions(n):` `    `  `    ``x ``=` `0` `    ``res ``=` `0` `    ``yCount ``=` `0`   `    ``# Find the count of different` `    ``# y values for x = 0.` `    ``while``(yCount ``*` `yCount < n):` `        ``yCount ``=` `yCount ``+` `1` `        `  `    ``# One by one increase value of` `    ``# x, and find yCount for current` `    ``# x. If yCount becomes 0, then ` `    ``# we have reached maximum ` `    ``# possible value of x.` `    ``while` `(yCount !``=` `0``):` `        ``# Add yCount (count of ` `        ``# different possible values` `        ``# of y for current x) to` `        ``# result` `        ``res ``=` `res ``+` `yCount`   `        ``# Increment x` `        ``x ``=` `x ``+` `1`   `        ``# Update yCount for current ` `        ``# x. Keep reducing yCount` `        ``# while the inequality is ` `        ``# not satisfied.` `        ``while` `(yCount !``=` `0` `and` `(x ``*` `x ` `                     ``+` `(yCount ``-` `1``) ``*` `                     ``(yCount ``-` `1``) >``=` `n)):` `            ``yCount ``=` `yCount ``-` `1` `        `  `    ``return` `res`   `# Driver program to test ` `# above function` `print` `(``"Total Number of distinct "``,` `           ``"Non-Negative pairs is "` `               ``, countSolutions(``6``))`   `# This code is contributed by Sam007.`

## C#

 `// An efficient C# program to ` `// find different (x, y) pairs ` `// that satisfy x*x + y*y < n.` `using` `System;`   `class` `GFG {` `    `  `    ``// This function counts number ` `    ``//of pairs (x, y) that satisfy` `    ``// the inequality x*x + y*y < n.` `    ``static` `int` `countSolutions(``int` `n)` `    ``{` `        ``int` `x = 0, yCount, res = 0;` `        `  `        ``// Find the count of different` `        ``// y values for x = 0.` `        ``for` `(yCount = 0; yCount * yCount < n;` `                                   ``yCount++) ;` `        `  `        ``// One by one increase value of x,` `        ``// and find yCount forcurrent x. If ` `        ``// yCount becomes 0, then we have` `        ``// reached maximum possible value` `        ``// of x.` `        ``while` `(yCount != 0)` `        ``{` `            `  `            ``// Add yCount (count of different` `            ``// possible values of y for` `            ``// current x) to result` `            ``res += yCount;` `            `  `            ``// Increment x` `            ``x++;` `            `  `            ``// Update yCount for current x.` `            ``// Keep reducing yCount while the` `            ``// inequality is not satisfied.` `            ``while` `(yCount != 0 && (x * x +` `                            ``(yCount - 1) *` `                         ``(yCount - 1) >= n))` `            ``yCount--;` `        ``}` `        `  `        ``return` `res;` `    ``}` `    `  `    ``// Driver program ` `    ``public` `static` `void` `Main()` `    ``{` `        ``Console.WriteLine( ``"Total Number of "` `          ``+ ``"distinct Non-Negative pairs is "` `                       ``+ countSolutions(6)) ;` `    ``}` `}`   `// This code is contributed by Sam007.`

## PHP

 `= ``\$n``))` `            ``\$yCount``--;` `    ``}` `    `  `    ``return` `\$res``;` `}`   `// Driver program to test above function` `echo` `"Total Number of distinct Non-Negative"``,` `        ``"pairs is "``, countSolutions(6) ,``"\n"``;`   `// This code is contributed by anuj_67.` `?>`

## Javascript

 ``

Output:

`Total Number of distinct Non-Negative pairs is 8`

Time Complexity of the above solution seems more but if we take a closer look, we can see that it is O(?n). In every step inside the inner loop, the value of yCount is decremented by 1. The value yCount can decrement at most O(?n) times as yCount is counted y values for x = 0. In the outer loop, the value of x is incremented. The value of x can also increment at most O(?n) times as the last x is for yCount equals 1.

Auxiliary Space: O(1)