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 picks all possible values of y for current value of x (picked by outer loop). Following is implementation of simple solution.

## C++

#include <iostream> 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)); } } // This article is contributed by vt_m.

Output:

Total Number of distinct Non-Negative pairs is 8

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

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 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 initial yCount, we can one by one increase 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 <iostream> 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)) ; } } // This article is contributed by vt_m.

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, value of yCount is decremented by 1. The value yCount can decrement at most O(√n) times as yCount is count 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 to 1.

This article is contributed by **Sachin Gupta**. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.