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# Probability of getting a perfect square when a random number is chosen in a given range

• Difficulty Level : Easy
• Last Updated : 16 Sep, 2022

Given two integers L and R that denote a range, the task is to find the probability of getting a perfect square number when a random number is chosen in the range L to R.

Examples:

Input: L = 6, R = 20
Output: 0.133333
Explanation:
Perfect squares in range [6, 20] = {9, 16} => 2 perfect squares
Total numbers in range [6, 20] = 15
Probability = 2 / 15 = 0.133333

Input: L = 16, R = 25
Output: 0.2

Approach: The key observation in this problem is the count of the perfect squares in the range from 0 to a number can be computed with the given formulae:

// Count of perfect squares in the range 0 to N is given as
Count of perfect squares = Floor(sqrt(N))

Similarly, the count of the perfect squares in the given range can be computed with the help of the above formulae as follows:

Count of perfect Squares[L, R] = floor(sqrt(R)) – ceil(sqrt(L)) + 1
Total numbers in the range = R – L + 1

Below is the implementation of the above approach:

## C++

 // C++ implementation to find the // probability of getting a // perfect square number   #include  using namespace std;   // Function to return the probability // of getting a perfect square // number in a range float findProb(int l, int r) {     // Count of perfect squares     float countOfPS = floor(sqrt(r)) - ceil(sqrt(l)) + 1;       // Total numbers in range l to r     float total = r - l + 1;       // Calculating probability     float prob = (float)countOfPS / (float)total;     return prob; }   // Driver Code int main() {     int L = 16, R = 25;     cout << findProb(L, R);       return 0; }

## Java

 // Java implementation to find the // probability of getting a // perfect square number   class GFG{   // Function to return the probability // of getting a perfect square // number in a range static float findProb(int l, int r) {       // Count of perfect squares     float countOfPS = (float) (Math.floor(Math.sqrt(r)) -                                Math.ceil(Math.sqrt(l)) + 1);       // Total numbers in range l to r     float total = r - l + 1;       // Calculating probability     float prob = (float)countOfPS / (float)total;     return prob; }   // Driver Code public static void main(String[] args) {     int L = 16, R = 25;     System.out.print(findProb(L, R)); } }   // This code is contributed by Amit Katiyar

## Python3

 # Python3 implementation to find   # the probability of getting a  # perfect square number  import math   # Function to return the probability  # of getting a perfect square  # number in a range  def findProb(l, r):           # Count of perfect squares      countOfPS = (math.floor(math.sqrt(r)) -                   math.ceil(math.sqrt(l)) + 1)           # Total numbers in range l to r      total = r - l + 1       # Calculating probability      prob = countOfPS / total           return prob        # Driver code  if __name__=='__main__':           L = 16     R = 25           print(findProb(L, R))       # This code is contributed by rutvik_56

## C#

 // C# implementation to find the probability  // of getting a perfect square number using System;   class GFG{   // Function to return the probability // of getting a perfect square // number in a range static float findProb(int l, int r) {           // Count of perfect squares     float countOfPS = (float)(Math.Floor(Math.Sqrt(r)) -                             Math.Ceiling(Math.Sqrt(l)) + 1);       // Total numbers in range l to r     float total = r - l + 1;       // Calculating probability     float prob = (float)countOfPS / (float)total;     return prob; }   // Driver Code public static void Main(String[] args) {     int L = 16, R = 25;           Console.Write(findProb(L, R)); } }   // This code is contributed by Amit Katiyar

## Javascript

 

Output:

0.2

Time Complexity: O(log(r) + log(l))
Auxiliary Space: O(1)

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