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

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 rangefloat 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 Codeint 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 rangestatic 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 Codepublic 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 numberimport math # Function to return the probability# of getting a perfect square# number in a rangedef 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 codeif __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 numberusing System; class GFG{ // Function to return the probability// of getting a perfect square// number in a rangestatic 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 Codepublic 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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