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 = 0.133333Input: 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++ implementation to find the // probability of getting a // perfect square number #include <bits/stdc++.h> 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 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 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# 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 |
<script> // Javascript implementation to find the // probability of getting a // perfect square number // Function to return the probability // of getting a perfect square // number in a range function findProb(l, r)
{ // Count of perfect squares
var countOfPS = (Math.floor(Math.sqrt(r)) -
Math.ceil(Math.sqrt(l)) + 1);
// Total numbers in range l to r
var total = r - l + 1;
// Calculating probability
var prob = countOfPS / total;
return prob;
} // Driver code var L = 16, R = 25;
// Function Call document.write(findProb(L, R)); // This code is contributed by Khushboogoyal499 </script> |
Output:
0.2
Time Complexity: O(log(r) + log(l))
Auxiliary Space: O(1)