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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 : 22 Jun, 2020

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

\text{Probability of getting perfect square} =\frac{floor(sqrt(R)) - ceil(sqrt(L)) + 1}{R - L + 1}

Below is the implementation of the above approach:

C++

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// 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;
}

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Java

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// 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

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Python3

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# 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    

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C#

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// 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

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Output:

0.2

Time Complexity: O(1)

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