Number of perfect squares between two given numbers

Given two given numbers a and b where 1<=a<=b, find the number of perfect squares between a and b (a and b inclusive).

Examples

Input :  a = 3, b = 8
Output : 1
The only perfect in given range is 4.

Input : a = 9, b = 25
Output : 3
The three squares in given range are 9, 
16 and 25

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

Method 1 : One naive approach is to check all the numbers between a and b (inclusive a and b) and increase count by one whenever we encounter a perfect square.

Below is the implementation of above idea :

C++

// A Simple Method to count squares between a and b 
#include <bits/stdc++.h> 
using namespace std; 
  
int countSquares(int a, int b) 
{ 
    int cnt = 0; // Initialize result 
  
    // Traverse through all numbers 
    for (int i = a; i <= b; i++) 
  
        // Check if current number 'i' is perfect 
        // square 
        for (int j = 1; j * j <= i; j++) 
            if (j * j == i) 
                cnt++; 
  
    return cnt; 
} 
  
// Driver code 
int main() 
{ 
    int a = 9, b = 25; 
    cout << "Count of squares is "
         << countSquares(a, b); 
    return 0; 
} 

Java

// Java program to count squares between a and b 
class CountSquares { 
  
    static int countSquares(int a, int b) 
    { 
        int cnt = 0; // Initialize result 
  
        // Traverse through all numbers 
        for (int i = a; i <= b; i++) 
  
            // Check if current number 'i' is perfect 
            // square 
            for (int j = 1; j * j <= i; j++) 
                if (j * j == i) 
                    cnt++; 
        return cnt; 
    } 
} 
  
// Driver Code 
public class PerfectSquares { 
    public static void main(String[] args) 
    { 
        int a = 9, b = 25; 
        CountSquares obj = new CountSquares(); 
        System.out.print("Count of squares is " + obj.countSquares(a, b)); 
    } 
} 

Python

# Python program to count squares between a and b 
  
def CountSquares(a, b): 
  
    cnt = 0 # initialize result 
  
    # Traverse through all numbers 
    for i in range (a, b + 1): 
        j = 1; 
        while j * j <= i: 
            if j * j == i: 
                 cnt = cnt + 1
            j = j + 1
        i = i + 1
    return cnt 
  
# Driver Code 
a = 9
b = 25
print "Count of squares is:", CountSquares(a, b) 

C#

// C# program to count squares 
// between a and b 
using System; 
  
class GFG { 
  
    // Function to count squares 
    static int countSquares(int a, int b) 
    { 
        // Initialize result 
        int cnt = 0; 
  
        // Traverse through all numbers 
        for (int i = a; i <= b; i++) 
  
            // Check if current number 
            // 'i' is perfect square 
            for (int j = 1; j * j <= i; j++) 
                if (j * j == i) 
                    cnt++; 
        return cnt; 
    } 
  
    // Driver Code 
    public static void Main() 
    { 
        int a = 9, b = 25; 
        Console.Write("Count of squares is " + countSquares(a, b)); 
    } 
} 
  
// This code is contributed by Sam007 

PHP

<?php 
// A Simple Method to count squares 
//between a and b 
  
function countSquares($a, $b) 
{ 
    $cnt = 0; // Initialize result 
  
    // Traverse through all numbers 
    for ($i = $a; $i <= $b; $i++) 
  
        // Check if current number 
        // 'i' is perfect square 
        for ($j = 1; $j * $j <= $i; 
                              $j++) 
            if ($j * $j == $i) 
                $cnt++; 
  
    return $cnt; 
} 
  
// Driver code 
  
    $a = 9; $b = 25; 
    echo "Count of squares is ". 
              countSquares($a, $b); 
  
// This code is contributed by ajit. 
?> 

Output :

Count of squares is 3

An upper bound on time Complexity of this solution is O((b-a) * sqrt(b)).

Method 2 (Efficient) We can simply take square root of ‘a’ and square root of ‘b’ and count the perfect squares between them using

floor(sqrt(b)) - ceil(sqrt(a)) + 1

We take floor of sqrt(b) because we need to consider 
numbers before b.

We take ceil of sqrt(a) because we need to consider 
numbers after a.


For example, let b = 24, a = 8.  floor(sqrt(b)) = 4, 
ceil(sqrt(a)) = 3.  And number of squares is 4 - 3 + 1
= 2. The two numbers are 9 and 16.

Below is the implementation of above idea :

C++

// An Efficient Method to count squares between a and b 
#include <bits/stdc++.h> 
using namespace std; 
  
// An efficient solution to count square between a 
// and b 
int countSquares(int a, int b) 
{ 
    return (floor(sqrt(b)) - ceil(sqrt(a)) + 1); 
} 
  
// Driver code 
int main() 
{ 
    int a = 9, b = 25; 
    cout << "Count of squares is "
         << countSquares(a, b); 
    return 0; 
} 

Java

// An Efficient method to count squares between 
// a and b 
class CountSquares { 
    double countSquares(int a, int b) 
    { 
        return (Math.floor(Math.sqrt(b)) - Math.ceil(Math.sqrt(a)) + 1); 
    } 
} 
  
// Driver Code 
public class PerfectSquares { 
    public static void main(String[] args) 
    { 
        int a = 9, b = 25; 
        CountSquares obj = new CountSquares(); 
        System.out.print("Count of squares is " + (int)obj.countSquares(a, b)); 
    } 
} 

Python

# An Efficient Method to count squares between a 
# and b 
import math 
def CountSquares(a, b): 
    return (math.floor(math.sqrt(b)) - math.ceil(math.sqrt(a)) + 1) 
  
# Driver Code 
a = 9
b = 25
print "Count of squares is:", int(CountSquares(a, b)) 

C#

// C# program for efficient method 
// to count squares between a & b 
using System; 
  
class GFG { 
  
    // Function to count squares 
    static double countSquares(int a, int b) 
    { 
        return (Math.Floor(Math.Sqrt(b)) - Math.Ceiling(Math.Sqrt(a)) + 1); 
    } 
  
    // Driver Code 
    public static void Main() 
    { 
        int a = 9, b = 25; 
        Console.Write("Count of squares is " + (int)countSquares(a, b)); 
    } 
} 
  
// This code is contributed by Sam007. 

PHP

<?php 
// An Efficient PHP code to count 
// squares between a and b 
  
// Method to count square 
// between a and b 
function countSquares($a, $b) 
{ 
    return (floor(sqrt($b)) -  
            ceil(sqrt($a)) + 1); 
} 
  
// Driver code 
{ 
    $a = 9; 
    $b = 25; 
    echo "Count of squares is ",  
           countSquares($a, $b); 
    return 0; 
} 
// This code is contributed by nitin mittal. 
?> 

Output :

Count of squares is 3

Time complexity of this solution is O(Log b). A typical implementation of square root for a number n takes time equal to O(Log n) [See this for a sample implementation of square root]

This article is contributed by Rahul Aggarwal. If you like GeeksforGeeks and would like to contribute, you can also write an article and mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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My Personal Notes

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

C++

Java

Python

C#

PHP

C++

Java

Python

C#

PHP

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