Skip to content
Related Articles

Related Articles

Print all perfect squares from the given range
  • Last Updated : 08 Apr, 2021

Given a range [L, R], the task is to print all the perfect squares from the given range.
Examples: 
 

Input: L = 2, R = 24 
Output: 4 9 16
Input: L = 1, R = 100 
Output: 1 4 9 16 25 36 49 64 81 100 
 

 

Naive approach: Starting from L to R check whether the current element is a perfect square or not. If yes then print it.
Below is the implementation of the above approach: 
 

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to print all the perfect
// squares from the given range
void perfectSquares(float l, float r)
{
 
    // For every element from the range
    for (int i = l; i <= r; i++) {
 
        // If current element is
        // a perfect square
        if (sqrt(i) == (int)sqrt(i))
            cout << i << " ";
    }
}
 
// Driver code
int main()
{
    int l = 2, r = 24;
 
    perfectSquares(l, r);
 
    return 0;
}

Java




//Java implementation of the approach
import java.io.*;
 
class GFG
{
     
// Function to print all the perfect
// squares from the given range
static void perfectSquares(int l, int r)
{
 
    // For every element from the range
    for (int i = l; i <= r; i++)
    {
 
        // If current element is
        // a perfect square
        if (Math.sqrt(i) == (int)Math.sqrt(i))
            System.out.print(i + " ");
    }
}
 
// Driver code
public static void main (String[] args)
{
    int l = 2, r = 24;
    perfectSquares(l, r);
}
}
 
// This code is contributed by jit_t

Python3




# Python3 implementation of the approach
 
# Function to print all the perfect
# squares from the given range
def perfectSquares(l, r):
 
    # For every element from the range
    for i in range(l, r + 1):
 
        # If current element is
        # a perfect square
        if (i**(.5) == int(i**(.5))):
            print(i, end=" ")
 
# Driver code
l = 2
r = 24
 
perfectSquares(l, r)
 
# This code is contributed by mohit kumar 29

C#




// C# implementation of the approach
using System;
 
class GFG
{
     
// Function to print all the perfect
// squares from the given range
static void perfectSquares(int l, int r)
{
 
    // For every element from the range
    for (int i = l; i <= r; i++)
    {
 
        // If current element is
        // a perfect square
        if (Math.Sqrt(i) == (int)Math.Sqrt(i))
            Console.Write(i + " ");
    }
}
 
// Driver code
public static void Main(String[] args)
{
    int l = 2, r = 24;
    perfectSquares(l, r);
}
}
 
// This code is contributed by 29AjayKumar

Javascript




<script>
// js implementation of the approach
 
// Function to print all the perfect
// squares from the given range
function perfectSquares(l, r){
 
    //For every element from the range
    for (let i = l; i <= r; i++)
    {
   
        // If current element is
        // a perfect square
        if (Math.sqrt(i) == parseInt(Math.sqrt(i)))
            document.write(i + " ");
    }
}
 
// Driver code
let l = 2;
let r = 24;
perfectSquares(l, r)
 
// This code is contributed by sravan.
</script>
Output: 



4 9 16

 

It is solution with O(n). moreover the use of number of square roots leads to computational expense.
Efficient approach: This method is based on the fact that the very first perfect square after number L will definitely be the square of ⌈sqrt(L)⌉. In very simple terms, the square root of L will be very close to the number whose square root we are trying to find. Therefore, the number will be pow(ceil(sqrt(L)), 2)
The very first perfect square is important for this method. Now the original answer is hidden over this pattern i.e. 0 1 4 9 16 25 
the difference between 0 and 1 is 1 
the difference between 1 and 4 is 3 
the difference between 4 and 9 is 5 and so on… 
which means that the difference between two perfect squares is always an odd number.
Now, the question arises what must be added to get the next number and the answer is (sqrt(X) * 2) + 1 where X is the already known perfect square.
Let the current perfect square be 4 then the next perfect square will definitely be 4 + (sqrt(4) * 2 + 1) = 9. Here, number 5 is added and the next number to be added will be 7 then 9 and so on… which makes a series of odd numbers.
Addition is computationally less expensive than performing multiplication or finding square roots of every number.
Below is the implementation of the above approach: 
 

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to print all the perfect
// squares from the given range
void perfectSquares(float l, float r)
{
 
    // Getting the very first number
    int number = ceil(sqrt(l));
 
    // First number's square
    int n2 = number * number;
 
    // Next number is at the difference of
    number = (number * 2) + 1;
 
    // While the perfect squares
    // are from the range
    while ((n2 >= l && n2 <= r)) {
 
        // Print the perfect square
        cout << n2 << " ";
 
        // Get the next perfect square
        n2 = n2 + number;
 
        // Next odd number to be added
        number += 2;
    }
}
 
// Driver code
int main()
{
    int l = 2, r = 24;
 
    perfectSquares(l, r);
 
    return 0;
}

Java




// Java implementation of the approach
class GFG
{
 
// Function to print all the perfect
// squares from the given range
static void perfectSquares(float l, float r)
{
 
    // Getting the very first number
    int number = (int) Math.ceil(Math.sqrt(l));
 
    // First number's square
    int n2 = number * number;
 
    // Next number is at the difference of
    number = (number * 2) + 1;
 
    // While the perfect squares
    // are from the range
    while ((n2 >= l && n2 <= r))
    {
 
        // Print the perfect square
        System.out.print(n2 + " ");
 
        // Get the next perfect square
        n2 = n2 + number;
 
        // Next odd number to be added
        number += 2;
    }
}
 
// Driver code
public static void main(String[] args)
{
    int l = 2, r = 24;
 
    perfectSquares(l, r);
 
}
}
 
// This code is contributed by 29AjayKumar

Python3




# Python3 implementation of the approach
 
from math import ceil, sqrt
 
# Function to print all the perfect
# squares from the given range
def perfectSquares(l, r) :
 
 
    # Getting the very first number
    number = ceil(sqrt(l));
 
    # First number's square
    n2 = number * number;
 
    # Next number is at the difference of
    number = (number * 2) + 1;
 
    # While the perfect squares
    # are from the range
    while ((n2 >= l and n2 <= r)) :
 
        # Print the perfect square
        print(n2, end= " ");
 
        # Get the next perfect square
        n2 = n2 + number;
 
        # Next odd number to be added
        number += 2;
 
# Driver code
if __name__ == "__main__" :
 
    l = 2; r = 24;
 
    perfectSquares(l, r);
 
# This code is contributed by AnkitRai01

C#




// C# implementation of the approach
using System;
 
class GFG
{
 
// Function to print all the perfect
// squares from the given range
static void perfectSquares(float l, float r)
{
 
    // Getting the very first number
    int number = (int) Math.Ceiling(Math.Sqrt(l));
 
    // First number's square
    int n2 = number * number;
 
    // Next number is at the difference of
    number = (number * 2) + 1;
 
    // While the perfect squares
    // are from the range
    while ((n2 >= l && n2 <= r))
    {
 
        // Print the perfect square
        Console.Write(n2 + " ");
 
        // Get the next perfect square
        n2 = n2 + number;
 
        // Next odd number to be added
        number += 2;
    }
}
 
// Driver code
public static void Main(String[] args)
{
    int l = 2, r = 24;
 
    perfectSquares(l, r);
}
}
 
// This code is contributed by Rajput Ji

Javascript




<script>
// Javascript implementation of the approach
 
// Function to print all the perfect
// squares from the given range
function perfectSquares(l, r)
{
 
    // Getting the very first number
    let number = Math.ceil(Math.sqrt(l));
 
    // First number's square
    let n2 = number * number;
 
    // Next number is at the difference of
    number = (number * 2) + 1;
 
    // While the perfect squares
    // are from the range
    while ((n2 >= l && n2 <= r)) {
 
        // Print the perfect square
        document.write(n2 + " ");
 
        // Get the next perfect square
        n2 = n2 + number;
 
        // Next odd number to be added
        number += 2;
    }
}
 
// Driver code
let l = 2, r = 24;
 
perfectSquares(l, r);
 
// This code is contributed by subhammahato348
</script>
Output: 
4 9 16

 

Attention reader! Don’t stop learning now. Get hold of all the important mathematical concepts for competitive programming with the Essential Maths for CP Course at a student-friendly price. To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

My Personal Notes arrow_drop_up
Recommended Articles
Page :