Number of perfect squares between two given numbers (original) (raw)

Last Updated : 23 Jul, 2025

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

**Example

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

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

Try It Yourself

Table of Content

• [Naive Approach] - Counting perfect Squared between A and B
• [Expected Appproach] - Using Square Root of a and b

[Naive Approach] - One by One Counting

The idea for this approach is to use a loop to iterate from a to b and count the numbers that are perfect squares.

C++ `

#include <bits/stdc++.h> using namespace std;

int countSquares(int a, int b) {
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;

}

int main() { int a = 9, b = 25; cout <<countSquares(a, b); return 0; }

Java

class CountSquares {

static int countSquares(int a, int b)
{   
    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;
}

} public class PerfectSquares { public static void main(String[] args) { int a = 9, b = 25; CountSquares obj = new CountSquares(); System.out.print(obj.countSquares(a, b)); } }

Python

def CountSquares(a, b):

cnt = 0 

# 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

a = 9 b = 25 print (CountSquares(a, b))

C#

using System; class GFG {

// Function to count squares
static int countSquares(int a, int b)
{
    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;
}
public static void Main()
{
    int a = 9, b = 25;
    Console.Write(countSquares(a, b));
}

}

JavaScript

function countSquares(a, b) { let cnt = 0;

// Traverse through all numbers
for (let i = a; i <= b; i++)

    // Check if current number
    // 'i' is perfect square
    for (let j = 1; j * j <= i; j++)
        if (j * j == i)
            cnt++;

return cnt;

}

let a = 9; let b = 25; console.log(countSquares(a, b));

`

Output

Count of squares is 3

Time Complexity: O((b-a) * sqrt(b)).
Auxiliary Space: O(1)

**[Expected Approach] - Using Square Roots of a and b

The idea for this approach is to simply take square root of ‘a’ and ‘b’ and count the numbers between them by using this formula **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.

Example: b = 24, a = 8.
floor(sqrt(b)) = 4, ceil(sqrt(a)) = 3.
And number of squares is 4 - 3 + 1 = 2.(9 and 16) .

C++ `

#include <bits/stdc++.h> using namespace std;

int countSquares(int a, int b) { return (floor(sqrt(b)) - ceil(sqrt(a)) + 1); }

int main() { int a = 9, b = 25; cout <<countSquares(a, b); return 0; }

Java

class CountSquares { double countSquares(int a, int b) { return (Math.floor(Math.sqrt(b)) - Math.ceil(Math.sqrt(a)) + 1); } }

public class PerfectSquares { public static void main(String[] args) { int a = 9, b = 25; CountSquares obj = new CountSquares(); System.out.print((int)obj.countSquares(a, b)); } }

Python

import math def CountSquares(a, b): return (math.floor(math.sqrt(b)) - math.ceil(math.sqrt(a)) + 1)

a = 9 b = 25 print (int(CountSquares(a, b)))

C#

using System; class GFG {

static double countSquares(int a, int b)
{
    return (Math.Floor(Math.Sqrt(b)) - Math.Ceiling(Math.Sqrt(a)) + 1);
}

public static void Main()
{
    int a = 9, b = 25;
    Console.Write((int)countSquares(a, b));
}

}

JavaScript

function countSquares(a, b) { return (Math.floor(Math.sqrt(b)) - Math.ceil(Math.sqrt(a)) + 1); }

let a = 9; 
let b = 25;
console.log(countSquares(a, b));

`

Output

Count of squares is 3

**Time Complexity: O(log b) as any typical implementation of square root for a number n takes time equal to O(Log n)
**Auxiliary Space: O(1)