# 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 square in given range is 4. Input : a = 9, b = 25 Output : 3 The three perfect squares in given range are 9, 16 and 25

**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));` ` ` `}` `}` |

## Python3

`# 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.` `?>` |

## Javascript

`<script>` `// A Simple Method to count squares` `//between a and b` `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;` `}` `// Driver code` ` ` `let a = 9;` ` ` `let b = 25;` ` ` `document.write( ` `"Count of squares is "` `,` ` ` `countSquares(a, b));` `// This code is contributed by sravan.` `</script>` |

**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)) + 1We 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));` ` ` `}` `}` |

## Python3

`# 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.` `?>` |

## Javascript

`<script>` `// A Simple Method to count squares` `//between a and b` `function` `countSquares(a, b)` `{` ` ` `return` `(Math.floor(Math.sqrt(b)) - Math.ceil(Math.sqrt(a)) + 1);` `}` `// Driver code` ` ` `let a = 9;` ` ` `let b = 25;` ` ` `document.write( ` `"Count of squares is "` `,` ` ` `countSquares(a, b));` `// This code is contributed by sravan.` `</script>` |

**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 review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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