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

**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; ` `} ` |

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## 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)); ` ` ` `} ` `} ` |

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## 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) ` |

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## 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 ` |

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## 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. ` `?> ` |

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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; ` `} ` |

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## 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)); ` ` ` `} ` `} ` |

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## 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)) ` |

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## 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. ` |

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## 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. ` `?> ` |

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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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