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; } |
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 = 9b = 25print "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 = 9b = 25print "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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