# Floor square root without using sqrt() function : Recursive

Given a number N, the task is to find the floor square root of the number N without using the built-in square root function. Floor square root of a number is the greatest whole number which is less than or equal to its square root.

Examples:

Input: N = 25
Output: 5
Explanation:
Square root of 25 = 5. Therefore 5 is the greatest whole number less than equal to Square root of 25.

Input: N = 30
Output: 5
Explanation:
Square root of 25 = 5.47
Therefore 5 is the greatest whole number less than equal to Square root of 25 (5.47)

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Naive Approach:
In the basic approach to find the floor square root of a number, find the square of numbers from 1 to N, till the square of some number K becomes greater than N. Hence the value of (K – 1) will be the floor square root of N.

Below is the algorithm to solve this problem using Naive approach:

• Iterate a loop from numbers 1 to N in K.
• For any K, if its square becomes greater than N, then K-1 is the floor square root of N.

Time Complexity: O(√N)

Efficient Approach:
From the Naive approach, it is clear that the floor square root of N will lie in the range [1, N]. Hence instead of checking each number in this range, we can efficiently search the required number in this range. Therefore, the idea is to use the binary search in order to efficiently find the floor square root of the number N in log N.

Below is the recursive algorithm to solve the above problem using Binary Search:

1. Implement the Binary Search in the range 0 to N.
2. Find the mid value of the range using formula:
mid = (start + end) / 2

3. Base Case: The recursive call will get executed till square of mid is less than or equal to N and the square of (mid+1) is greater than equal to N.
(mid2 ≤ N) and ((mid + 1)2 > N)

4. If the base case is not satisfied, then the range will get changed accordingly.
• If the square of mid is less than equal to N, then the range gets updated to [mid + 1, end]
if(mid2 ≤ N)
updated range = [mid + 1, end]

• If the square of mid is greater than N, then the range gets updated to [low, mid + 1]
if(mid2 > N)
updated range = [low, mid - 1]

Below is the implmentation of the above approach:

## C++

 // C++ implementation to find the // square root of the number N // without using sqrt() function    #include using namespace std;    // Function to find the square // root of the number N using BS int sqrtSearch(int low, int high, int N) {        // If the range is still valid     if (low <= high) {            // Find the mid-value of the range         int mid = (low + high) / 2;            // Base Case         if ((mid * mid <= N)             && ((mid + 1) * (mid + 1) > N)) {             return mid;         }            // Condition to check if the         // left search space is useless         else if (mid * mid < N) {             return sqrtSearch(mid + 1, high, N);         }         else {             return sqrtSearch(low, mid - 1, N);         }     }     return low; }    // Driver Code int main() {     int N = 25;     cout << sqrtSearch(0, N, N)          << endl;     return 0; }

## Java

 // Java implementation to find the // square root of the number N // without using sqrt() function class GFG {            // Function to find the square     // root of the number N using BS     static int sqrtSearch(int low, int high, int N)     {                // If the range is still valid         if (low <= high) {                    // Find the mid-value of the range             int mid = (int)(low + high) / 2;                    // Base Case             if ((mid * mid <= N)                 && ((mid + 1) * (mid + 1) > N)) {                 return mid;             }                    // Condition to check if the             // left search space is useless             else if (mid * mid < N) {                 return sqrtSearch(mid + 1, high, N);             }             else {                 return sqrtSearch(low, mid - 1, N);             }         }         return low;     }            // Driver Code     public static void main (String[] args)     {         int N = 25;         System.out.println(sqrtSearch(0, N, N));     } }    // This code is contributed by Yash_R

## Python3

 # Python3 implementation to find the  # square root of the number N  # without using sqrt() function     # Function to find the square  # root of the number N using BS  def sqrtSearch(low, high, N) :         # If the range is still valid      if (low <= high) :            # Find the mid-value of the range          mid = (low + high) // 2;             # Base Case          if ((mid * mid <= N) and ((mid + 1) * (mid + 1) > N)) :             return mid;             # Condition to check if the          # left search space is useless          elif (mid * mid < N) :              return sqrtSearch(mid + 1, high, N);                 else :             return sqrtSearch(low, mid - 1, N);         return low;     # Driver Code  if __name__ == "__main__" :         N = 25;      print(sqrtSearch(0, N, N))     # This code is contributed by Yash_R

## C#

 // C# implementation to find the // square root of the number N // without using sqrt() function using System;    class GFG {             // Function to find the square     // root of the number N using BS     static int sqrtSearch(int low, int high, int N)     {                 // If the range is still valid         if (low <= high) {                     // Find the mid-value of the range             int mid = (int)(low + high) / 2;                     // Base Case             if ((mid * mid <= N)                 && ((mid + 1) * (mid + 1) > N)) {                 return mid;             }                     // Condition to check if the             // left search space is useless             else if (mid * mid < N) {                 return sqrtSearch(mid + 1, high, N);             }             else {                 return sqrtSearch(low, mid - 1, N);             }         }         return low;     }             // Driver Code     public static void Main(String[] args)     {         int N = 25;         Console.WriteLine(sqrtSearch(0, N, N));     } }    // This code is contributed by PrinciRaj1992

Output:

5

Performance Analysis:

• Time Complexity: As in the above approach, there is Binary Search used over the search space of 0 to N which takes O(log N) time in worst case, Hence the Time Complexity will be O(log N).
• Space Complexity: As in the above approach, taking consideration of the stack space used in the recursive calls which can take O(logN) space in worst case, Hence the space complexity will be O(log N)

GeeksforGeeks has prepared a complete interview preparation course with premium videos, theory, practice problems, TA support and many more features. Please refer Placement 100 for details

My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Improved By : Yash_R, princiraj1992