Given an integer x, find it’s square root. If x is not a perfect square, then return floor(√x).

Examples :

Input: x = 4
Output: 2
Explanation:  The square root of 4 is 2.

Input: x = 11
Output: 3
Explanation:  The square root of 11 lies in between
3 and 4 so floor of the square root is 3.

There can be many ways to solve this problem. For example Babylonian Method is one way.

Simple Approach: To find the floor of the square root, try with all-natural numbers starting from 1. Continue incrementing the number until the square of that number is greater than the given number.

• Algorithm:
  1. Create a variable (counter) i and take care of some base cases, i.e when the given number is 0 or 1.
  2. Run a loop until i*i <= n , where n is the given number. Increment i by 1.
  3. The floor of the square root of the number is i – 1
• Implementation:
  

  C++

  
  

  
  

  filter_none

  edit
  close

  play_arrow

  link
  brightness_4
  code

  // A C++ program to find floor(sqrt(x) #include<bits/stdc++.h> using namespace std;    // Returns floor of square root of x int floorSqrt(int x) {     // Base cases     if (x == 0 || x == 1)     return x;        // Staring from 1, try all numbers until     // i*i is greater than or equal to x.     int i = 1, result = 1;     while (result <= x)     {       i++;       result = i * i;     }     return i - 1; }    // Driver program int main() {     int x = 11;     cout << floorSqrt(x) << endl;     return 0; }

  chevron_right

  filter_none

  Java

  filter_none

  edit
  close

  play_arrow

  link
  brightness_4
  code

  // A Java program to find floor(sqrt(x))    class GFG {            // Returns floor of square root of x     static int floorSqrt(int x)     {         // Base cases         if (x == 0 || x == 1)             return x;            // Staring from 1, try all numbers until         // i*i is greater than or equal to x.         int i = 1, result = 1;                    while (result <= x) {             i++;             result = i * i;         }         return i - 1;     }        // Driver program     public static void main(String[] args)     {         int x = 11;         System.out.print(floorSqrt(x));     } }    // This code is contributed by Smitha Dinesh Semwal.

  chevron_right

  filter_none

  Python3

  filter_none

  edit
  close

  play_arrow

  link
  brightness_4
  code

  # Python3 program to find floor(sqrt(x)    # Returns floor of square root of x def floorSqrt(x):        # Base cases     if (x == 0 or x == 1):         return x        # Staring from 1, try all numbers until     # i*i is greater than or equal to x.     i = 1; result = 1     while (result <= x):                i += 1         result = i * i            return i - 1    # Driver Code x = 11 print(floorSqrt(x))    # This code is contributed by Smitha Dinesh Semwal.

  chevron_right

  filter_none

  C#

  filter_none

  edit
  close

  play_arrow

  link
  brightness_4
  code

  // A C# program to  // find floor(sqrt(x)) using System;    class GFG {     // Returns floor of     // square root of x     static int floorSqrt(int x)     {         // Base cases         if (x == 0 || x == 1)             return x;            // Staring from 1, try all         // numbers until i*i is          // greater than or equal to x.         int i = 1, result = 1;                    while (result <= x)          {             i++;             result = i * i;         }         return i - 1;     }        // Driver Code     static public void Main ()     {         int x = 11;         Console.WriteLine(floorSqrt(x));     } }    // This code is contributed by ajit

  chevron_right

  filter_none

  PHP

  filter_none

  edit
  close

  play_arrow

  link
  brightness_4
  code

  <?php // A PHP program to find floor(sqrt(x)    // Returns floor of square root of x function floorSqrt($x) {     // Base cases     if ($x == 0 || $x == 1)     return $x;        // Staring from 1, try all      // numbers until i*i is      // greater than or equal to x.     $i = 1;     $result = 1;     while ($result <= $x)     {         $i++;         $result = $i * $i;     }     return $i - 1; }    // Driver Code $x = 11; echo floorSqrt($x), "\n";    // This code is contributed by ajit ?>

  chevron_right

  filter_none

  
  Output :

  3

• Complexity Analysis:
  • Time Complexity: O(√ n).
    Only one traversal of the solution is needed, so the time complexity is O(√ n).
  • Space Complexity: O(1).
    Constant extra space is needed.

Thanks Fattepur Mahesh for suggesting this solution.

Better Approach: The idea is to find the largest integer i whose square is less than or equal to the given number. The idea is to use Binary Search to solve the problem. The values of i * i is monotonically increasing, so the problem can be solved using binary search.

• Algorithm:
  1. Take care of some base cases, i.e when the given number is 0 or 1.
  2. Create some variables, lowerbound l = 0, upperbound r = n, where n is the given number, mid and ans to store the answer.
  3. Run a loop until l <= r , the search space vanishes
  4. Check if the square of mid (mid = (l + r)/2 ) is less than or equal to n, If yes then search for a larger value in second half oF search space, i.e l = mid + 1, update ans = mid
  5. Else if the square of mid is less than n then search for a smaller value in first half oF search space, i.e r = mid – 1
  6. Print the value of answer ( ans)
• Implementation:
  

  C/C++

  filter_none

  edit
  close

  play_arrow

  link
  brightness_4
  code

  // A C++ program to find floor(sqrt(x) #include<bits/stdc++.h> using namespace std;    // Returns floor of square root of x          int floorSqrt(int x)  {         // Base cases     if (x == 0 || x == 1)         return x;        // Do Binary Search for floor(sqrt(x))     int start = 1, end = x, ans;        while (start <= end)      {                 int mid = (start + end) / 2;            // If x is a perfect square         if (mid*mid == x)             return mid;            // Since we need floor, we update answer when mid*mid is          // smaller than x, and move closer to sqrt(x)         if (mid*mid < x)          {             start = mid + 1;             ans = mid;         }          else // If mid*mid is greater than x             end = mid-1;             }     return ans; }     // Driver program int main()  {          int x = 11;     cout << floorSqrt(x) << endl;     return 0;    }

  chevron_right

  filter_none

  Java

  filter_none

  edit
  close

  play_arrow

  link
  brightness_4
  code

  // A Java program to find floor(sqrt(x) public class Test {     public static int floorSqrt(int x)     {         // Base Cases         if (x == 0 || x == 1)             return x;            // Do Binary Search for floor(sqrt(x))         int start = 1, end = x, ans=0;         while (start <= end)         {             int mid = (start + end) / 2;                // If x is a perfect square             if (mid*mid == x)                 return mid;                // Since we need floor, we update answer when mid*mid is             // smaller than x, and move closer to sqrt(x)             if (mid*mid < x)             {                 start = mid + 1;                 ans = mid;             }             else   // If mid*mid is greater than x                 end = mid-1;         }         return ans;     }        // Driver Method     public static void main(String args[])     {         int x = 11;         System.out.println(floorSqrt(x));     } } // Contributed by InnerPeace

  chevron_right

  filter_none

  Python3

  filter_none

  edit
  close

  play_arrow

  link
  brightness_4
  code

  # Python 3 program to find floor(sqrt(x)    # Returns floor of square root of x          def floorSqrt(x) :        # Base cases     if (x == 0 or x == 1) :         return x         # Do Binary Search for floor(sqrt(x))     start = 1     end = x        while (start <= end) :         mid = (start + end) // 2                    # If x is a perfect square         if (mid*mid == x) :             return mid                        # Since we need floor, we update          # answer when mid*mid is smaller         # than x, and move closer to sqrt(x)         if (mid * mid < x) :             start = mid + 1             ans = mid                        else :                            # If mid*mid is greater than x             end = mid-1                    return ans    # driver code     x = 11 print(floorSqrt(x))        # This code is contributed by Nikita Tiwari.

  chevron_right

  filter_none

  C#

  filter_none

  edit
  close

  play_arrow

  link
  brightness_4
  code

  // A C# program to  // find floor(sqrt(x) using System;    class GFG {     public static int floorSqrt(int x)     {         // Base Cases         if (x == 0 || x == 1)             return x;            // Do Binary Search          // for floor(sqrt(x))         int start = 1, end = x, ans = 0;         while (start <= end)         {             int mid = (start + end) / 2;                // If x is a              // perfect square             if (mid * mid == x)                 return mid;                // Since we need floor, we              // update answer when mid *              // mid is smaller than x,              // and move closer to sqrt(x)             if (mid * mid < x)             {                 start = mid + 1;                 ans = mid;             }                            // If mid*mid is              // greater than x             else                  end = mid-1;         }         return ans;     }        // Driver Code     static public void Main ()     {         int x = 11;         Console.WriteLine(floorSqrt(x));     } }    // This code is Contributed by m_kit

  chevron_right

  filter_none

  PHP

  filter_none

  edit
  close

  play_arrow

  link
  brightness_4
  code

  <?php // A PHP program to find floor(sqrt(x)    // Returns floor of  // square root of x          function floorSqrt($x)  {      // Base cases     if ($x == 0 || $x == 1)      return $x;        // Do Binary Search      // for floor(sqrt(x))     $start = 1; $end = $x; $ans;      while ($start <= $end)      {          $mid = ($start + $end) / 2;            // If x is a perfect square         if ($mid * $mid == $x)             return $mid;            // Since we need floor, we update          // answer when mid*mid is  smaller         // than x, and move closer to sqrt(x)         if ($mid * $mid < $x)          {             $start = $mid + 1;             $ans = $mid;         }                     // If mid*mid is          // greater than x         else              $end = $mid-1;          }     return $ans; }    // Driver Code $x = 11; echo floorSqrt($x), "\n";    // This code is contributed by ajit ?>

  chevron_right

  filter_none

  
  Output :

  3

• Complexity Analysis:
  • Time complexity: O(log n).
    The time complexity of binary search is O(log n).
  • Space Complexity: O(1).
    Constant extra space is needed.

Thanks to Gaurav Ahirwar for suggesting above method.

Note: The Binary Search can be further optimized to start with ‘start’ = 0 and ‘end’ = x/2. Floor of square root of x cannot be more than x/2 when x > 1.

Thanks to vinit for suggesting above optimization.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.