Square root of an integer

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

C++

#includevàlt;bits/stdc++.hvàgt;

using namespace std;

 

int floorSqrt(int x)

{

    

    if (x == 0 || x == 1)

    return x;

 

    

    

    int i = 1, result = 1;

    while (result <= x)

    {

      i++;

      result = i * i;

    }

    return i - 1;

}

 

int main()

{

    int x = 11;

    cout << floorSqrt(x) << endl;

    return 0;

}



Java

 

class GFG {

     

    

    static int floorSqrt(int x)

    {

        

        if (x == || x == 1)

            return x;

 

        

        

        int i = 1, result = 1;

         

        while (result <= x) {

            i++;

            result = i * i;

        }

        return i - 1;

    }

 

    

    public static void main(String[] args)

    {

        int x = 11;

        System.out.print(floorSqrt(x));

    }

}

 



Python3

 

def floorSqrt(x):

 

    

    if (x == or x == 1):

        return x

 

    

    

    i = 1; result = 1

    while (result <= x):

     

        i += 1

        result = i * i

     

    return i - 1

 

x = 11

print(floorSqrt(x))

 



C#

using System;

 

class GFG

{

    

    

    static int floorSqrt(int x)

    {

        

        if (x == 0 || x == 1)

            return x;

 

        

        

        

        int i = 1, result = 1;

         

        while (result <= x)

        {

            i++;

            result = i * i;

        }

        return i - 1;

    }

 

Xem Thêm  Cách phát video trực tuyến bằng HTML - mã html để phát video

    

    static public void Main ()

    {

        int x = 11;

        Console.WriteLine(floorSqrt(x));

    }

}

 



PHP

<?php

 

function floorSqrt($x)

{

    

    if ($x == 0 || $x == 1)

    return $x;

 

    

    

    

    $i = 1;

    $result = 1;

    while ($result <= $x)

    {

        $i++;

        $result = $i * $i;

    }

    return $i - 1;

}

 

$x = 11;

echo floorSqrt($x), "n";

 

?>



Javascript

<scriptvàgt;

 

 

function floorSqrt(x)

{

     

    

    if (x == 0 || x == 1)

        return x;

 

    

    

    

    let i = 1;

    let result = 1;

     

    while (result <= x)

    {

        i++;

        result = i * i;

    }

    return i - 1;

}

 

let x = 11;

document.write(floorSqrt(x));

 

 

</scriptvàgt;



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.

 
Better Approach: The idea is to find the largest integer 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 , upperbound , where n is the given number, and to store the answer.
    3. Run a loop until , the search space vanishes
    4. Test if the square of mid () 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, cập nhật ans = mid
    5. Else if the square of mid is more 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 ( )
  • Implementation:

C++

#include <iostreamvàgt;

 

using namespace std;

int floorSqrt(int x)

{

    

    if (x == 0 || x == 1)

        return x;

 

    

    int start = 1, end = x/2, ans;

    while (start <= end) {

        int mid = (start + end) / 2;

 

        

        int sqr = mid * mid;

        if (sqr == x)

            return mid;

 

Xem Thêm  Cách tạo bảng từ truy vấn SQL - tạo bảng từ truy vấn chọn

        

        

        

 

        

           

                   

                           

                           

                   

            

           

           

           

            

           

 

           

        if (sqr <= x) {

            start = mid + 1;

            ans = mid;

        }

        else

            end = mid - 1;

    }

    return ans;

}

 

int main()

{

    int x = 20221;

    cout << floorSqrt(x) << endl;

    return 0;

}



Java

public class Check

{

    public static int floorSqrt(int x)

    {

        

        if (x == || x == 1)

            return x;

 

 

        

        long start = 1, end = x, ans=;

        while (start <= end)

        {

            int mid = (start + end) / 2;

 

            

            if (mid*mid == x)

                return (int)mid;

 

            

            

            if (mid*mid < x)

            {

                start = mid + 1;

                ans = mid;

            }

            else  

                end = mid-1;

        }

        return (int)ans;

    }

 

    

    public static void main(String args[])

    {

        int x = 11;

        System.out.println(floorSqrt(x));

    }

}



Python3

 

def floorSqrt(x) :

 

    

    if (x == or x == 1) :

        return x

  

    

    start = 1

    end = x  

    while (start <= end) :

        mid = (start + end) // 2

         

        

        if (mid*mid == x) :

            return mid

             

        

        

        

        if (mid * mid < x) :

            start = mid + 1

            ans = mid

             

        else :

             

            

            end = mid-1

             

    return ans

 

x = 11

print(floorSqrt(x))

     



C#

using System;

 

class GFG

{

    public static int floorSqrt(int x)

    {

        

        if (x == 0 || x == 1)

            return x;

 

        

        

        int start = 1, end = x, ans = 0;

        while (start <= end)

        {

            int mid = (start + end) / 2;

 

            

            

            if (mid * mid == x)

                return mid;

 

            

            

            

            

            if (mid * mid < x)

            {

                start = mid + 1;

                ans = mid;

            }

             

            

            

            else

                end = mid-1;

        }

        return ans;

    }

 

    

    static public void Main ()

    {

        int x = 11;

        Console.WriteLine(floorSqrt(x));

    }

}

 

Xem Thêm  Pandas For Loop - vòng lặp while ở gấu trúc



PHP

<?php

 

function floorSqrt($x)

{

    

    if ($x == 0 || $x == 1)

    return $x;

 

    

    

    $start = 1; $end = $x; $ans;

    while ($start <= $end)

    {

        $mid = ($start + $end) / 2;

 

        

        if ($mid * $mid == $x)

            return $mid;

 

        

        

        

        if ($mid * $mid < $x)

        {

            $start = $mid + 1;

            $ans = $mid;

        }

         

        

        

        else

            $end = $mid-1;    

    }

    return $ans;

}

 

$x = 11;

echo floorSqrt($x), "n";

 

?>



Javascript

<scriptvàgt;

    

 

function floorSqrt(x)

{

    

    if (x == 0 || x == 1)

    return x;

 

    

    

    let start = 1;

    let end = x;

    let ans;

    while (start <= end)

    {

        let mid = (start + end) / 2;

 

        

        if (mid * mid == x)

            return mid;

 

        

        

        

        if (mid * mid < x)

        {

            start = mid + 1;

            ans = mid;

        }

         

        

        

        else

            end = mid-1;    

    }

    return ans;

}

 

let x = 11;

document.write(floorSqrt(x) +  "<br>");

 

</scriptvàgt;



Output :

142
  • 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.
chú ý: 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.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

My Personal Notes

Viết một bình luận