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:
- Create a variable (counter) i and take care of some base cases, i.e when the given number is 0 or 1.
- Run a loop until i*i <= n , where n is the given number. Increment i by 1.
- The floor of the square root of the number is i – 1
- Implementation:
C++
// 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; } |
Java
// 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. |
Python3
# 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. |
C#
// 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 |
PHP
<?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 ?> |
Javascript
<script> // A Javascript 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. let i = 1; let result = 1; while (result <= x) { i++; result = i * i; } return i - 1; } // Driver Code let x = 11; document.write(floorSqrt(x)); // This code is contributed by mohan </script> |
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.
- Time Complexity: O(√ n).
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:
- Take care of some base cases, i.e when the given number is 0 or 1.
- Create some variables, lowerbound l = 0, upperbound r = n, where n is the given number, mid and ans to store the answer.
- Run a loop until l <= r , the search space vanishes
- 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
- 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
- Print the value of answer ( ans)
- Implementation:
C++
// 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; } |
Java
// 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 |
Python3
# 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. |
C#
// 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 |
PHP
<?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 ?> |
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.
- Time complexity: O(log n).
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.
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