Given an integer X, find its 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.
Naive 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.
Follow the steps below to implement the above idea
- 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
Below is the implementation of the above approach:
C++
// A C++ program to find floor(sqrt(x)#include <bits/stdc++.h>using namespace std;// Returns floor of square root of xint floorSqrt(int x){ // Base cases if (x == 0 || x == 1) return x; // Starting 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 programint main(){ int x = 11; cout << floorSqrt(x) << endl; return 0;} |
C
#include <stdio.h>// Returns floor of square root of xint floorSqrt(int x){ // Base cases if (x == 0 || x == 1) return x; // Starting 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 programint main(){ int x = 11; printf("%d\n", floorSqrt(x)); return 0;}// code is contributed by lalith kumar.g |
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; // Starting 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 xdef floorSqrt(x): # Base cases if (x == 0 or x == 1): return x # Starting 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 Codex = 11print(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; // Starting 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 xfunction floorSqrt($x){ // Base cases if ($x == 0 || $x == 1) return $x; // Starting 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 xfunction floorSqrt(x){ // Base cases if (x == 0 || x == 1) return x; // Starting 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 Codelet x = 11;document.write(floorSqrt(x));// This code is contributed by mohan</script> |
Output
3
Complexity Analysis:
- Time Complexity: O(√X). Only one traversal of the solution is needed, so the time complexity is O(√X).
- Auxiliary Space: O(1).
Thanks, Fattepur Mahesh for suggesting this solution.
Square root an integer using Binary search:
The idea is to find the largest integer i whose square is less than or equal to the given number. The values of i * i is monotonically increasing, so the problem can be solved using binary search.
Below is the implementation of the above idea:
- Base cases for the given problem are when the given number is 0 or 1, then return X;
- Create some variables, for storing the lower bound say l = 0, and for upper bound r = X / 2 (i.e, The floor of the square root of x cannot be more than x/2 when x > 1).
- 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 X, If yes then search for a larger value in the second half of the search space, i.e l = mid + 1, update ans = mid
- Else if the square of mid is more than X then search for a smaller value in the first half of the search space, i.e r = mid – 1
- Finally, Return the ans
Below is the implementation of the above approach:
C++
#include <iostream>using namespace std;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 / 2, ans; while (start <= end) { int mid = (start + end) / 2; // If x is a perfect square int sqr = mid * mid; if (sqr == 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; } Here basically if we multiply mid with itself so there will be integer overflow which will throw tle for larger input so to overcome this situation we can use long or we can just divide the number by mid which is same as checking mid*mid < x */ if (sqr <= x) { start = mid + 1; ans = mid; } else // If mid*mid is greater than x end = mid - 1; } return ans;}// Driver programint main(){ int x = 11; cout << floorSqrt(x) << endl; return 0;} |
C
#include <math.h>#include <stdio.h>int floorSqrt(int x){ // Base Cases if (x == 0 || x == 1) return x; // Do Binary Search for floor(sqrt(x)) long start = 1, end = x / 2, ans = 0; while (start <= end) { int mid = (start + end) / 2; // If x is a perfect square if (mid * mid == x) return (int)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 (int)ans;}// Driver Methodint main(){ int x = 11; printf("%d\n", floorSqrt(x));}// Contributed by lalith kumar.g |
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)) long start = 1, end = x / 2, ans = 0; while (start <= end) { long mid = (start + end) / 2; // If x is a perfect square if (mid * mid == x) return (int)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 (int)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 xdef floorSqrt(x): # Base cases if (x == 0 or x == 1): return x # Do Binary Search for floor(sqrt(x)) start = 1 end = x//2 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 codex = 11print(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 / 2, 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/2; $ans; while ($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?> |
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; // Do Binary Search // for floor(sqrt(x)) let start = 1; let end = x/2; let ans; while (start <= end) { let 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 Codelet x = 11;document.write(floorSqrt(x) + "<br>");// This code is contributed by _saurabh_jaiswal</script> |
Output
3
Complexity Analysis:
- Time Complexity: O(log(X)).
- Auxiliary Space: O(1).
Thanks to Gaurav Ahirwar for suggesting the above method.
Square root an integer using built-in functions:
Below is the implementation for finding the square root using the built-in function.
C++
#include <bits/stdc++.h>using namespace std;int countSquares(int x){ int sqr = sqrt(x); int result = (int)(sqr); return result;}int main(){ int x = 9; cout << (countSquares(x)); return 0;}// This code is contributed by Rajput-Ji |
C
#include <math.h>#include <stdio.h>static int countSquares(int x){ int sqr = (int)sqrt(x); int result = (int)(sqr); return result;}int main(){ int x = 9; printf("%d\n", countSquares(x));}// This code is contributed by lalith kumar.g |
Java
import java.util.*;class GFG { static int countSquares(int x) { int sqr = (int)Math.sqrt(x); int result = (int)(sqr); return result; } public static void main(String[] args) { int x = 9; System.out.print(countSquares(x)); }}// This code is contributed by Rajput-Ji |
Python3
def countSquares(x): sqrt = x**0.5 result = int(sqrt) return resultx = 9print(countSquares(x)) |
C#
using System;public class GFG { static int countSquares(int x) { int sqr = (int)Math.Sqrt(x); int result = (int)(sqr); return result; } public static void Main(String[] args) { int x = 9; Console.Write(countSquares(x)); }}// This code is contributed by Rajput-Ji |
Javascript
<script> function countSquares(x) { var sqr = parseInt( Math.sqrt(x)); var result = parseInt(sqr); return result; } var x = 9; document.write(countSquares(x));// This code is contributed by Rajput-Ji</script> |
Output
3
Time Complexity: O(log(X))
Auxiliary Space: O(1)
There can be many ways to solve this problem. For example, the Babylonian Method is one way.
Another Approach to Solve This Problem Using Exponential Function:
The basic idea behind the method is to calculate the exponential of the logarithm of the integer divided by two.
Below are steps to implement the above approach:
- Take the integer value as input and save it in a variable.
- Use the exponential function exp() and the logarithmic function log() from the <cmath> library to calculate the square root of the integer. exp(log(x) / 2) will give the square root of x.
- Use the floor() function to get the integer part of the result.
- Check whether the square of the floor result is equal to the input x.
- If the square of the floor result is equal to the input x, then return the floor result as it is the square root of x.
- If the square of the floor result is not equal to the input x, then return the floor result as the floor of the square root
Below is the implementation of the above approach:
C++
// C++ code to implement the above approach#include <iostream>// header file for math functions#include <cmath>using namespace std;int findSquareRoot(int x){ // using exponential and logarithmic function to // calculate square root of x double result = exp(log(x) / 2); // floor function to get integer part of the result int floorResult = floor(result); // If the integer square of the floor result is equal to // the input x, // then x is a perfect square, and floor result is the // square root. if (floorResult * floorResult == x) { return floorResult; } else { // If not, then x is not a perfect square, and // floor result is the floor of the square root. return floorResult; }}// Driver codeint main(){ int x = 11; int squareRoot = findSquareRoot( x); // calling the findSquareRoot function to // calculate the square root cout << squareRoot << endl; // printing the result return 0;}// This code is contributed by Veerendra_Singh_Rajpoot |
Java
// java code to implement the above approachimport java.lang.Math;public class Main { public static int findSquareRoot(int x) { // using exponential and logarithmic function to // calculate square root of x double result = Math.exp(Math.log(x) / 2); // floor function to get integer part of the result int floorResult = (int)Math.floor(result); // If the integer square of the floor result is // equal to the input x, then x is a perfect square, // and floor result is the square root. if (floorResult * floorResult == x) { return floorResult; } else { // If not, then x is not a perfect square, // and floor result is the floor of the // square root. return floorResult; } } // Driver code public static void main(String[] args) { int x = 11; int squareRoot = findSquareRoot( x); // calling the findSquareRoot function to // calculate the square root System.out.println( squareRoot); // printing the result }}// This code is contributed by Veerendra_Singh_Rajpoot |
Python3
# Python code to implement the above approachimport mathdef findSquareRoot(x): # using exponential and logarithmic function to # calculate square root of x result = math.exp(math.log(x) / 2) # floor function to get integer part of the result floorResult = math.floor(result) # If the integer square of the floor result is equal to # the input x, # then x is a perfect square, and floor result is the # square root. if floorResult * floorResult == x: return floorResult else: # If not, then x is not a perfect square, and # floor result is the floor of the square root. return floorResult# Driver codex = 11# calling the findSquareRoot function to calculate the square rootsquareRoot = findSquareRoot(x)print(squareRoot) # printing the result |
C#
using System;public class Program{ public static int FindSquareRoot(int x) { // using exponential and logarithmic function to // calculate square root of x double result = Math.Exp(Math.Log(x) / 2); // floor function to get integer part of the result int floorResult = (int)Math.Floor(result); // If the integer square of the floor result is equal to // the input x, // then x is a perfect square, and floor result is the // square root. if (floorResult * floorResult == x) { return floorResult; } else { // If not, then x is not a perfect square, and // floor result is the floor of the square root. return floorResult; } } public static void Main() { int x = 11; int squareRoot = FindSquareRoot(x); // calling the FindSquareRoot function to calculate the square root Console.WriteLine(squareRoot); // printing the result }} |
Javascript
function findSquareRoot(x) { // using exponential and logarithmic function to // calculate square root of x let result = Math.exp(Math.log(x) / 2); // floor function to get integer part of the result let floorResult = Math.floor(result); // If the integer square of the floor result is equal to // the input x, // then x is a perfect square, and floor result is the // square root. if (floorResult * floorResult == x) { return floorResult; } else { // If not, then x is not a perfect square, and // floor result is the floor of the square root. return floorResult; }}// Driver codelet x = 11;let squareRoot = findSquareRoot(x);// calling the findSquareRoot function to // calculate the square rootconsole.log(squareRoot); // printing the result |
Output
3
Time Complexity: O(1), The time complexity of the given approach is O(1) since it uses only one mathematical formula exp(log(x) / 2) which is constant time, and a few arithmetic operations, comparisons, and function calls that take constant time as well. Therefore, the time complexity of this algorithm is constant or O(1).
Auxiliary Space: O(1), The space complexity of the given approach is O(1) as it only uses a constant amount of extra space. It declares two integer variables, result and floorResult, which each take constant space, and there is no dynamic memory allocation or recursive calls. Therefore, the space complexity of this algorithm is constant or O(1).
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