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) and take care of some base cases, i.e when the given number is 0 or 1.
- Run a loop until , where n is the given number. Increment i by 1.
- 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;
}
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.
- Time Complexity: O(√ n).
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:
- Take care of some base cases, i.e when the given number is 0 or 1.
- Create some variables, lowerbound , upperbound , where n is the given number, and to store the answer.
- Run a loop until , the search space vanishes
- 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
- 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
- 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;
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));
}
}
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.
- Time complexity: O(log n).
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.
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