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;
}
chevron_rightfilter_noneJava
// 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_rightfilter_nonePython3
# 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_rightfilter_noneC#
// 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_rightfilter_nonePHP
<?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_rightfilter_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.
- 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/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;
}
chevron_rightfilter_noneJava
// 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_rightfilter_nonePython3
# 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_rightfilter_noneC#
// 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_rightfilter_nonePHP
<?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_rightfilter_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.
- 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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