 Open in App
Not now

# Square root of an integer

• Difficulty Level : Easy
• Last Updated : 15 Sep, 2022

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.

Recommended Practice

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

1. Create a variable (counter) i and take care of some base cases, (i.e when the given number is 0 or 1).
2. Run a loop until i*i <= n, where n is the given number. Increment i by 1.
3. 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 ``using` `namespace` `std;` `// Returns floor of square root of x``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``int` `main()``{``    ``int` `x = 11;``    ``cout << floorSqrt(x) << endl;``    ``return` `0;``}`

## C

 `#include ` `// Returns floor of square root of x``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``int` `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 x`  `def` `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 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;` `        ``// 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

 ``

## Javascript

 ``

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:

1. Base cases for the given problem are when the given number is 0 or 1, then return X;
2. 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).
3. Run a loop until l <= r, the search space vanishes
4. 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
5. 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
6. Finally, Return the ans

Below is the implementation of the above approach:

## C++

 `#include ` `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 program``int` `main()``{``    ``int` `x = 11;``    ``cout << floorSqrt(x) << endl;``    ``return` `0;``}`

## C

 `#include ``#include ` `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 Method``int` `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 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``/``/``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 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 / 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

 ``

## Javascript

 ``

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 ``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 ``#include ` `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` `result`  `x ``=` `9``print``(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

 ``

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.

My Personal Notes arrow_drop_up