# Find square root of number upto given precision using binary search

Given a positive number n and precision p, find the square root of number upto p decimal places using binary search.
Note : Prerequisite : Binary search
Examples:

```Input : number = 50, precision = 3
Output : 7.071

Input : number = 10, precision = 4
Output : 3.1622
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

We have discussed how to compute integral value of square root in Square Root using Binary Search

Approach :
1) As the square root of number lies in range 0 <= squareRoot <= number, therefore, initialize start and end as : start = 0, end = number.
2) Compare the square of mid integer with the given number. If it is equal to the number, then we found our integral part, else look for the same in left or right side depending upon the scenario.
3) Once we are done with finding the integral part, start computing the fractional part.
4) Initialize the increment variable by 0.1 and iteratively compute the fractional part upto p places. For each iteration, increment changes to 1/10th of it’s previous value.
5) Finally return the answer computed.

Below is the implementation of above approach :

## C++

 `// C++ implementation to find ` `// square root of given number ` `// upto given precision using ` `// binary search. ` `#include ` `using` `namespace` `std; ` ` `  `// Function to find square root ` `// of given number upto given ` `// precision ` `float` `squareRoot(``int` `number, ``int` `precision) ` `{ ` `    ``int` `start = 0, end = number; ` `    ``int` `mid; ` ` `  `    ``// variable to store the answer ` `    ``float` `ans; ` ` `  `    ``// for computing integral part ` `    ``// of square root of number ` `    ``while` `(start <= end) { ` `        ``mid = (start + end) / 2; ` `        ``if` `(mid * mid == number) { ` `            ``ans = mid; ` `            ``break``; ` `        ``} ` ` `  `        ``// incrementing start if integral ` `        ``// part lies on right side of the mid ` `        ``if` `(mid * mid < number) { ` `            ``start = mid + 1; ` `            ``ans = mid; ` `        ``} ` ` `  `        ``// decrementing end if integral part ` `        ``// lies on the left side of the mid ` `        ``else` `{ ` `            ``end = mid - 1; ` `        ``} ` `    ``} ` ` `  `    ``// For computing the fractional part ` `    ``// of square root upto given precision ` `    ``float` `increment = 0.1; ` `    ``for` `(``int` `i = 0; i < precision; i++) { ` `        ``while` `(ans * ans <= number) { ` `            ``ans += increment; ` `        ``} ` ` `  `        ``// loop terminates when ans * ans > number ` `        ``ans = ans - increment; ` `        ``increment = increment / 10; ` `    ``} ` `    ``return` `ans; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``// Function calling ` `    ``cout << squareRoot(50, 3) << endl; ` ` `  `    ``// Function calling ` `    ``cout << squareRoot(10, 4) << endl; ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation to find ` `// square root of given number ` `// upto given precision using ` `// binary search. ` `import` `java.io.*; ` ` `  `class` `GFG { ` `     `  `    ``// Function to find square root ` `    ``// of given number upto given ` `    ``// precision ` `    ``static` `float` `squareRoot(``int` `number, ``int` `precision) ` `    ``{ ` `        ``int` `start = ``0``, end = number; ` `        ``int` `mid; ` ` `  `        ``// variable to store the answer ` `        ``double` `ans = ``0.0``; ` ` `  `        ``// for computing integral part ` `        ``// of square root of number ` `        ``while` `(start <= end)  ` `        ``{ ` `            ``mid = (start + end) / ``2``; ` `             `  `            ``if` `(mid * mid == number)  ` `            ``{ ` `                ``ans = mid; ` `                ``break``; ` `            ``} ` ` `  `            ``// incrementing start if integral ` `            ``// part lies on right side of the mid ` `            ``if` `(mid * mid < number) { ` `                ``start = mid + ``1``; ` `                ``ans = mid; ` `            ``} ` ` `  `            ``// decrementing end if integral part ` `            ``// lies on the left side of the mid ` `            ``else` `{ ` `                ``end = mid - ``1``; ` `            ``} ` `        ``} ` ` `  `        ``// For computing the fractional part ` `        ``// of square root upto given precision ` `        ``double` `increment = ``0.1``; ` `        ``for` `(``int` `i = ``0``; i < precision; i++) { ` `            ``while` `(ans * ans <= number) { ` `                ``ans += increment; ` `            ``} ` ` `  `            ``// loop terminates when ans * ans > number ` `            ``ans = ans - increment; ` `            ``increment = increment / ``10``; ` `        ``} ` `        ``return` `(``float``)ans; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``// Function calling ` `        ``System.out.println(squareRoot(``50``, ``3``)); ` ` `  `        ``// Function calling ` `        ``System.out.println(squareRoot(``10``, ``4``)); ` `    ``} ` `} ` ` `  `// This code is contributed by vt_m. `

## Python3

 `# Python3 implementation to find ` `# square root of given number ` `# upto given precision using ` `# binary search. ` ` `  `# Function to find square root of ` `# given number upto given precision ` `def` `squareRoot(number, precision): ` ` `  `    ``start ``=` `0` `    ``end ``=` `number ` ` `  `    ``# For computing integral part ` `    ``# of square root of number ` `    ``while` `(start <``=` `end) : ` `        ``mid ``=` `int``((start ``+` `end) ``/` `2``) ` `         `  `        ``if` `(mid ``*` `mid ``=``=` `number) : ` `            ``ans ``=` `mid ` `            ``break` `         `  `        ``# incrementing start if integral ` `        ``# part lies on right side of the mid ` `        ``if` `(mid ``*` `mid < number) : ` `            ``start ``=` `mid ``+` `1` `            ``ans ``=` `mid ` `         `  `        ``# decrementing end if integral part ` `        ``# lies on the left side of the mid ` `        ``else` `: ` `            ``end ``=` `mid ``-` `1` `         `  `    ``# For computing the fractional part ` `    ``# of square root upto given precision ` `    ``increment ``=` `0.1` `    ``for` `i ``in` `range``(``0``, precision):  ` `        ``while` `(ans ``*` `ans <``=` `number): ` `            ``ans ``+``=` `increment ` `         `  `        ``# loop terminates when ans * ans > number ` `        ``ans ``=` `ans ``-` `increment ` `        ``increment ``=` `increment ``/` `10` `     `  `    ``return` `ans ` ` `  `# Driver code ` `print``(``round``(squareRoot(``50``, ``3``), ``4``)) ` `print``(``round``(squareRoot(``10``, ``4``), ``4``)) ` `     `  `# This code is contributed by Smitha Dinesh Semwal.  `

## C#

 `// C# implementation to find ` `// square root of given number ` `// upto given precision using ` `// binary search. ` `using` `System; ` ` `  `class` `GFG { ` `     `  `    ``static` `int` `mod = 1000000007; ` `     `  `    ``// Finding number of possible number with ` `    ``// n digits excluding a particular digit ` `    ``static` `int` `digitNumber(``long` `n) { ` `     `  `        ``// Checking if number of digits is zero ` `        ``if` `(n == 0) ` `            ``return` `1; ` `     `  `        ``// Checking if number of digits is one ` `        ``if` `(n == 1) ` `            ``return` `9; ` `     `  `        ``// Checking if number of digits is odd ` `        ``if` `(n % 2 != 0) { ` `     `  `            ``// Calling digitNumber function ` `            ``// with (digit-1)/2 digits ` `            ``int` `temp = digitNumber((n - 1) / 2) % mod; ` `             `  `            ``return` `(9 * (temp * temp) % mod) % mod; ` `        ``}  ` `        ``else` `{ ` `     `  `            ``// Calling digitNumber function ` `            ``// with n/2 digits ` `            ``int` `temp = digitNumber(n / 2) % mod; ` `             `  `            ``return` `(temp * temp) % mod; ` `        ``} ` `    ``} ` `     `  `    ``static` `int` `countExcluding(``int` `n, ``int` `d) { ` `         `  `        ``// Calling digitNumber function ` `        ``// Checking if excluding digit is ` `        ``// zero or non-zero ` `        ``if` `(d == 0) ` `            ``return` `(9 * digitNumber(n - 1)) % mod; ` `        ``else` `            ``return` `(8 * digitNumber(n - 1)) % mod; ` `    ``} ` `     `  `    ``// Driver function to run above program ` `    ``public` `static` `void` `Main() { ` `         `  `        ``// Initializing variables ` `        ``int` `d = 9; ` `        ``int` `n = 3; ` `         `  `        ``Console.WriteLine(countExcluding(n, d)); ` `    ``} ` `} ` ` `  `// This code is contributed by vt_m. `

## PHP

 ` number ` `        ``\$ans` `= ``\$ans` `- ``\$increment``; ` `        ``\$increment` `= ``\$increment` `/ 10; ` `    ``} ` `    ``return` `\$ans``; ` `} ` ` `  `    ``// Driver code ` `    ``// Function calling ` `    ``echo` `squareRoot(50, 3),``"\n"``; ` ` `  `    ``// Function calling ` `    ``echo` `squareRoot(10, 4),``"\n"``; ` ` `  `// This code is contributed by ajit. ` `?> `

Output :

```7.071
3.1622
```

Time Complexity : The time required to compute the integral part is O(log(number)) and constant i.e, = precision for computing the fractional part. Therefore, overall time complexity is O(log(number) + precision) which is approximately equal to O(log(number)).

My Personal Notes arrow_drop_up In love with a semicolon because sometimes i miss it so badly)

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Improved By : jit_t