Find square root of number upto given precision using binary search - GeeksforGeeks

24 Jun, 2021
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
 
We have discussed how to compute the integral value of square root in Square Root using Binary SearchApproach : 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 the mid integer with the given number. If it is equal to the number,  the square root is found. Else look for the same in the left or right side depending upon the scenario. 3) Once we are done with finding an integral part, start computing the fractional part. 4) Initialize the increment variable by 0.1 and iteratively compute the fractional part up to P places. For each iteration, the increment changes to 1/10th of its previous value. 5) Finally return the answer computed.Below is the implementation of above approach :  
C++
Java
Python3
C#
PHP
Javascript
// C++ implementation to find// square root of given number// upto given precision using// binary search.#include <bits/stdc++.h>using namespace std; // Function to find square root// of given number upto given// precisionfloat 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 codeint main(){    // Function calling    cout << squareRoot(50, 3) << endl;     // Function calling    cout << squareRoot(10, 4) << endl;     return 0;}
// 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 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, ans = number, 1     # 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 codeprint(round(squareRoot(50, 3), 4))print(round(squareRoot(10, 4), 4)) # This code is contributed by Smitha Dinesh Semwal.
// C# implementation to find// square root of given number// upto given precision using// binary search.using System;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()    {        // Function calling        Console.WriteLine(squareRoot(50, 3));         // Function calling        Console.WriteLine(squareRoot(10, 4));    }} // This code is contributed by Sheharaz Sheikh
<?php// PHP implementation to find// square root of given number// upto given precision using// binary search. // Function to find square root// of given number upto given// precisionfunction squareRoot($number, $precision){    $start=0;    $end=$number;    $mid;     // variable to store    // the answer    $ans;     // for computing integral part    // of square root of number    while ($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    $increment = 0.1;    for ($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    // Function calling    echo squareRoot(50, 3),"\n";     // Function calling    echo squareRoot(10, 4),"\n"; // This code is contributed by ajit.?>
<script> // JavaScript program 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    function squareRoot(number, precision)    {        let start = 0, end = number;        let mid;           // variable to store the answer        let 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        let increment = 0.1;        for (let 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          // Function calling        document.write(squareRoot(50, 3) + "<br/>");           // Function calling        document.write(squareRoot(10, 4) + "<br/>");                                </script>
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)). 
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Binary Search
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Binary Search
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