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
We have discussed how to compute the 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 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++
// 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
// 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 precisiondef 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#
// 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// 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.?> |
Javascript
<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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