Find square root of number upto given precision using binary search
Last Updated :
14 Sep, 2024
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
// 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 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 code
print(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
<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>
<?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
// precision
function 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.
?>
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)).
Auxiliary Space: O(1) since it is using constant space for variables
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