Root of number - JavaScript code - 10:00 am mock interview Feb 25, 2018

rootOfNumber_Feb25_2018.js

function root(x, n) {
  // your code goes here
  let lowerBound = 0;
  let higherBound = x;
  let candidate = ((higherBound + lowerBound) / 2);
  
  while (Math.abs(Math.pow(candidate, n) - x) >  0.001) {
    if (Math.pow(candidate, n) > x) {
      higherBound = candidate;
    } else {
      lowerBound = candidate;
    }
    
    candidate = ((higherBound + lowerBound) / 2);
  }
  
  return parseInt(candidate * 1000 + 0.5) / 1000.0;  // (int)(d * 1000)/ 1000.0
}

/*
const x = 9;
const p = 3;

console.log(root(x, p));
*/

/*
input: number => integer, power
output: number => float

error < 0.001

iterate starting from 2 => candidate:
  until 
   multiply candiate by itself times power < target (X)

candiate = x
keep on repeating candiate = candiate / 2  until candidate time to the power is less than X

x = 9, power = 3

2 * 2 * 2 = 8

3 * 3 * 3 = 27

between 2 and 3

current step / 10

2.1 2.2 2.3

currentStep / 10

until currentStep < 0.001

Julia's hint
Lower bound is 2, and upper bound is 3 -> how many numbers between 2 and 3 you have to find? 
  
  every step 0.1, you have 10 number, 2.1, 2.2, ..., 3.0 
  0.001 error range

 1000 from 2 to 3, 2, 2.001, 2.002, ,,,,3.0 

how to expedite the search? 
*/