Below question was asked to me in interview for PTC Software & I made certain mistakes by not asking the follow-up
​questions & not optimizing the solution further.

Question : Find the square root of the given number without using the Math api in Java. If no complete square root exists then
​                   return 0.

Solution : Below is not the solution I gave in interview, in interview I wrote very naive solution without using Math.sqrt()
     Please let me know how can I correct it further as it is giving many wrong results & how we can optimize it further.
/**
 * Below is one sample to find the square root in Java.
 * It is not efficient one, neither a complete one. It just finds the square roots in whole numbers.
 * It returns only integer values if possible else will return 0, if no such square root exists for the
 * given number. So in that case, it works faster than Math.sqrt(), as shown.
 * @author nitin
 *
 */
public class SquareRoot {
 public static void main(String[] args) {
  int num = 99980001;
  long start = System.nanoTime();
  System.out.println(find(num));
  System.out.println("Time taken : " + (System.nanoTime() - start) + " ns");
  start = System.nanoTime();
  System.out.println(Math.sqrt(num));
  System.out.println("Time taken : " + (System.nanoTime() - start) + " ns");
 }
 public static int find(int num) {
  int digits = 0;
  String temp = Integer.toString(num);
  digits = temp.length()/2;
  int start = 1;
  while(digits > 0) {
   start *= 10;
   digits--;
  }
  int i = start;
  if(i*i <= num) {
   int res = i*i;
   while(res < num/4) {
    i = i * 2;
    res = i * i;
   }
   while(res < num) {
    i++;
    res = i * i;
   }
   if(res == num) {
    return i;
   } else {
    return 0;
   }
  } else {
   
   i = --start;
   while(i >= 1) {
    if(i*i == num) {
     return i;
    }
    i--;
   }
   return 0;
  }
 }
}

​Below is some improved version of above method & one can tweak the breakpoint as per the understanding, but for now it gave mostly correct results.
​Tested below code for numbers 1 to 10000000000 to get the perfect square root. And seems it is giving all correct answers.

​​/**
  * This method finds square root of the given number.
  * It will give the result if perfect square root exists, else
  * will return 0.
  * Till now no errors found & has been around 100 times faster
  * than Math.sqrt() for such requirements.
  * Though can skip break condition but have put here to give the option
  * to get the values in decimals also.
  * @param num
  * @return
  */
 public static long findSqrRoot(long num) {
  
  if(num < 0)
   return 0;
  if(num == 1)
   return 1;
  String str = Long.toString(num);
  int len = str.length();
  int proc = 0;
  if(len > 7) {
   proc = len/2-1;
  }
  long startMin = 1;
  long startMax = num/2l;
  
  long min = startMin;
  long max = startMax;
  long ans = 0;
  while(proc > 0) {
   startMax = startMax/10;
   proc--;
  }
  
  long start = (startMax-startMin)/2l;
  long i = start;
  while(ans == 0 && max > min) {
   if(max-min < 10) {
    break;
   } else if(i*i == num) {
    ans = I;
    break;
   } else if(i*i > 0 && i*i < num) {
    min = i+1;
    i = min + (max-min)/2l;
   } else if(i*i > num || i*i < 0) {
    max = i-1;
    i = max - (max-min)/2l;
   }
  }
  for(long k = min; k <= max; k++) {
   if(k*k == num) {
    ans = k;
    break;
   }
  }
  return ans;
 }

​A faster way and it worked faster than Math.sqrt(), though it is not responding for Integer.MAX_VALUE:-