Java Coding Interview Practice on Pramp

In this session, the asked questions are:
- Root of number
- Smallest substring of all Character

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[ТюзМайский]

I got hired yesterday after the same interview. Im flying to Kanzas city on March 29th as a Big data middle.

[colorfulcodes]

One of my favorite sites, some questions are insanely hard though. Like number of islands, sudoku checker etc. I need to study more.

[mks]

I did it by finding the permutations of the array and check if the target string contains such permutation, check heap algorith

[helikopeter]

A more efficient way to solve the root question assuming we want to find-by-approximating, and not just using Math.pow(x, 1/(double)n), would be:

static double root(double x, int n) {
double guess = x;
while (Math.abs(Math.pow(guess, n) - x) >
guess -= (Math.pow(guess, n)-x)/(n*Math.pow(guess, n-1));
return guess;
}

this converges faster towards the actual root. :)

[nemesisc]

As far as live coding interviews please don't stay quiet at any time. Let the interviewer know what you are thinking. Let them know your thought process . Read aloud. Yes you are answering a question, but you have the interviewer to help you.

[killakev]

+Pramp - Live Coding Interview Practice

[nihadogresevic]

why is ti called java coding interview its problem solving interview.

[leobruekers]

The best solution would be pow of 1.0/n.

Anyway, your root solution doesn't solve the problem, you should think on special cases. Follow some observations:
- It will solve root of 1 but will take much longer than needed.
- In case of root of a number between 0 and 1 it will never find the result because the result will be higher than the number.

[markvolkov]

static double root(double x, int n) {
return Math.pow(x, 1.0/n);
}
This would of been easier since u were allowed to use the power function

[kenn]

What kind of questions would be asked for a user with their settings set to less than a year of experience as an engineer? Would we need to discuss the Big-O Notation?
I'm using the Pramp platform to prepare for a junior front-end role, just so you know my background. Thanks in advance!

[foottounsi]

Solution for the Smallest Substring of all Characters:

public static boolean evaluate(String s, char[] tab)
{
boolean result = true;
boolean tmp = false;
if (s.length() == tab.length) {
for (int j = 0; j < tab.length; j++) {
tmp = false;
for (int i = 0; i < tab.length; i++) {

tmp = (tmp ^ (tab[j] == s.charAt(i)));
}
result = result && tmp;
if (result ==false) return result;

}
return result;
}
return false;
}


private static List<Integer> getOccurrences(String s, char c)
{
List<Integer> result= new ArrayList();
System.out.println(s +" lenght: " +s.length() );
for (int i = 0; i < s.length(); i++)
{ "+s.charAt(i));
if ((s.charAt(i) == c))
{
result.add(i);
}
}

return result;
}

private static String arr, String str) {

String strResult ="";
int k = arr.length ;
String tmp;
System.out.println(arr[0]);
List<Integer> Occurrences = getOccurrences(str, arr[0]);
for (int i=0; i<Occurrences.size(); i++) {
" + Occurrences.get(i));
int l = -1;
for (int r = k - 1; r >=0; r--) {
l++;
if ((Occurrences.get(i) - r >= 0) && (Occurrences.get(i) + l < str.length())) {
tmp = - r, Occurrences.get(i) + l+1);
" + tmp);
if (evaluate(tmp, arr))
return tmp;
}
" + Occurrences.get(i));
}
}
return strResult;
}

[RabbitConfirmed]

I just realised, Im not made to be a programmer..
I will probably do the System Engineering Degree first and go into Web Development later.
Since I really think, programming is really hard and rn I don't think I can handle all the pressure.

[ManishKumar-upyj]

There is very less questions on pramp.com, I am running out of questions

[leonardvictorio]

Interesting approach of testing, but this is not a good way of testing a developer/programmer. This test is very uncomfortable and annoying, at least for me.

[vaghii]

I am confused ..who shares the screen interviewer or inteviewee

[erikalves]

It is a horrible way to test developers.

[shellysmith]

public static String arr, String str) {
String result = "";
//str must be longer than unique character set
if (arr.length > str.length()) {
return result;
}
//For each possible substring
//Store arr elements in this hash
//Remove them as you find them in str
HashSet<Character> punchList;

int subStrSize = arr.length;
int strSize = str.length();
//for each size of substring arr.len to str.len
//We start with arr.len because we want to find the shortest substring
for (int currSize = subStrSize; currSize < strSize; currSize++) {
//test for currSize length subStr
//iterate all possible currSize strings
//
for (int strStart = 0; strStart <= strSize - currSize; strStart++) {
punchList = buildPunchList(arr);
int end = strStart + currSize;
StringBuilder subStr = new StringBuilder();
for (int k = strStart; k < end; k++) {
char c = str.charAt(k);
subStr.append(c);
//remove chars I've seen
punchList.remove(c);
}
//punchList is empty, so I saw all available chars
//therefore we have the first shortest solution
if (punchList.size() == 0) {
return subStr.toString();
}
}
}
return result;
}

[Calisby]

do these people know that they are being uploaded to pramp's youtube channel? I'm worried, I have done many interviews naked on pramp and I don't want to appear on their youtube videos

[dorokei]

Why is it that the first person can see the solution while answering, but the second person can't?

[killakev]

Hey, nice job guys. I solved the second problem in Swift. I only used one while loop so I believe it's a linear O(n) solution. Let me know what you guys think.

func [Character], str: String)->String{
    guard !str.isEmpty || !uniqueCharacters.isEmpty else {return ""}
    var index = 0
    let sorted = uniqueCharacters.sorted()
    let stringArray = Array(str)
    let lastRange = stringArray.count - sorted.count
    
    while index < lastRange{
        if sorted == + index].sorted() {
            return + index])
        }
      index += 1
    }
  
    if == sorted{
        return
    }
    else{
        return ""
    }
}
["x", "y", "z"], str: "xyyzyyxzx")      //Should return "yxz"