Strings Common Prefix

[tender panther]

public class Solution {
    public static String longestCommonPrefix(String[] strs) {
        int length_string = strs[0].length();
        String product = "";
        char reference = strs[0].charAt(0);

        for(int i = 0; i<length_string; i++){
            // one letter = 0th letter has been found
            for(int j = 0; j<strs.length; j++){
                System.out.println(reference);
                System.out.println(strs[j].charAt(i));
                if(reference != strs[j].charAt(i)){
                    return product;
                }
                product += reference;
                reference = strs[0].charAt(i+1);
                System.out.println(reference);
            }
        }
        
        return product;
    }
    
    public static void main(String[] args) {
        String[] strs = {"flower","flow","floght"};
        longestCommonPrefix(strs);
    }
}

[lucid skiff]

Detected code, here are some useful tools:

Formatted code

public class Solution {
  public static String longestCommonPrefix(String[] strs) {
    int length_string = strs[0] .length();
    String product = "";
    char reference = strs[0] .charAt(0);
    for (int i = 0; i < length_string; i++) {
      // one letter = 0th letter has been found
      for (int j = 0; j < strs.length; j++) {
        System.out.println(reference);
        System.out.println(strs[j] .charAt(i));
        if (reference != strs[j] .charAt(i)) {
          return product;
        }
        product += reference;
        reference = strs[0] .charAt(i + 1);
        System.out.println(reference);
      }
    }
    return product;
  }
  public static void main(String[] args) {
    String[] strs = {
      "flower", "flow", "floght"};
    longestCommonPrefix(strs);
  }
}

<@&987246399047479336> please have a look, thanks.

[lucid skiff]

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[tender panther]

Above is the following code - where I'm trying to find what all prefixes are matching in each string. Keyword: Prefix

So the easy part is that all of it starts from the beginning of each string. Because the prefix needs to be same in all strings, I took the length of first string as my constant value, because further down the line, if there's a bigger word, it doesn't matter ... I'm realizing that I should've found the smallest word first ummm ... still the logic holds, even though I'll fix this part

Now, then I set the reference point of the code to be the first char of the first word... my output when I'm printing is the following ...

f
f
l
l
f

Idk why the reference is jumping back to an 'f'

[tender panther]

Someone please please or the channel will just close

[quartz urchin]

you're printing reference twice, aren't you?

[tender panther]

Yeah

Wait I just came up with an update

package practice;

public class Solution {
    public static String longestCommonPrefix(String[] strs) {
        int length_string = strs[0].length();
        String product = "";
        char reference = strs[0].charAt(0);

        for(int i = 0; i<length_string; i++){
            // one letter = 0th letter has been found
            for(int j = 0; j<strs.length; j++){
                if(reference != strs[j].charAt(i)){
                    return product;
                }
            }
            product += reference;
            reference = strs[0].charAt(i+1);
        }
        
        return product;
    }
    
    public static void main(String[] args) {
        String[] strs = {"flower","flow","flight"};
        String solution = longestCommonPrefix(strs);
        System.out.println(solution);
    }
}

[lucid skiff]

Detected code, here are some useful tools:

[tender panther]

I've been working on it and the issue is that I was changing the reference and the product inside the inner loop even though it should've been changed in the outer loop

SO that all SEEMS to be working fine, but I get an ArrayOutofBound error which clearly means, that the first word wasn't the smallest word, and because I run the program word1.length() times, it went to some word further down in the array and fucked it up

[quartz urchin]

yeah I was trying to solve the shortest string thing

[tender panther]

So see @quartz urchin I can run another loop prior to this whole saga to just determine the shortest word in this whole thing ... but like I don't wanna. Effiency-wise it's ASS SHIT

I'm so confused. Like I'm an engineering student. I can't just be brute-forcing my way through life...

So how can I tackle this efficiency stuff?! Please please help out!

[tender panther]

Asymptotic analysis wise, it's still O(n^2) total, but like I don't wanna make computer do extra stuff

Hurts my ego as a programmer in terms of efficiency 😂 😂

Sorry for the rant, but I think you get the point

[quartz urchin]

I can't figure out another way to do it

[tender panther]

Yeah same

[quartz urchin]

you can try with the comparator interface

    String smallest = "";
    System.out.println("");
    for(String string :collegeArray){
        //if input-string is null or empty
        if(string == null || string.trim() == "" || string.trim().length()==0){
            System.out.println("Emtpy String Encountered");
            continue;
        }
        if(smallest == ""){
            smallest = string;
            continue;
        }
        if(string.length() < smallest.length()){
            smallest = string;
        }
    }
    return smallest;
}```

[lucid skiff]

Detected code, here are some useful tools:

Formatted code

public static String SmallestString(ArrayList<String> collegeArray) {
  String smallest = "";
  System.out.println("");
  for (String string : collegeArray) {
    //if input-string is null or empty
    if (string == null  || string.trim() == "" || string.trim().length() == 0) {
      System.out.println("Emtpy String Encountered");
      continue ;
    }
    if (smallest == "") {
      smallest = string;
      continue ;
    }
    if (string.length() < smallest.length()) {
      smallest = string;
    }
  }
  return smallest;
}

[quartz urchin]

This code is pretty good, uses a for cycle, which's the only way I can think of haha

just change the ArrayList for a normal String[]