Java > AP-1 > wordsWithout (CodingBat Solution)

Problem:

Given an array of strings, return a new array without the strings that are equal to the target string. One approach is to count the occurrences of the target string, make a new array of the correct length, and then copy over the correct strings.

wordsWithout({"a", "b", "c", "a"}, "a") → {"b", "c"}
wordsWithout({"a", "b", "c", "a"}, "b") → {"a", "c", "a"}
wordsWithout({"a", "b", "c", "a"}, "c") → {"a", "b", "a"}


Solution:

public String[] wordsWithout(String[] words, String target) {
  int found = 0;
  
  for (int i = 0; i < words.length; i++) {
    if (words[i].equals(target))
      found++;
  }
  found = words.length - found;
  int place = 0;
  
  String[] str = new String[found];
  for (int j = 0; j < words.length; j++) {
    if (!words[j].equals(target)) {
      str[place] = words[j];
      place++;
    }
  }
  return str;
}

12 comments:

  1. UnknownOctober 27, 2015 at 12:58 PM

    public String[] wordsWithout(String[] words, String target) {
    ArrayList list = new ArrayList();

    for (int i = 0; i < words.length; i++)
    if (words[i] != target) list.add(words[i]);

    return (String[])list.toArray(new String[list.size()]);
    }

    ReplyDelete
  2. UnknownSeptember 23, 2017 at 1:34 AM

    public String[] wordsWithout(String[] words, String target) {
    int count=0;

    for (int i = 0; i < words.length; i++) {
    if(words[i]!=target){
    count++;
    }
    }
    int chk=0;
    String[] arr=new String[count];
    for (int i = 0; i < words.length; i++) {
    if(words[i]!=target){
    arr[chk]=words[i];
    chk++;
    }
    }
    return arr;
    }

    ReplyDelete
  3. UnknownFebruary 11, 2019 at 10:36 AM

    public String[] wordsWithout(String[] words, String target) {
    return Arrays.stream(words)
    .filter(n -> !n.equals(target))
    .toArray(String[]::new);
    }

    ReplyDelete
  4. AnonymousMay 28, 2019 at 5:28 AM

    Easiest and simplest way via Java 8 streams:

    return java.util.Arrays.stream(words).filter(x -> !x.equals(target)).toArray(String[]::new);

    ReplyDelete
  5. SamarthJuly 26, 2019 at 5:14 AM

    public String[] wordsWithout(String[] words, String target) {
    int len = words.length;
    int counter = 0;
    for(int i=0; i<len; i++) {
    if(words[i].equals(target)) {
    continue;
    }
    else {
    counter++;
    }
    }
    String[] ar = new String[counter];
    int b = 0;
    for(int i=0; i<len; i++) {
    if(words[i].equals(target)) {
    continue;
    }
    else {
    ar[b] = words[i];
    b++;
    }
    }
    return ar;
    }

    ReplyDelete
    Replies
    1. SamarthJuly 26, 2019 at 5:16 AM

      I need to find a faster and more efficient way to do this. Because I did the for loop twice which is not good.

      Delete
  6. UnknownJanuary 12, 2020 at 5:38 AM

    public String[] wordsWithout(String[] words, String target) {
    ArrayList list = new ArrayList ();
    for(String str : words ){
    if( !(str.equals(target)) ) list.add(str);
    }
    String [] arr = list.toArray(new String[list.size()]);
    return arr;
    }

    ReplyDelete
  7. AnonymousSeptember 20, 2020 at 4:19 PM

    public class WordsWithoutList {
    public List wordsWithoutList(String[] words, int len) {
    final List list = new ArrayList<>(words.length);
    for (final String s: words){
    if(s.length() != len) {
    list.add(s);
    }
    }
    return list;
    }
    }

    ReplyDelete
  8. AnonymousSeptember 20, 2020 at 4:19 PM


    public List wordsWithoutList(String[] words, int len) {
    final List list = new ArrayList<>(words.length);
    for (final String s: words){
    if(s.length() != len) {
    list.add(s);
    }
    }
    return list;
    }

    ReplyDelete
  9. Benjamin NguyenJanuary 2, 2021 at 7:19 PM

    Here is mine:
    public String[] wordsWithout(String[] words, String target) {

    ArrayList list = new ArrayList<>();

    for (int i = 0; i < words.length; i++){
    list.add(words[i]);
    if (words[i] == target){
    list.remove(words[i]);
    }
    }
    String[] arr = new String[list.size()];
    return list.toArray(arr);
    }

    ReplyDelete