How Can We Optimize a For Loop to Find Characters Around a Word in a String?

[Arnoldjavs3]

Homework Statement

Given a string and a non-empty word string, return a string made of each char just before and just after every appearance of the word in the string. Ignore cases where there is no char before or after the word, and a char may be included twice if it is between two words.

Homework Equations

The Attempt at a Solution

public String wordEnds(String str, String word) {
  String newString = "";
 
  for(int i=0; i<str.length()-word.length(); i++) {
    int finalCheck = str.length()-word.length();
    if(str.substring(i, i+word.length()).equals(word)) {
      if(i!=0) {
        newString += "" + str.charAt(i-1);
      }
      newString += "" + str.charAt(i+word.length());
     
    }
  }
  return newString;
}

This code works in all cases unless the word is at the last index(e.g: wordEnds("abc1xyz1abc", "abc") → "11" got "1"
wordEnds("abc1abc1abc", "abc") → "1111" got "111") It isn't adding the last char before the last index.)
So I made this code which works but it's not incorporated for inside the loop which I'd like to do

public String wordEnds(String str, String word) {
  String newString = "";
 
  for(int i=0; i<str.length()-word.length(); i++) {
    int finalCheck = str.length()-word.length();
    if(str.substring(i, i+word.length()).equals(word)) {
      if(i!=0) {
        newString += "" + str.charAt(i-1);
      }
      newString += "" + str.charAt(i+word.length());
     
    }
  }
  int finalOccurence = str.length()-word.length();
  if(str.length()>word.length()&&str.substring(finalOccurence).equals(word)) {
    newString+="" + str.charAt(finalOccurence-1);
  }
 
  return newString;
}

 

[Mark44]

Homework Statement

Given a string and a non-empty word string, return a string made of each char just before and just after every appearance of the word in the string. Ignore cases where there is no char before or after the word, and a char may be included twice if it is between two words.

Homework Equations

The Attempt at a Solution

public String wordEnds(String str, String word) {
  String newString = "";
 
  for(int i=0; i<str.length()-word.length(); i++) {
    int finalCheck = str.length()-word.length();
    if(str.substring(i, i+word.length()).equals(word)) {
      if(i!=0) {
        newString += "" + str.charAt(i-1);
      }
      newString += "" + str.charAt(i+word.length());
    
    }
  }
  return newString;
}

This code works in all cases unless the word is at the last index(e.g: wordEnds("abc1xyz1abc", "abc") → "11" got "1"
wordEnds("abc1abc1abc", "abc") → "1111" got "111") It isn't adding the last char before the last index.)
So I made this code which works but it's not incorporated for inside the loop which I'd like to do

public String wordEnds(String str, String word) {
  String newString = "";
 
  for(int i=0; i<str.length()-word.length(); i++) {
    int finalCheck = str.length()-word.length();
    if(str.substring(i, i+word.length()).equals(word)) {
      if(i!=0) {
        newString += "" + str.charAt(i-1);
      }
      newString += "" + str.charAt(i+word.length());
    
    }
  }
  int finalOccurence = str.length()-word.length();
  if(str.length()>word.length()&&str.substring(finalOccurence).equals(word)) {
    newString+="" + str.charAt(finalOccurence-1);
  }
 
  return newString;
}

Without looking that closely at your code, it seems to me that you could put the logic of checking for the last word as an if statement inside the for loop. I would structure it so that the if clause executes if the word isn't the last word, and the else clause handles the case when the word is the last word. In other words, the part right after if is set up to execute most often.

Also, as far as I can tell, since str and word are passed as parameters to your wordEnds() function, your code is using length() unnecessarily. I would calculate str.length() and word.length() once at the top of the function, and store these values in local variables. It's very inefficient to call a function multiple times in each loop iteration if the lengths aren't actually changing.