Sort String based on a pattern array of String

Question

I need to sort a string of words in an order those words are placed in the pattern array:

String[] pattern = {"Burger", "Fries", "Chicken", "Pizza", "Sandwich", "Onionrings", "Milkshake", "Coke"};

String s = "Chicken Coke Chicken Chicken Onionrings Coke Burger Milkshake Milkshake Burger Coke Milkshake Burger Onionrings Onionrings Coke Chicken Burger Fries Milkshake"; // input string

Expect

"Burger Burger Burger Burger Fries Chicken Chicken Chicken Chicken Onionrings Onionrings Onionrings Milkshake Milkshake Milkshake Milkshake Coke Coke Coke Coke" // output string

Do you have any code snippets of your attempt at this problem? — shinjw
– shinjw, Commented Aug 30, 2020 at 7:40
Also, what should happen if s does contain a word that is not in the pattern? — Turing85
– Turing85, Commented Aug 30, 2020 at 7:46
@Turing85 I've handled it in another method. I only need to sort the string in the order given in the pattern — Arthur
– Arthur, Commented Aug 30, 2020 at 7:48

azro · Accepted Answer · 2020-08-30 07:53:33Z

5

You may use a List to be able to get index from word, then sort the words using these indexes

Get a List os the words to get them indexable
split your String into an array to be able to sort the words
sort them based on their indices in the List
join the words to get one String

// INPUTS
String s = "Chicken Coke Chicken Chicken Onionrings Coke Burger Milkshake Milkshake Burger Coke Milkshake Burger Onionrings Onionrings Coke Chicken Burger Fries Milkshake";
String expect = "Burger Burger Burger Burger Fries Chicken Chicken Chicken Chicken Onionrings Onionrings Onionrings Milkshake Milkshake Milkshake Milkshake Coke Coke Coke Coke";
String[] pattern = {"Burger", "Fries", "Chicken", "Pizza", "Sandwich", "Onionrings",
    "Milkshake", "Coke"};

// Get the patterns a list to get the indexOf method
List<String> listPatterns = Arrays.asList(pattern);

// Split in words
String[] values = s.split(" ");

//Sort base on index in list
Arrays.sort(values, Comparator.comparing(listPatterns::indexOf));

// Rebuild a string by joining
String result = String.join(" ", values);

// true
System.out.println(result.equals(expect));

answered Aug 30, 2020 at 7:53

azro

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2 Comments

Natesh Over a year ago

I have given boiler plate code snippet, both solution are correct it depends on @Arther what he needs.

azro Over a year ago

@Natesh you can given an answer to something that haven't been asked, you may delete your post, that is not relevant to the question

Turing85 · Accepted Answer · 2020-08-30 07:55:05Z

1

Since the question does not contain any code, I will only post a rough sketch of a solution.

I would approach the problem by splitting the String in a list of words, then use a Comparator to sort those words and re-join them with separating blanks.

A common way to write Comparators is to assign the both objects compared against each other an int-value, subtract one from the other and return that as the comparison result.

In the Comparator, I would use the index of the two words compared against each other in the pattern-array as base for comparison.

answered Aug 30, 2020 at 7:55

Turing85

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Comments

user4910279 · Accepted Answer · 2020-08-30 08:17:37Z

Try this.

String[] pattern = {"Burger", "Fries", "Chicken", "Pizza", "Sandwich", "Onionrings", "Milkshake", "Coke"};

// make pattern to index map. {Burger=0, Fries=1, Chicken=2, ...}
Map<String, Integer> patternMap = IntStream.range(0, pattern.length).boxed()
    .collect(Collectors.toMap(i -> pattern[i], Function.identity()));

// sort words using patternMap
String s = "Chicken Coke Chicken Chicken Onionrings Coke Burger Milkshake Milkshake Burger Coke Milkshake Burger Onionrings Onionrings Coke Chicken Burger Fries Milkshake"; // input string
String[] result = Arrays.stream(s.split("\\s+"))
    .sorted(Comparator.comparing(word -> patternMap.get(word)))
    .toArray(String[]::new);

System.out.println(Arrays.toString(result));

output:

[Burger, Burger, Burger, Burger, Fries, Chicken, Chicken, Chicken, Chicken, Onionrings, Onionrings, Onionrings, Milkshake, Milkshake, Milkshake, Milkshake, Coke, Coke, Coke, Coke]

Natesh · Accepted Answer · 2020-08-30 07:48:44Z

-1

here is code to sort alphabetically

public class MainProgram {

    public static void main(String[] args) {
        String temp;
        String[] pattern = {"Burger", "Fries", "Chicken", "Pizza", "Sandwich", "Onionrings", "Milkshake", "Coke"};
          //Sorting the strings
        for (int i = 0; i < pattern.length; i++) 
        {
            for (int j = i + 1; j < pattern.length; j++) { 
                if (pattern[i].compareTo(pattern[j])>0) 
                {
                    temp = pattern[i];
                    pattern[i] = pattern[j];
                    pattern[j] = temp;
                }
            }
        }
      //Displaying the strings after sorting them based on alphabetical order
        System.out.print("Strings in Sorted Order:");
        for (int i = 0; i <= pattern.length - 1; i++) 
        {
            System.out.print(pattern[i] + ", ");
        }
        
    }

}

answered Aug 30, 2020 at 7:48

Natesh

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4 Comments

azro Over a year ago

That is not what is asked, at all

Natesh Over a year ago

This code return your expected output.. anything else you want

azro Over a year ago

This is not my post. And the expected output is

"Burger Burger Burger Burger Fries Chicken Chicken Chicken Chicken Onionrings Onionrings Onionrings Milkshake Milkshake Milkshake Milkshake Coke Coke Coke Coke"

Turing85 Over a year ago

No it does not. The code sorts pattern. That is not the task at hand

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