Kotlin Language Features Related to Null Handling

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Any software engineer with a Java background would find the null handling features in the Kotlin language interesting. Let's summarize this topic with some examples. Nullable types: In Kotlin, types are non-nullable by default. If you want a variable to be able to hold a null value, you need to explicitly declare its type as nullable using the Type? syntax. For example, String? denotes a nullable string, while String represents a non-nullable string. Safe calls (?.): Kotlin introduces the safe call operator (?.) for handling nullable types. It allows you to safely invoke a method or access a property on a nullable object. If the object is null, the expression returns null instead of throwing a NullPointerException. Example: data class Person(val name: String, val age: Int, val address: String?) fun main() {     // Create a person with a nullable address     val person1 = Person("John Doe", 25, "123 Main Street")     val person2 = Person("Jane Doe", 30,...
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swapLexOrder: Finding lexicographically largest string

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The following coding question is really amazing:

Given a string str and array of pairs that indicates which indices in the string can be swapped, return the lexicographically largest string that results from doing the allowed swaps. You can swap indices any number of times.

Example

For str = "abdc" and pairs = [[1, 4], [3, 4]], the output should be
swapLexOrder(str, pairs) = "dbca".

By swapping the given indices, you get the strings: "cbda", "cbad", "dbac", "dbca". The lexicographically largest string in this list is "dbca".

Input/Output

  • [execution time limit] 3 seconds (java)

  • [input] string str

    A string consisting only of lowercase English letters.

    Guaranteed constraints:
    1 ≤ str.length ≤ 104.

  • [input] array.array.integer pairs

    An array containing pairs of indices that can be swapped in str (1-based). This means that for each pairs[i], you can swap elements in str that have the indices pairs[i][0] and pairs[i][1].

    Guaranteed constraints:
    0 ≤ pairs.length ≤ 5000,
    pairs[i].length = 2.


Solution: 

Consider indices as vertices.
Connect two vertices with an edge if swapping is allowed between them

Now for each connected component in graph sort the characters represented by the indices(vertex) and place them from highest value to lowest value in those indices.

Result would be a lexicographically largest string.


Example :

"acxrabdz"
Swaps allowed(0 based index) :-
(0,2)(5,7)(2,7),(1,6)

There are two connected components
(0,2,5,7) and (1,6)
Sort (0,2,5,7) values that is a,x,b,z you get zxba place them in respective indices (z->0,x->2,b->5,a->7)
Similarly sort (1,6) -> c,d - > d,c

final output :- "zcxrabca" lexicographically sorted string


Java Implementation: 

import java.util.Arrays;

import java.util.HashMap;

import java.util.Map;


public class Main {


    private int[] connectedComponents;


    String swapLexOrder(String str, int[][] pairs) {


        //Use a union-find algorithm to find connected indexes (components)

        findConnectedComponents(str, pairs);


        //convert implicit connected indexes to map

        Map<Integer, Integer[]> componentsMap = getConnectedComponents();


        for (Integer key : componentsMap.keySet()) {


            Integer[] components = componentsMap.get(key);


            str = sort(str, components);

        }


        return str;

    }


    private void findConnectedComponents(String str, int[][] pairs) {

        connectedComponents = new int[str.length() + 1];

        for (int i = 0; i < connectedComponents.length; i++) {

            connectedComponents[i] = i;

        }


        for (int[] pair : pairs) {

            connect(pair[0], pair[1]);

        }

    }


    public Map<Integer, Integer[]> getConnectedComponents() {


        Map<Integer, Integer[]> map = new HashMap<>();


        for (int i = 0; i < connectedComponents.length; i++) {


            if (map.get(connectedComponents[i]) == null) {

                map.put(connectedComponents[i], new Integer[]{i});

            } else {

                Integer[] value = map.get(connectedComponents[i]);

                Integer[] newValue = Arrays.copyOf(value, value.length + 1);


                newValue[newValue.length - 1] = i;


                map.put(connectedComponents[i], newValue);

            }

        }


        map.remove(0);


        return map;


    }


    private void connect(int a, int b) {


        for (int i = 0; i < connectedComponents.length; i++) {


            if (connectedComponents[i] == connectedComponents[a] && i != a) {

                connectedComponents[i] = connectedComponents[b];

            }

        }


        connectedComponents[a] = connectedComponents[b];

    }


    public String sort(String s, Integer[] connectedIndexes) {


        Integer[] calibratedIndexes = Arrays.copyOf(connectedIndexes, connectedIndexes.length);


        for (int i = 0; i < calibratedIndexes.length; i++) {

            calibratedIndexes[i] -= 1;

        }


        StringBuilder sb = new StringBuilder(s);


        class Item {


            public Item(int index, String value) {

                this.index = index;

                this.value = value;

            }


            int index;

            String value;

        }


        Item[] values = new Item[calibratedIndexes.length];


        for (int i = 0; i < calibratedIndexes.length; i++) {

            values[i] = new Item(calibratedIndexes[i], s.substring(calibratedIndexes[i], calibratedIndexes[i] + 1));

        }


        Arrays.sort(values, (o1, o2) -> o2.value.compareTo(o1.value));


        for (int i = 0; i < values.length; i++) {

            sb.replace(calibratedIndexes[i], calibratedIndexes[i] + 1, values[i].value);

        }


        return sb.toString();

    }


    public static void main(String[] args) {


        String str = "lvvyfrbhgiyexoirhunnuejzhesylojwbyatfkrv";


        String result = new Main().swapLexOrder(str, new int[][]

                {{13, 23},

                        {13, 28},

                        {15, 20},

                        {24, 29},

                        {6, 7},

                        {3, 4},

                        {21, 30},

                        {2, 13},

                        {12, 15},

                        {19, 23},

                        {10, 19},

                        {13, 14},

                        {6, 16},

                        {17, 25},

                        {6, 21},

                        {17, 26},

                        {5, 6},

                        {12, 24}}

        );


        System.out.println(result);

    }

}





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