Fixing 2 methods that are supposed to remove duplicate and reverse order pairs

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Just learning and hoping to change careers...I need help correcting the 2nd and 3rd methods so they print correctly! I was able to hack this together(which took a very long time), what code do I need to fix the 2nd and 3rd methods?

I'd like to get it working and then go back and teach myself as that's my learning style!

Both the 2nd and 3rd methods are incorrectly printing these pairs: (5,5)(5,5)(6,4)(9,1)

import java.util.*;

public class TenPairs {

    public static void main(String[] args) {
        int a[] = { 1, 1, 2, 4, 4, 5, 5, 5, 6, 7, 9 };
        findAllPairs(a, 10);
        findUniquePairs(a, 10);
        findComboPairs(a, 10);
    }

    // Method 1 - output all pairs would output: [1,9], [1,9], [4,6], [4,6], [5,5],
    // [5,5], [5,5], [5,5], [5,5],[5,5],[6,4],[6,4][9,1],[9,1]

    static void findAllPairs(int[] array, int sum) {

        System.out.println("All pairs(including duplicates and reverse order pairs) whose sum is " + sum + ":");

        for (int i = 0; i < array.length; i++) {
            for (int j = i + 1; j < array.length; j++) {
                if (array[i] + array[j] == sum) {
                    System.out.println("(" + array[i] + "," + array[j] + ")" + "(" + array[j] + "," + array[i] + ")");
                }
            }
        }
    }

    // Method2 - output unique pairs only once would output: [1,9], [4,6], [5,5],
    // [6,4], [9,1]
    static void findUniquePairs(int[] array, int sum) {

        System.out.println("All pairs only once(including reverse order pairs but excluding duplicates) whose sum is "
                + sum + ":");

        Map<Integer, Integer> pairs = new HashMap<Integer, Integer>();
        for (int i = 0; i < array.length; i++) {
            if (pairs.containsKey(sum - array[i]))
                System.out.println("(" + array[i] + "," + (sum - array[i]) + ")");
            else
                pairs.put(array[i], 0);
        }
    }

    // Method3 - output the same combo pair only once would output: [1,9], [4,6],
    // [5,5]
    static void findComboPairs(int[] array, int sum) {

        System.out
                .println("All pairs only once(excluding reverse order pairs and duplicates) whose sum is " + sum + ":");

        Map<Integer, Integer> pairs = new HashMap<Integer, Integer>();
        for (int i = 0; i < array.length; i++) {
            if (pairs.containsKey(sum - array[i]))
                System.out.println("(" + array[i] + "," + (sum - array[i]) + ")");
            else
                pairs.put(array[i], 0);
        }
    }
}
duplicates arrays java
1 Answers

There can be many ways to do it. One of the ways is to create a List out of the array and then set the used elements to null as you process the list.

Do it as follows:

import java.util.Arrays;
import java.util.List;
import java.util.stream.Collectors;

public class TenPairs {

    public static void main(String[] args) {
        int a[] = { 1, 1, 2, 4, 4, 5, 5, 5, 6, 7, 9 };
        findAllPairs(a, 10);
        findUniquePairs(a, 10);
        findComboPairs(a, 10);
    }

    static void findAllPairs(int[] array, int sum) {

        System.out.println("All pairs(including duplicates and reverse order pairs) whose sum is " + sum + ":");

        for (int i = 0; i < array.length; i++) {
            for (int j = i + 1; j < array.length; j++) {
                if (array[i] + array[j] == sum) {
                    System.out.println("(" + array[i] + "," + array[j] + ")" + "(" + array[j] + "," + array[i] + ")");
                }
            }
        }
    }

    static void findUniquePairs(int[] array, int sum) {

        System.out.println("All pairs only once(including reverse order pairs but excluding duplicates) whose sum is "
                + sum + ":");
        List<Integer> list = Arrays.stream(array).boxed().collect(Collectors.toList());
        int x = 0, y = 0;
        for (int i = 0; i < array.length; i++) {
            for (int j = i + 1; j < array.length; j++) {
                if (list.get(i) != null) {
                    x = list.get(i);
                }
                if (list.get(j) != null) {
                    y = list.get(j);
                }
                if (x + y == sum) {
                    if (x != y) {
                        System.out.println("(" + x + "," + y + ")");
                    }
                    System.out.println("(" + y + "," + x + ")");
                    list.set(i, null);
                    list.set(j, null);
                    x = 0;
                    y = 0;
                }
            }
        }
    }

    static void findComboPairs(int[] array, int sum) {

        System.out.println(
                "All pairs only once(excluding reverse order pairs and duplicates) whose sum is " + sum + ": ");
        List<Integer> list = Arrays.stream(array).boxed().collect(Collectors.toList());
        int x = 0, y = 0;
        for (int i = 0; i < array.length; i++) {
            for (int j = i + 1; j < array.length; j++) {
                if (list.get(i) != null) {
                    x = list.get(i);
                }
                if (list.get(j) != null) {
                    y = list.get(j);
                }
                if (x + y == sum) {
                    System.out.println("(" + x + "," + y + ")");
                    list.set(i, null);
                    list.set(j, null);
                    x = 0;
                    y = 0;
                }
            }
        }
    }
}

Output:

All pairs(including duplicates and reverse order pairs) whose sum is 10:
(1,9)(9,1)
(1,9)(9,1)
(4,6)(6,4)
(4,6)(6,4)
(5,5)(5,5)
(5,5)(5,5)
(5,5)(5,5)
All pairs only once(including reverse order pairs but excluding duplicates) whose sum is 10:
(1,9)
(9,1)
(4,6)
(6,4)
(5,5)
All pairs only once(excluding reverse order pairs and duplicates) whose sum is 10: 
(1,9)
(4,6)
(5,5)

The main difference between the 2nd and the 3rd method is the number of times you print the pairs. In the 2nd method, you have to print the pairs twice (once in the order of occurrence and another in the reverse order) except when the pair has same elements. Feel free to comment in case of any doubt/issue.

user_108195730
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