Find out n numbers of missing elements from an array in java

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I have an array which carry some integer numbers. Say,numbers={3,0,1} or say, numbers={9,6,4,2,3,5,7,0,1}. Now I have to find out the missing numbers from the array. As per this example there is only one missing number in each set. 1st one miss 2 and the 2nd one miss 8.

I have already code it. My code not only find out one missing number from specified set it can also find out more than 1 missing numbers from a given set.

But if two consecutive numbers are missing from same set it fail to find out.

My code import java.util.Arrays;  public class Missing_number  {     public static void main( String args[] )     {         int numbers[]={9,6,4,5,7,0,1};         Arrays.sort(numbers);         int i=1;          while ( i < numbers.length )          {             if ( numbers[i] - numbers[i-1] == 1 )              {             }              else              {                 System.out.println( "Missing number is " + ( numbers[i-1] + 1 ) );             }             i++;         }     } } 

I am thinking like that if I am able to append the 1st missing number in the array and then start searching then how's the code look like? numbers={9,6,4,5,7,0,1} Now, 8 is already missing from this set. Now I have terminated two more element (2,3) from the list. Output: as per my code: 2,8 But 3 is also missing but that does not display.

I am thinking like if I am able to append 2 in number array then it may be little bit easier. But as we all know Java array is immutable so we cannot increase it's length.

So, maybe I will use List. But in list this type of indexing number[0]=somethingnot supported. So how could I proceed then. Am I using list or still stuck into array?

So I take a attempt to create it with an arraylist.

Mycode(modified version from array)   public class T1 {  public static void main(String args[]){     List<Integer> numbers=new ArrayList<>();     numbers.add(9);     numbers.add(6);     numbers.add(4);     numbers.add(5);     numbers.add(7);     numbers.add(0);     numbers.add(1);     Collections.sort(numbers);     int i=1;     while(i< numbers.size()) {         if (numbers.get(i) - numbers.get(i-1) == 1) {          } else {             System.out.println("Missing number is " + (numbers.get(i-1) + 1));             numbers.add((numbers.get(i-1)+1));             Collections.sort(numbers);         }         i++;     }      } } 

Arraylist can solve my problem. But is there any possibility that a simple array can solve this problem?

 


This code uses a HashSet:

public static void main(String[] args) {     int[] numbers = {9, 6, 4, 5, 7, 0, 1};     Arrays.sort(numbers);     HashSet<Integer> set = new HashSet<>();      for (int i = numbers[0]; i < numbers[numbers.length - 1]; i++) {         set.add(i);     }      for (int i = 0; i < numbers.length; i++) {         set.remove(numbers[i]);     }      for (int x : set) {         System.out.print(x + " ");     } } 

will print:

2 3 8  

Here is how it works:
1. Adds all numbers from the minimum number of the array to the maximum number of the array to the set.
2. Iterates through the array and removes every item of the array from the set.
3. Prints the remaining items in the set, which are all the missing items of the array.

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