Java Program to Find the smallest missing number

Given a sorted array of n distinct integers where each integer is in the range from 0 to m-1 and m > n. Find the smallest number that is missing from the array. 

Examples 

Input: {0, 1, 2, 6, 9}, n = 5, m = 10 
Output: 3

Input: {4, 5, 10, 11}, n = 4, m = 12 
Output: 0

Input: {0, 1, 2, 3}, n = 4, m = 5 
Output: 4

Input: {0, 1, 2, 3, 4, 5, 6, 7, 10}, n = 9, m = 11 
Output: 8

Thanks to Ravichandra for suggesting following two methods.

Method 1 (Use Binary Search)
For i = 0 to m-1, do binary search for i in the array. If i is not present in the array then return i.
Time Complexity: O(m log n) 

Method 2 (Linear Search) 
If arr[0] is not 0, return 0. Otherwise traverse the input array starting from index 0, and for each pair of elements a[i] and a[i+1], find the difference between them. if the difference is greater than 1 then a[i]+1 is the missing number. 
Time Complexity: O(n)

Method 3 (Use Modified Binary Search)
Thanks to yasein and Jams for suggesting this method. 
In the standard Binary Search process, the element to be searched is compared with the middle element and on the basis of comparison result, we decide whether to search is over or to go to left half or right half. 
In this method, we modify the standard Binary Search algorithm to compare the middle element with its index and make decision on the basis of this comparison.

• If the first element is not same as its index then return first index
• Else get the middle index say mid
  • If arr[mid] greater than mid then the required element lies in left half.
  • Else the required element lies in right half.

Java

class SmallestMissing
{
    int findFirstMissing(int array[], int start, int end)
    {
        if (start > end)
            return end + 1;

        if (start != array[start])
            return start;

        int mid = (start + end) / 2;

        // Left half has all elements from 0 to mid
        if (array[mid] == mid)
            return findFirstMissing(array, mid+1, end);

        return findFirstMissing(array, start, mid);
    }

    // Driver program to test the above function
    public static void main(String[] args)
    {
        SmallestMissing small = new SmallestMissing();
        int arr[] = {0, 1, 2, 3, 4, 5, 6, 7, 10};
        int n = arr.length;
        System.out.println("First Missing element is : "
                + small.findFirstMissing(arr, 0, n - 1));
    }
}

Output

Smallest missing element is 8

Note: This method doesn’t work if there are duplicate elements in the array.
Time Complexity: O(log n), as we are using binary search to find the first missing element.
Auxiliary Space: O(1), as we are only using a few variables and no extra space is used.

Another Method: The idea is to use Recursive Binary Search to find the smallest missing number. Below is the illustration with the help of steps:

• If the first element of the array is not 0, then the smallest missing number is 0.
• If the last elements of the array is N-1, then the smallest missing number is N.
• Otherwise, find the middle element from the first and last index and check if the middle element is equal to the desired element. i.e. first + middle_index.
  • If the middle element is the desired element, then the smallest missing element is in the right search space of the middle.
  • Otherwise, the smallest missing number is in the left search space of the middle.

Below is the implementation of the above approach:

Java

// Java Program for above approach
import java.io.*;

class GFG
{

    // Program to find Smallest
    // Missing in Sorted Array
    int findSmallestMissinginSortedArray(
                            int[] arr)
    {
    // Check if 0 is missing
    // in the array
    if(arr[0] != 0)
        return 0;

    // Check is all numbers 0 to n - 1
    // are present in array
    if(arr[arr.length-1] == arr.length - 1)
        return arr.length;

    int first = arr[0];

    return findFirstMissing(arr,0,
                    arr.length-1,first);
    }

    // Program to find missing element
    int findFirstMissing(int[] arr , int start ,
                            int end, int first)
    {

    if (start < end)
    {
        int mid = (start+end)/2;

        /** Index matches with value
        at that index, means missing
        element cannot be upto that point */
        if (arr[mid] != mid+first)
        return findFirstMissing(arr, start,
                                    mid , first);
        else
        return findFirstMissing(arr, mid+1,
                                    end , first);
    }
    return start+first;

    }

    // Driver program to test the above function
    public static void main(String[] args)
    {
        GFG small = new GFG();
        int arr[] = {0, 1, 2, 3, 4, 5, 7};
        int n = arr.length;

        // Function Call
        System.out.println("First Missing element is : "
            + small.findSmallestMissinginSortedArray(arr));
    }
}

Output

First Missing element is : 6

Time Complexity: O(log n) 
Auxiliary Space: O(log n) where log n is the size of the recursive stack tree

Please refer complete article on Find the smallest missing number for more details!