Find the smallest missing number

2.6

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.

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

C

// C program to find the smallest elements missing
// in a sorted array.
#include<stdio.h>

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 above function
int main()
{
    int arr[] = {0, 1, 2, 3, 4, 5, 6, 7, 10};
    int n = sizeof(arr)/sizeof(arr[0]);
    printf("Smallest missing element is %d",
           findFirstMissing(arr, 0, n-1));
    return 0;
}

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));
    }
}

Note: This method doesn’t work if there are duplicate elements in the array.

Time Complexity: O(Logn)

Please write comments if you find the above codes/algorithms incorrect, or find other ways to solve the same problem.

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