[GeeksForGeeks] 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. 

Analysis:

Solution 1. O(n), linear scan;

Solution 2. O(log n), binary search: if arr[mid] > mid, then the first missing number must be in the left half; otherwise, it must be in the right half. 

Solution 2 does not work if the given array can have duplicated integers.

public class SmallestMissingNumber {
    public static int SmallestMissingNumberLinear(int[] arr) {
        int idx = 0;
        if(arr[idx] != idx) {
            return idx;
        }
        while(idx < arr.length - 1) {
            if(arr[idx + 1] - arr[idx] > 1) {
                break;
            }
            idx++;
        }
        return arr[idx] + 1;
    }
    public static int SmallestMissingNumberBinary(int[] arr) {
        int start = 0, end = arr.length - 1;
        if(arr[start] != start) {
            return start;
        }
        while(end >= start) {
            int mid = start + (end - start) / 2;        
            if(arr[mid] > mid) {
                end = mid - 1;
            }
            else {
                start = mid + 1;
            }
        }
        return start;
    }
    public static int SmallestMissingNumberWithDuplicates(int[] arr) {
        int idx = 0;
        if(arr[idx] != idx) {
            return idx;
        }
        int nextVal = 1;
        for(idx = 1; idx < arr.length; idx++) {
            if(arr[idx] == arr[idx - 1]) {
                continue;
            }
            else if(arr[idx] == nextVal) {
                nextVal++;
            }
            else {
                break;
            }
        }
        return nextVal;
    }
    public static void main(String[] args) {
        int[] arr1 = {0, 1, 2, 6, 9};
        System.out.println(SmallestMissingNumberBinary(arr1));
        int[] arr2 = {4, 5, 10, 11};
        System.out.println(SmallestMissingNumberBinary(arr2));        
        int[] arr3 = {0, 1, 2, 3, 4};
        System.out.println(SmallestMissingNumberBinary(arr3));    
        int[] arr4 = {1, 2, 3, 4};
        System.out.println(SmallestMissingNumberBinary(arr4));    
        int[] arr5 = {0, 0, 0, 0};
        System.out.println(SmallestMissingNumberWithDuplicates(arr5));
        int[] arr6 = {0, 1, 2, 3, 3};
        System.out.println(SmallestMissingNumberWithDuplicates(arr6));        
    }
}