sample interview question and solution(s) because there's definitely more than one way to solve a problem

Interview.java

import java.io.*;
import java.util.*;

public class Interview {
	
	private static int findMissing(int[] array) {
		int length = array.length;
	
		// This three-line implementation is better but possibly less easy to understand.
		// We use some math here: sum the numbers from 1 to (length + 1)
		// This is what the value would be if there were no missing numbers
		// Next, sum the values that *are* present in the array.
		// Subtracting the values present from the maximum returns the number that is missing.
    // This requires only 1 iteration through the array (to calculate the sum of numbers present)
		int OneToN = length * (length+1)/2;
		int sum = Arrays.stream(array).sum();		
		System.out.println("missing " + (OneToN - sum));
		
		// Below is the initial implementation, which requires (n^2)+n comparisons
		// make an array of 0..n elements to hold which ones are present
		Boolean[] present = new Boolean[array.length+1];
		// set them all to false
		Arrays.fill(present, false);
		
		// for each position in the array we were passed in,
		for (int position = 0; position < length; position ++) {
			// determine if array[position] is any of the possible values of 0..n
			for (int check = 0; check <= length; check++) {
				// if it is,
				if (array[position] == check) {
					// Mark the value as present and break out of the inner loop (where we check against all values),
					// and start checking the next position in the outer loop
					present[check] = true;
					break;
				}
			}
		}
		
		// There should be one element that is still false in the array of possible values
		// because we have set all the other values to true
		for (int i = 0; i <= length; i++) {
			// If the value is false, return it: it is our missing element
			if (!present[i]) {
				return i;
			}
		}
		
		return -1;
	}

	public static void main(String[] args) {
		
		System.out.println(findMissing(new int[] {5, 0, 1, 2, 4}));
		System.out.println(findMissing(new int[] {3, 2, 1}));
		System.out.println(findMissing(new int[] {0, 1, 2}));
	}

}