How to Find Missing Number in a Sorted Array in Java

Today's coding problem is not very new, it's an age-old classic Programming interview Question. You have a sorted array containing n - 1 unique number starting from 0 to n - 1. There is only one number missing in this range and you need to find that out. I mean you need to write a Java method to find the missing number and print its value in the console. Some of you might have seen this question before, but if you have not been asked this question before, what is the first approach comes into your mind to solve this question? Since only one number is missing, many programmers come up with the approach of iterating over the array, and comparing each element with the expected one like the first element should be 0, the second element should be 1, and so on.

Though this will solve the problem, it will cost you O(n) time. I mean time complexity of your solution would be O(n) which is not good for a big array like with 100 million entries. What can you do to improve performance?

The key here is that you already have a sorted array, do you think our earlier solution is taking full advantage of this knowledge, well it is but not fully. 

What it is doing is performing a linear search which is costing O(n), but if you do a binary search, which of course needs a sorted array, we can reduce the time taken in the range of O(logN).

Since numbers are in the range from 0 to n - 1 and are sorted, the first number till the missing one should be the same as their indexes. I mean if 0 is not missing, then it must be in the first index, I mean at 0.

If you generalize this, you will find out that if the missing number is k then all numbers less than k is located in an array with indexes the same as their value.

Also, number k + 1 will be located at index k, and number k + 2 will be located at index k + 1. What does this mean? Well, it means that the missing number is the first cell whose value is not the same as its index. So our problem reduces to search in an array to find the first cell, whose value is not the same as its index.

You can easily find out this by using the binary search algorithm in O(logN) time. Our solution implements this logic to find the missing integer in a sorted array in Java. You can use this solution to find the missing number in an array of numbers 1-1000 or 1 -100.

This problem also shows that having a good knowledge of fundamental data structure is essential to solve any coding problems. Therefore, you must spend some time brushing up your Data Structure skills before you go for an interview. If you need a course, I highly recommend Data Structure and  Algorithms in Java: Deep Dive on Udemy. It's both comprehensive and enjoyable and also very affordable. You can buy it in just $10 on Udemy sale.

How to Find Missing Number in Sorted Array- Solution 

Here is our complete solution to this problem. As discussed in the first paragraph, the solution is based upon a binary search algorithm and that's why it's complexity is in logarithmic order. If you asked this question in the interview, you must write production-quality code, what this means is handling invalid input and boundary conditions.

In our method, we are checking whether the array is not null and empty before proceeding further. If you are not familiar with the binary search algorithm than this diagram will help you how does it work. In binary search, instead of starting from the front, we start from the middle element.

If the middle element is greater than the expected number is on the left-hand side of a middle element (lower array), otherwise, it is on the right-hand side (higher array). So in each iteration, we end up reducing our problem set by half.

So in the start, if you have 16 elements in the array, next iteration you only need to search in 8 elements and subsequently 4 and 2, this is how we get O(logN) complexity. This problem also shows that knowledge of coding patterns is very important for coding interviews.

If you know the pattern you can solve many unseen problems and that's why you should spend some time-solving coding problems to build that pattern recognition logic in your mind. A course like Grokking the Coding Interview: Patterns for Coding Questions is really a god send course for someone who wants to mater these patterns. It will teach you 15 popular coding patterns to interview questions which means you can tackle most of the unseen problems during interviews.

Java Program to find missing number in a sorted integer array in Java

Java Program to find the missing number in a sorted Integer array

And, here is our code example to find the missing integer in a sorted array or series.

import java.util.Arrays;
 * Java Program to find the missing number in a sorted array 
 * with integers in range 0 to n -1
 * @author Javin Paul
public class MissingNumberFinder {

    public static void main(String args[]) {

        System.out.println("Test #1 : Missing number in sorted array ");
        int[] input = new int[]{1, 2, 3, 4, 6};
        int missing = missingNumberFromSortedArray(input);
        System.out.println("Missing number from array : " 
                  + Arrays.toString(input) + " is : " + missing);


    public static int missingNumberFromSortedArray(int[] numbers) {
        if (numbers == null || numbers.length <= 0) {
            throw new IllegalArgumentException("Null or Empty array not permitted");

        int left = 0;
        int right = numbers.length - 1;
        while (left <= right) {
            int middle = (right + left) >> 1;
            if (numbers[middle] != middle) {
                if (middle == 0 || numbers[middle - 1] == middle - 1) {
                    return middle;
                right = middle - 1;
            } else {
                left = middle + 1;
        throw new RuntimeException("No missing number");

Test #1 : Missing number in sorted array
Missing number from array : [1, 2, 3, 4, 6] is : 0

That's all about how do you find the missing integer in an array of 0 to n - 1. Remember you can also solve this problem with linear search and there is nothing wrong if you first come up with that solution, in fact, it's very natural, but you must pay attention to the sorted word. Binary search and sorted array go hand in hand. When it comes to improving the search algorithms, binary search always better than linear search.

Further Learning
Data Structures and Algorithms: Deep Dive Using Java
Algorithms and Data Structures - Part 1 and 2
The Coding Interview Bootcamp: Algorithms + Data Structures
Grokking the Coding Interview: Patterns for Coding Questions 

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Thanks for reading this article so far. If you like this article then please share it with your friends and colleagues. If you have any questions or doubt then please let us know and I'll try to find an answer for you. As always suggestions, comments, innovative and better answers are most welcome.

P. S. - If you are looking for some Free Algorithms courses to improve your understanding of Data Structure and Algorithms, then you should also check the Data Structure in Java free course on Udemy. It's completely free and all you need to do is create a free Udemy account to enroll in this course.

P. P. S. - If you want more of such questions from tech interviews, please see Cracking the Coding Interview 6th Edition, it contains over 190 coding questions from different software companies, startups, investment banks, and service-based companies.


  1. if (numbers[middle] != middle) does not seem to be correct... should test for less or greater. Plus the print out finds 0, while you seem to be trying to find 5.
    Great post otherwise!

  2. Isn't it just to find Sum of original expected vs. sum of actual resulted? Something like below program can solve very easily?

    public class MissingNumberFinder {

    public static int findMissing(int[] intArr){
    int last = intArr[intArr.length-1];
    int first = intArr[0];

    int actualSum = 0, resultedSum = 0;
    for ( int i = first; i < last; i++)
    actualSum +=i;

    for (int i = 0; i < intArr.length-1;i++){
    resultedSum += intArr[i];

    return (actualSum-resultedSum);

    public static void main(String args[]){
    int[] arr = {1,2,3,4,5,6,7,9,10};
    int result = findMissing(arr);
    if ( result != 0 ){
    System.out.println("Missing integer in the sorted array is: " + result);
    } else
    System.out.println("No missing integers");

    1. If 0 is missing, actual sum and expected sum will be same rite??

    2. Umesh you are still iterating the array and that too twice so your time complexity becomes 2*O(n). On the other hand the above stated method solves the problem with O(log N).

    3. This is a solution for NON sorted array. For sorted it's is not optimal.
      Btw, it can be improved to O(n). No need to calc sum of i. Can get the sum of 1+2+3+...+n = n*(n+1)/2 without useless iteration.

  3. it must return 0 na??how cum 5??

  4. How do you find the missing number in an un-sorted array? I was asked, You are given a list of n-1 integers and these integers are in the range of 1 to n. There are no duplicates in the array . One of the integers is missing in the array. Write an efficient code to find the missing integer in Java?

    can I use the first solution in this case?

    1. Yes, it's even simpler, if there is one missing element then number would sum of n elements - actual sum

  5. This also works
    int sortedArray[] = { 1, 2, 4, 7, 13, 17, 18, 24 };
    int naturalSeq = 0;
    System.out.println("mising numbers are : ");
    for (int index = 0; index < sortedArray.length; index++) {
    if (naturalSeq != sortedArray[index]) {
    int diff = sortedArray[index] - naturalSeq;
    for (int i = naturalSeq; i < naturalSeq + diff; i++) {
    naturalSeq = sortedArray[index] + 1;
    } else {
    naturalSeq = naturalSeq + 1;

  6. i was do that if not is sorted, that is the most high problem, if you try sorted that will grow complexity. also i use auxiliar structure and whit order of complexity O(2N):
    public int solution(int[] A) {
    boolean[] seen = new boolean[A.length];
    Arrays.fill(seen, false);
    for (int value : A) {
    if (0 < value && value <= A.length) {
    seen[value - 1] = true;

    for (int i = 0; i < seen.length; i++) {
    if (seen[i] == false) {
    return i + 1;

    return A.length + 1;

    also tan can be more performance in space if we use BitSet.

  7. x=sumList1();


  8. Sort array and then use java 8 stream API
    e.g. This is sorted array.
    int [] names = {2,4,5,6,7,8,9,10};

    IntStream.range(names[0], names[names.length-1]+1)
    .filter(i -> i != names[i-names[0]])