Find Kth Missing Positive Number in Sorted Array Binary Search Approach

Problem Statement❯

You're given a sorted array of unique positive integers and an integer k. The array is strictly increasing, but it may be missing some positive numbers. Your task is to find the kth missing positive number in the sequence.

The array only contains some of the positive integers starting from 1, and some numbers are missing. You must figure out what the kth missing number would be if we continued counting all the positive numbers in order.

If the array is empty, then the answer is simply k because all numbers are missing.

Examples❯

Input Array	k	Output	Description
[2, 3, 4, 7, 11]	5	9	Missing numbers: [1, 5, 6, 8, 9, 10], 5th one is 9
[1, 2, 3, 4]	2	6	No missing until 4, missing numbers are [5, 6], 2nd is 6
[1, 3, 5, 10]	4	6	Missing: [2, 4, 6, 7, 8, 9], 4th one is 6
[2, 3, 4]	1	1	1 is missing before array starts
[10, 11, 12]	3	3	Missing: [1, 2, 3, 4...], 3rd one is 3
[1, 2, 3, 4]	0	0	No missing required
[]	4	4	Empty array: all numbers are missing, return k = 4
[2]	1	1	1 is missing before the first element
[5]	2	2	Missing: [1, 2, 3, 4], 2nd is 2

Visualization Player

Solution❯

To solve the problem of finding the kth missing positive number, we need to understand where and how numbers go missing in the given array.

Understanding the Missing Numbers

Suppose we have an array like [2, 3, 4, 7]. Normally, if no numbers were missing, we would expect the first 4 numbers to be [1, 2, 3, 4]. But here, 1 is missing at the beginning. Then after 4, we expect 5 and 6, but we jump directly to 7, so 5 and 6 are missing too. This way, we can count how many numbers are missing up to any point.

Different Scenarios to Consider

Missing numbers at the beginning: If the array starts from a number greater than 1, then everything before that number is missing. For example, in [4, 5], the missing numbers are [1, 2, 3].
Missing in the middle: If numbers jump from one to a much bigger next number, then there are several missing numbers in between.
Empty array: If no numbers are given at all, then we assume all numbers are missing. So the kth missing is just k itself.
All initial values are continuous: If the array starts at 1 and continues in sequence (e.g., [1, 2, 3, 4]), then missing numbers only start after the last element.

How We Find the Kth Missing

As we go through the array, we can check how many numbers are missing up to each index. For example, at index i, the expected value is i + 1, but the actual value is arr[i]. So the number of missing values up to this point is arr[i] - (i + 1).

We continue this process until the number of missing values becomes at least k. If that happens, we know the kth missing value must be before or at that point.

If we reach the end of the array and still haven't found enough missing numbers, it means the missing number lies beyond the last element. In that case, the answer is arr.length + k if we’ve found no missing elements, or more accurately, last array value + (k - total missing).

Efficiency

This can be solved efficiently using binary search because the missing count increases with the index. We search for the first position where the number of missing elements becomes at least k and calculate the answer accordingly.

This approach ensures we avoid scanning all numbers one-by-one and still get the correct answer even for large arrays.

Algorithm Steps❯

Initialize low = 0 and high = arr.length - 1.
While low ≤ high:
→ Calculate mid = low + (high - low) / 2.
→ Compute missing = arr[mid] - (mid + 1).
→ If missing < k, search right side by setting low = mid + 1.
→ Else, search left side by setting high = mid - 1.
Finally, return low + k as the answer.

Code

Java

Python

JavaScript

C++

Kotlin

Swift

Php

public class KthMissingNumber {
  public static int findKthPositive(int[] arr, int k) {
    int low = 0;
    int high = arr.length - 1;

    while (low <= high) {
      int mid = low + (high - low) / 2;
      int missing = arr[mid] - (mid + 1); // How many numbers are missing before index mid

      if (missing < k) {
        low = mid + 1; // We need to move right to find more missing numbers
      } else {
        high = mid - 1; // Too many missing, move left
      }
    }

    return low + k; // kth missing number is after low numbers in total
  }

  public static void main(String[] args) {
    int[] arr = {2, 3, 4, 7, 11};
    int k = 5;
    System.out.println("Kth missing number is: " + findKthPositive(arr, k));
  }
}

Time Complexity❯

Case	Time Complexity	Explanation
Best Case	O(1)	The target is found in the first binary search step.
Average Case	O(log n)	Binary search halves the search space at each step, making it efficient even for large arrays.
Worst Case	O(log n)	At most, the search checks log(n) elements until the missing count reaches or exceeds k.

Space Complexity❯

O(1)

Explanation: The algorithm only uses a constant number of variables regardless of input size.

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