Java Program for k-th missing element in sorted array

Given an increasing sequence a[], we need to find the K-th missing contiguous element in the increasing sequence which is not present in the sequence. If no k-th missing element is there output -1.

Examples :

Input : a[] = {2, 3, 5, 9, 10};   
        k = 1;
Output : 1
Explanation: Missing Element in the increasing 
sequence are {1,4, 6, 7, 8}. So k-th missing element
is 1

Input : a[] = {2, 3, 5, 9, 10, 11, 12};       
        k = 4;
Output : 7
Explanation: missing element in the increasing 
sequence are {1, 4, 6, 7, 8}  so k-th missing 
element is 7

Approach 1: Start iterating over the array elements, and for every element check if the next element is consecutive or not, if not, then take the difference between these two, and check if the difference is greater than or equal to given k, then calculate ans = a[i] + count, else iterate for next element.

Java

// Java program to check for
// even or odd
import java.io.*;
import java.util.*;
  
public class GFG {
      
    // Function to find k-th
    // missing element
    static int missingK(int []a, int k,
                                 int n)
    {
        int difference = 0,
            ans = 0, count = k;
        boolean flag = false;
          
        // iterating over the array
        for(int i = 0 ; i < n - 1; i++)
        {
            difference = 0;
              
            // check if i-th and
            // (i + 1)-th element
            // are not consecutive
            if ((a[i] + 1) != a[i + 1])
            {
                  
                // save their difference
                difference +=
                    (a[i + 1] - a[i]) - 1;
                  
                // check for difference
                // and given k
                if (difference >= count)
                    {
                        ans = a[i] + count;
                        flag = true;
                        break;
                    }
                else
                    count -= difference;
            }
        }
          
        // if found
        if(flag)
            return ans;
        else
            return -1;
    }
      
    // Driver code
    public static void main(String args[])
    {
          
        // Input array
        int []a = {1, 5, 11, 19};
          
        // k-th missing element
        // to be found in the array
        int k = 11;
        int n = a.length;
          
        // calling function to
        // find missing element
        int missing = missingK(a, k, n);
          
        System.out.print(missing);
    }
  
}
  
// This code is contributed by
// Manish Shaw (manishshaw1)

Output

Time Complexity :O(n), where n is the number of elements in the array.

Space Complexity: O(1) as no extra space has been used.

Approach 2:

Apply a binary search. Since the array is sorted we can find at any given index how many numbers are missing as arr[index] – (index+1). We would leverage this knowledge and apply binary search to narrow down our hunt to find that index from which getting the missing number is easier.

Below is the implementation of the above approach:

Java

// Java program for above approach
public class GFG
{
 
  // Function to find
  // kth missing number
  static int missingK(int[] arr, int k)
  {
    int n = arr.length;
    int l = 0, u = n - 1, mid;   
    while(l <= u)
    {
      mid = (l + u)/2;       
      int numbers_less_than_mid = arr[mid] -
        (mid + 1);
 
      // If the total missing number
      // count is equal to k we can iterate
      // backwards for the first missing number
      // and that will be the answer.
      if(numbers_less_than_mid == k)
      {
 
        // To further optimize we check
        // if the previous element's
        // missing number count is equal
        // to k. Eg: arr = [4,5,6,7,8]
        // If you observe in the example array,
        // the total count of missing numbers for all
        // the indices are same, and we are
        // aiming to narrow down the
        // search window and achieve O(logn)
        // time complexity which
        // otherwise would've been O(n).
        if(mid > 0 && (arr[mid - 1] - (mid)) == k)
        {
          u = mid - 1;
          continue;
        }
 
        // Else we return arr[mid] - 1.
        return arr[mid] - 1;
      }
 
      // Here we appropriately
      // narrow down the search window.
      if(numbers_less_than_mid < k)
      {
        l = mid + 1;
      }
      else if(k < numbers_less_than_mid)
      {
        u = mid - 1;
      }
    }
 
    // In case the upper limit is -ve
    // it means the missing number set
    // is 1,2,..,k and hence we directly return k.
    if(u < 0)
      return k;
 
    // Else we find the residual count
    // of numbers which we'd then add to
    // arr[u] and get the missing kth number.
    int less = arr[u] - (u + 1);
    k -= less;
 
    // Return arr[u] + k
    return arr[u] + k;
  }
 
  // Driver code
  public static void main(String[] args)
  {
    int[] arr = {2,3,4,7,11};
    int k = 5;
 
    // Function Call
    System.out.println("Missing kth number = "+ missingK(arr, k));
  }
}
 
// This code is contributed by divyesh072019.