k-th missing element in increasing sequence which is not present in a given sequence

Given two sequences, one is increasing sequence a[] and another a normal sequence b[], find the K-th missing element in the increasing sequence which is not present in the given sequence. If no k-th missing element is there output -1

Examples:

Input:  a[] = {0, 2, 4, 6, 8, 10, 12, 14, 15};
        b[] = {4, 10, 6, 8, 12};
          k = 3
Output: 14
Explanation : The numbers from increasing sequence that
are not present in the given sequence are 0, 2, 14, 15.
The 3rd missing number is 14.

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

n1 Number of elements on increasing sequence a[].
n2 Number of elements in given sequence b[].

A naive approach is to iterate for every element in the increasing sequence and check if it is present in the given sequence or not, and keep a counter of not present elements, and print the k-th non present element. This will not be efficient enough as it has two nested for loops which will take O(n2).
Time complexity: O(n1 * n2)
Auxiliary space: O(1)

An efficient approach is to use hashing. We store all elements of given sequence in a hash table. Then we iterate through all elements of increasing sequence. For every element, we search it in the hash table. If element is present in not hash table, then we increment count of missing elements. If count becomes k, we return the missing element.

Below is the implementation of the above approach

C++

// C++ program to find the k-th missing element 
// in a given sequence 
#include <bits/stdc++.h> 
using namespace std; 
  
// Returns k-th missing element. It returns -1 if 
// no k is more than number of missing elements. 
int find(int a[], int b[], int k, int n1, int n2) 
{ 
    // Insert all elements of givens sequence b[]. 
    unordered_set<int> s; 
    for (int i = 0; i < n2; i++) 
        s.insert(b[i]); 
  
    // Traverse through increasing sequence and  
    // keep track of count of missing numbers. 
    int missing = 0; 
    for (int i = 0; i < n1; i++) { 
        if (s.find(a[i]) == s.end()) 
            missing++; 
        if (missing == k) 
            return a[i]; 
    } 
  
    return -1; 
} 
  
// driver program to test the above function 
int main() 
{ 
    int a[] = { 0, 2, 4, 6, 8, 10, 12, 14, 15 }; 
    int b[] = { 4, 10, 6, 8, 12 }; 
    int n1 = sizeof(a) / sizeof(a[0]); 
    int n2 = sizeof(b) / sizeof(b[0]); 
  
    int k = 3; 
    cout << find(a, b, k, n1, n2); 
    return 0; 
} 

Java

// Java program to find the k-th missing element 
// in a given sequence 
import java.util.*; 
  
class GFG 
{ 
      
// Returns k-th missing element. It returns -1 if 
// no k is more than number of missing elements. 
static int find(int a[], int b[], int k, int n1, int n2) 
{ 
    // Insert all elements of givens sequence b[]. 
    LinkedHashSet<Integer> s = new LinkedHashSet<>(); 
    for (int i = 0; i < n2; i++) 
        s.add(b[i]); 
  
    // Traverse through increasing sequence and  
    // keep track of count of missing numbers. 
    int missing = 0; 
    for (int i = 0; i < n1; i++)  
    { 
        if(!s.contains(a[i]) )  
            missing++; 
        if (missing == k) 
            return a[i]; 
    } 
  
    return -1; 
} 
  
// Driver code 
public static void main(String[] args)  
{ 
    int a[] = { 0, 2, 4, 6, 8, 10, 12, 14, 15 }; 
    int b[] = { 4, 10, 6, 8, 12 }; 
    int n1 = a.length; 
    int n2 = b.length; 
  
    int k = 3; 
    System.out.println(find(a, b, k, n1, n2)); 
} 
}  
  
// This code has been contributed by 29AjayKumar 

Python3

# Python3 program to find the k-th  
# missing element in a given sequence  
  
  
# Returns k-th missing element. It returns -1 if  
# no k is more than number of missing elements.  
def find(a, b, k, n1, n2): 
  
    # insert all elements of  
    # given sequence b[]. 
    s = set() 
      
    for i in range(n2): 
        s.add(b[i]) 
  
    # Traverse through increasing sequence and  
    # keep track of count of missing numbers.  
    missing = 0
    for i in range(n1): 
        if a[i] not in s: 
            missing += 1
        if missing == k: 
            return a[i] 
    return -1
  
# Driver code 
a = [0, 2, 4, 6, 8, 10, 12, 14, 15] 
b = [4, 10, 6, 8, 12] 
n1 = len(a) 
n2 = len(b) 
k = 3
print(find(a, b, k, n1, n2)) 
  
# This code is contributed by Shrikant13 

C#

// C# program to find the k-th missing element  
// in a given sequence  
using System; 
using System.Collections.Generic; 
  
class GFG  
{  
      
// Returns k-th missing element. It returns -1 if  
// no k is more than number of missing elements.  
static int find(int []a, int []b, int k, int n1, int n2)  
{  
    // Insert all elements of givens sequence b[].  
    HashSet<int> s = new HashSet<int>();  
    for (int i = 0; i < n2; i++)  
        s.Add(b[i]);  
  
    // Traverse through increasing sequence and  
    // keep track of count of missing numbers.  
    int missing = 0;  
    for (int i = 0; i < n1; i++)  
    {  
        if(!s.Contains(a[i]) )  
            missing++;  
        if (missing == k)  
            return a[i];  
    }  
  
    return -1;  
}  
  
// Driver code  
public static void Main(String[] args)  
{  
    int []a = { 0, 2, 4, 6, 8, 10, 12, 14, 15 };  
    int []b = { 4, 10, 6, 8, 12 };  
    int n1 = a.Length;  
    int n2 = b.Length;  
  
    int k = 3;  
    Console.WriteLine(find(a, b, k, n1, n2));  
}  
}  
  
/* This code contributed by PrinciRaj1992 */

Output:

Time complexity: O(n1 + n2)
Auxiliary Space: O(n2)

This article is contributed by Raja Vikramaditya. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

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