Find first k natural numbers missing in given array

Given an array of size n and a number k, we need to print first k natural numbers that are not there in the given array.

Examples:

Input : [2 3 4] 
         k = 3
Output : [1 5 6]

Input  : [-2 -3 4] 
          k = 2
Output : [1 2]

1. Sort the given array. 
2. After sorting, we find the position of the first positive number in the array. 
3. Now we traverse the array and keep printing elements in gaps between two consecutive array elements. 
4. If gaps don’t cover k missing numbers, we print numbers greater than the largest array element. 

Implementation:

C++

// C++ program to find missing k numbers
// in an array.
#include <bits/stdc++.h>
using namespace std;

// Prints first k natural numbers in
// arr[0..n-1]
void printKMissing(int arr[], int n, int k)
{
    sort(arr, arr + n);

    // Find first positive number
    int i = 0;
    while (i < n && arr[i] <= 0)
        i++;

    // Now find missing numbers
    // between array elements
    int count = 0, curr = 1;
    while (count < k && i < n) {
        if (arr[i] != curr) {
            cout << curr << " ";
            count++;
        }
        else
            i++;

        curr++;
    }

    // Find missing numbers after
    // maximum.
    while (count < k) {
        cout << curr << " ";
        curr++;
        count++;
    }
}

// Driver code
int main()
{
    int arr[] = { 2, 3, 4 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int k = 3;
    printKMissing(arr, n, k);
    return 0;
}

Java

// Java program to find missing k numbers
// in an array.
import java.util.Arrays;

class GFG {
    // Prints first k natural numbers in
    // arr[0..n-1]
    static void printKMissing(int[] arr, int n, int k)
    {
        Arrays.sort(arr);

        // Find first positive number
        int i = 0;
        while (i < n && arr[i] <= 0)
            i++;

        // Now find missing numbers
        // between array elements
        int count = 0, curr = 1;
        while (count < k && i < n) {
            if (arr[i] != curr) {
                System.out.print(curr + " ");
                count++;
            }
            else
                i++;
            curr++;
        }

        // Find missing numbers after
        // maximum.
        while (count < k) {
            System.out.print(curr + " ");
            curr++;
            count++;
        }
    }

    // Driver code
    public static void main(String[] args)
    {
        int[] arr = { 2, 3, 4 };
        int n = arr.length;
        int k = 3;
        printKMissing(arr, n, k);
    }
}
/* This code is contributed by Mr. Somesh Awasthi */

Python3

# Python3 program to find missing
# k numbers in an array.

# Prints first k natural numbers
# in arr[0..n-1]
def printKMissing(arr, n, k) :

    arr.sort()

    # Find first positive number
    i = 0
    while (i < n and arr[i] <= 0) :
        i = i + 1

    # Now find missing numbers
    # between array elements
    count = 0
    curr = 1
    while (count < k and i < n) :
        if (arr[i] != curr) :
            print(str(curr) + " ", end = '')
            count = count + 1
        else:
            i = i + 1

        curr = curr + 1

    # Find missing numbers after
    # maximum.
    while (count < k) :
        print(str(curr) + " ", end = '')
        curr = curr + 1
        count = count + 1

# Driver code
arr = [ 2, 3, 4 ]
n = len(arr)
k = 3
printKMissing(arr, n, k);

# This code is contributed
# by Yatin Gupta

C#

// C# program to find missing
// k numbers in an array.
using System;

class GFG {
    // Prints first k natural numbers
    // in arr[0..n-1]
    static void printKMissing(int[] arr,
                              int n,
                              int k)
    {
        Array.Sort(arr);

        // Find first positive number
        int i = 0;
        while (i < n && arr[i] <= 0)
            i++;

        // Now find missing numbers
        // between array elements
        int count = 0, curr = 1;
        while (count < k && i < n) {
            if (arr[i] != curr) {
                Console.Write(curr + " ");
                count++;
            }
            else
                i++;
            curr++;
        }

        // Find missing numbers
        // after maximum.
        while (count < k) {
            Console.Write(curr + " ");
            curr++;
            count++;
        }
    }

    // Driver code
    public static void Main()
    {
        int[] arr = {2, 3, 4};
        int n = arr.Length;
        int k = 3;
        printKMissing(arr, n, k);
    }
}

// This code is contributed by Nitin Mittal

PHP

<?php
// PHP program to find missing k numbers
// in an array.

// Prints first k natural numbers in
// arr[0..n-1]
function printKMissing($arr, $n, $k)
{
    sort($arr); sort($arr , $n);

    // Find first positive number
    $i = 0;
    while ($i < $n && $arr[$i] <= 0)
        $i++;

    // Now find missing numbers
    // between array elements
    $count = 0; $curr = 1;
    while ($count < $k && $i < $n) {
        if ($arr[$i] != $curr) {
            echo $curr , " ";
            $count++;
        }
        else
            $i++;

        $curr++;
    }

    // Find missing numbers after
    // maximum.
    while ($count < $k) {
        echo $curr , " ";
        $curr++;
        $count++;
    }
}

    // Driver code
    $arr =array ( 2, 3, 4 );
    $n = sizeof($arr);
    $k = 3;
    printKMissing($arr, $n, $k);

// This code is contributed by Nitin Mittal.
?>

Javascript

<script>

// JavaScript program to find missing k numbers
// in an array.

// Prints first k natural numbers in
// arr[0..n-1]
function printKMissing(arr, n, k) {
    arr.sort((a, b) => a - b);

    // Find first positive number
    let i = 0;
    while (i < n && arr[i] <= 0)
        i++;

    // Now find missing numbers
    // between array elements
    let count = 0, curr = 1;
    while (count < k && i < n) {
        if (arr[i] != curr) {
            document.write(curr + " ");
            count++;
        }
        else
            i++;

        curr++;
    }

    // Find missing numbers after
    // maximum.
    while (count < k) {
        document.write(curr, " ");
        curr++;
        count++;
    }
}

// Driver code
let arr = new Array(2, 3, 4);
let n = arr.length;
let k = 3;
printKMissing(arr, n, k);

// This code is contributed by gfgking

</script>

Output

1 5 6 

Time Complexity: O(n Log n)
Auxiliary Space: O(1)

Alternative Method:

1. We can use hashmap to search in O(1) time.
2. Use a dictionary to store values in the array. 
3. We run a loop from 1 to n+k and check whether they are in hashmap.
4. If they are not present print the number. 
5. if all k elements are found break the loop. 

Implementation:

C++

// C++ code for
// the above approach
#include <bits/stdc++.h>
using namespace std;

// Program to print first k
// missing number
void printmissingk(int arr[],
                   int n, int k)
{
  // Creating a hashmap
  map<int, int> d;

  // Iterate over array
  for (int i = 0; i < n; i++)
    d[arr[i]] = arr[i];

  int cnt = 1;
  int fl = 0;

  // Iterate to find missing
  // element
  for (int i = 0; i < (n + k); i++)
  {
    if (d.find(cnt) == d.end())
    {
      fl += 1;
      cout << cnt << " ";
      if (fl == k)
        break;
    }
    cnt += 1;
  }
}

// Driver Code
int main()
{
  int arr[] = {1, 4, 3};
  int n = sizeof(arr) /
          sizeof(arr[0]);
  int k = 3;;
  printmissingk(arr, n, k);
}

// This code is contributed by Chitranayal

Java

// Java code for
// the above approach
import java.io.*;
import java.util.HashMap;

class GFG
{

    // Program to print first k
    // missing number
    static void printmissingk(int arr[], int n, int k)
    {
        // Creating a hashmap
        HashMap<Integer, Integer> d = new HashMap<>();

        // Iterate over array
        for (int i = 0; i < n; i++)
            d.put(arr[i], arr[i]);

        int cnt = 1;
        int fl = 0;

        // Iterate to find missing
        // element
        for (int i = 0; i < (n + k); i++) {
            if (!d.containsKey(cnt)) {
                fl += 1;
                System.out.print(cnt + " ");
                if (fl == k)
                    break;
            }
            cnt += 1;
        }
    }

    // Driver Code
    public static void main(String[] args)
    {
        int arr[] = { 1, 4, 3 };
        int n = arr.length;
        int k = 3;
        printmissingk(arr, n, k);
    }
}

// This code is contributed by subhammahato348.

Python3

# Python3 code for above approach

# Program to print first k
# missing number
def printmissingk(arr,n,k):

    #creating a hashmap
    d={}

    # Iterate over array
    for i in range(len(arr)):
        d[arr[i]]=arr[i]

    cnt=1
    fl=0

    # Iterate to find missing
    # element
    for i in range(n+k):
        if cnt not in d:
            fl+=1
            print(cnt,end=" ")
            if fl==k:
                break
        cnt+=1
    print()

# Driver Code
arr=[1,4,3]
n=len(arr)
k=3
printmissingk(arr,n,k)

#This code is contributed by Thirumalai Srinivasan

C#

// C# code for
// the above approach
using System;
using System.Collections.Generic;

public class GFG
{

  // Program to print first k
  // missing number
  static void printmissingk(int[] arr, int n, int k)
  {

    // Creating a hashmap
    Dictionary<int,int> d = new Dictionary<int,int>();

    // Iterate over array
    for (int i = 0; i < n; i++)
    {
      d.Add(arr[i], arr[i]);
    }
    int cnt = 1;
    int fl = 0;

    // Iterate to find missing
    // element
    for (int i = 0; i < (n + k); i++)
    {
      if (!d.ContainsKey(cnt))
      {
        fl += 1;
        Console.Write(cnt + " ");
        if (fl == k)
          break;
      }
      cnt += 1;
    }
  }

  // Driver Code
  static public void Main (){
    int[] arr = { 1, 4, 3 };
    int n = arr.Length;
    int k = 3;
    printmissingk(arr, n, k);
  }
}

// This code is contributed by avanitrachhadiya2155

Javascript

<script>
// Javascript code for
// the above approach

// Program to print first k
    // missing number
function printmissingk(arr,n,k)
{
    // Creating a hashmap
        let d = new Map();

        // Iterate over array
        for (let i = 0; i < n; i++)
            d.set(arr[i], arr[i]);

        let cnt = 1;
        let fl = 0;

        // Iterate to find missing
        // element
        for (let i = 0; i < (n + k); i++) {
            if (!d.has(cnt)) {
                fl += 1;
                document.write(cnt + " ");
                if (fl == k)
                    break;
            }
            cnt += 1;
        }
}

// Driver Code
let arr=[1, 4, 3];
let  n = arr.length;
let k = 3;
printmissingk(arr, n, k);


// This code is contributed by ab2127
</script>

Output

2 5 6 

Time complexity: O(n+k)
Auxiliary Space: O(n)

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