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]- Sort the given array.
- After sorting, we find the position of the first positive number in the array.
- Now we traverse the array and keep printing elements in gaps between two consecutive array elements.
- If gaps don’t cover k missing numbers, we print numbers greater than the largest array element.
Implementation:
C++
#include <bits/stdc++.h>
using namespace std;
void printKMissing(int arr[], int n, int k)
{
sort(arr, arr + n);
int i = 0;
while (i < n && arr[i] <= 0)
i++;
int count = 0, curr = 1;
while (count < k && i < n) {
if (arr[i] != curr) {
cout << curr << " ";
count++;
}
else
i++;
curr++;
}
while (count < k) {
cout << curr << " ";
curr++;
count++;
}
}
int main()
{
int arr[] = { 2, 3, 4 };
int n = sizeof(arr) / sizeof(arr[0]);
int k = 3;
printKMissing(arr, n, k);
return 0;
}
|
Java
import java.util.Arrays;
class GFG {
static void printKMissing(int[] arr, int n, int k)
{
Arrays.sort(arr);
int i = 0;
while (i < n && arr[i] <= 0)
i++;
int count = 0, curr = 1;
while (count < k && i < n) {
if (arr[i] != curr) {
System.out.print(curr + " ");
count++;
}
else
i++;
curr++;
}
while (count < k) {
System.out.print(curr + " ");
curr++;
count++;
}
}
public static void main(String[] args)
{
int[] arr = { 2, 3, 4 };
int n = arr.length;
int k = 3;
printKMissing(arr, n, k);
}
}
|
Python3
def printKMissing(arr, n, k) :
arr.sort()
i = 0
while (i < n and arr[i] <= 0) :
i = i + 1
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
while (count < k) :
print(str(curr) + " ", end = '')
curr = curr + 1
count = count + 1
arr = [ 2, 3, 4 ]
n = len(arr)
k = 3
printKMissing(arr, n, k);
|
C#
using System;
class GFG {
static void printKMissing(int[] arr,
int n,
int k)
{
Array.Sort(arr);
int i = 0;
while (i < n && arr[i] <= 0)
i++;
int count = 0, curr = 1;
while (count < k && i < n) {
if (arr[i] != curr) {
Console.Write(curr + " ");
count++;
}
else
i++;
curr++;
}
while (count < k) {
Console.Write(curr + " ");
curr++;
count++;
}
}
public static void Main()
{
int[] arr = {2, 3, 4};
int n = arr.Length;
int k = 3;
printKMissing(arr, n, k);
}
}
|
PHP
<?php
function printKMissing($arr, $n, $k)
{
sort($arr); sort($arr , $n);
$i = 0;
while ($i < $n && $arr[$i] <= 0)
$i++;
$count = 0; $curr = 1;
while ($count < $k && $i < $n) {
if ($arr[$i] != $curr) {
echo $curr , " ";
$count++;
}
else
$i++;
$curr++;
}
while ($count < $k) {
echo $curr , " ";
$curr++;
$count++;
}
}
$arr =array ( 2, 3, 4 );
$n = sizeof($arr);
$k = 3;
printKMissing($arr, $n, $k);
?>
|
Javascript
<script>
function printKMissing(arr, n, k) {
arr.sort((a, b) => a - b);
let i = 0;
while (i < n && arr[i] <= 0)
i++;
let count = 0, curr = 1;
while (count < k && i < n) {
if (arr[i] != curr) {
document.write(curr + " ");
count++;
}
else
i++;
curr++;
}
while (count < k) {
document.write(curr, " ");
curr++;
count++;
}
}
let arr = new Array(2, 3, 4);
let n = arr.length;
let k = 3;
printKMissing(arr, n, k);
</script>
|
Time Complexity: O(n Log n)
Auxiliary Space: O(1)
Alternative Method:
- We can use hashmap to search in O(1) time.
- Use a dictionary to store values in the array.
- We run a loop from 1 to n+k and check whether they are in hashmap.
- If they are not present print the number.
- if all k elements are found break the loop.
Implementation:
C++
#include <bits/stdc++.h>
using namespace std;
void printmissingk(int arr[],
int n, int k)
{
map<int, int> d;
for (int i = 0; i < n; i++)
d[arr[i]] = arr[i];
int cnt = 1;
int fl = 0;
for (int i = 0; i < (n + k); i++)
{
if (d.find(cnt) == d.end())
{
fl += 1;
cout << cnt << " ";
if (fl == k)
break;
}
cnt += 1;
}
}
int main()
{
int arr[] = {1, 4, 3};
int n = sizeof(arr) /
sizeof(arr[0]);
int k = 3;;
printmissingk(arr, n, k);
}
|
Java
import java.io.*;
import java.util.HashMap;
class GFG
{
static void printmissingk(int arr[], int n, int k)
{
HashMap<Integer, Integer> d = new HashMap<>();
for (int i = 0; i < n; i++)
d.put(arr[i], arr[i]);
int cnt = 1;
int fl = 0;
for (int i = 0; i < (n + k); i++) {
if (!d.containsKey(cnt)) {
fl += 1;
System.out.print(cnt + " ");
if (fl == k)
break;
}
cnt += 1;
}
}
public static void main(String[] args)
{
int arr[] = { 1, 4, 3 };
int n = arr.length;
int k = 3;
printmissingk(arr, n, k);
}
}
|
Python3
def printmissingk(arr,n,k):
d={}
for i in range(len(arr)):
d[arr[i]]=arr[i]
cnt=1
fl=0
for i in range(n+k):
if cnt not in d:
fl+=1
print(cnt,end=" ")
if fl==k:
break
cnt+=1
print()
arr=[1,4,3]
n=len(arr)
k=3
printmissingk(arr,n,k)
|
C#
using System;
using System.Collections.Generic;
public class GFG
{
static void printmissingk(int[] arr, int n, int k)
{
Dictionary<int,int> d = new Dictionary<int,int>();
for (int i = 0; i < n; i++)
{
d.Add(arr[i], arr[i]);
}
int cnt = 1;
int fl = 0;
for (int i = 0; i < (n + k); i++)
{
if (!d.ContainsKey(cnt))
{
fl += 1;
Console.Write(cnt + " ");
if (fl == k)
break;
}
cnt += 1;
}
}
static public void Main (){
int[] arr = { 1, 4, 3 };
int n = arr.Length;
int k = 3;
printmissingk(arr, n, k);
}
}
|
Javascript
<script>
function printmissingk(arr,n,k)
{
let d = new Map();
for (let i = 0; i < n; i++)
d.set(arr[i], arr[i]);
let cnt = 1;
let fl = 0;
for (let i = 0; i < (n + k); i++) {
if (!d.has(cnt)) {
fl += 1;
document.write(cnt + " ");
if (fl == k)
break;
}
cnt += 1;
}
}
let arr=[1, 4, 3];
let n = arr.length;
let k = 3;
printmissingk(arr, n, k);
</script>
|
Time complexity: O(n+k)
Auxiliary Space: O(n)
This article is contributed by Biswajit Mohapatra. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.