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++
// 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 codeint 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 = 3printKMissing(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 codelet 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:
- 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++
// C++ code for // the above approach#include <bits/stdc++.h>using namespace std;// Program to print first k// missing numbervoid 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 Codeint 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 approachimport 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 numberdef 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 Codearr=[1,4,3]n=len(arr)k=3printmissingk(arr,n,k)#This code is contributed by Thirumalai Srinivasan |
C#
// C# code for// the above approachusing 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 numberfunction 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 Codelet 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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