Given an array arr[] of size N, where arr[i] is natural numbers less than or equal to N, the task is to find all the numbers in the range [1, N] that are not present in the given array.
Examples:
Input: arr[ ] = {5, 5, 4, 4, 2}
Output: 1 3
Explanation:
For all numbers in the range [1, 5], 1 and 3 are not present in the array.Input: arr[ ] = {3, 2, 3, 1}
Output: 4
Naive Approach: The simplest approach is to hash every array element using any data structure like the dictionary and then iterate over the range [1, N] and print all numbers not present in the hash.
d
Time Complexity: O(N)
Auxiliary Space: O(N)
Approach: The above approach can be optimized further by marking the number at position arr[i] – 1, negative to mark i is present in the array. Then print all positions of the array elements that are positive as they are missing. Follow the steps below to solve the problem:
-
Iterate over the array, arr[] and for each current element, num perform the following steps:
- Update arr[abs(num)-1] to -abs(arr[abs(num)-1]).
- Iterate over the array, arr[] using the variable i, and print the i+1 if arr[i] is positive.
Below is the implementation of the above approach:
// C++ program for above approach #include <iostream> using namespace std;
// Function to find the missing numbers void getMissingNumbers( int arr[], int N)
{ // traverse the array arr[]
for ( int i = 0; i < N; i++) {
// Update
arr[ abs (arr[i]) - 1] = -( abs (arr[ abs (arr[i]) - 1]));
}
// Traverse the array arr[]
for ( int i = 0; i < N; i++) {
// If Num is not present
if (arr[i] > 0)
cout << i + 1 << " " ;
}
} // Driver Code int main()
{ // Given Input
int N = 5;
int arr[] = { 5, 5, 4, 4, 2 };
// Function Call
getMissingNumbers(arr, N);
return 0;
} // This codeis contributed by dwivediyash |
// Java program for the above approach import java.io.*;
class GFG
{ // Function to find the missing numbers
static void getMissingNumbers( int arr[], int N)
{
// traverse the array arr[]
for ( int i = 0 ; i < N; i++)
{
// Update
arr[(Math.abs(arr[i]) - 1 )]
= -(Math.abs(arr[(Math.abs(arr[i]) - 1 )]));
}
// Traverse the array arr[]
for ( int i = 0 ; i < N; i++)
{
// If Num is not present
if (arr[i] > 0 )
System.out.print(i + 1 + " " );
}
}
// Driver Code
public static void main(String[] args)
{
// Given Input
int N = 5 ;
int arr[] = { 5 , 5 , 4 , 4 , 2 };
// Function Call
getMissingNumbers(arr, N);
}
} // This code is contributed by Potta Lokesh |
# Python program for the above approach # Function to find the missing numbers def getMissingNumbers(arr):
# Traverse the array arr[]
for num in arr:
# Update
arr[ abs (num) - 1 ] = - ( abs (arr[ abs (num) - 1 ]))
# Traverse the array arr[]
for pos, num in enumerate (arr):
# If Num is not present
if num > 0 :
print (pos + 1 , end = ' ' )
# Given Input arr = [ 5 , 5 , 4 , 4 , 2 ]
# Function Call getMissingNumbers(arr) |
// C# program for above approach using System;
using System.Collections.Generic;
class GFG{
// Function to find the missing numbers static void getMissingNumbers( int []arr, int N)
{ // traverse the array arr[]
for ( int i = 0; i < N; i++)
{
// Update
arr[(Math.Abs(arr[i]) - 1)] = -(Math.Abs(arr[(Math.Abs(arr[i]) - 1)]));
}
// Traverse the array arr[]
for ( int i = 0; i < N; i++)
{
// If Num is not present
if (arr[i] > 0)
Console.Write(i + 1 + " " );
}
} // Driver Code public static void Main()
{ // Given Input
int N = 5;
int []arr = { 5, 5, 4, 4, 2 };
// Function Call
getMissingNumbers(arr, N);
} } // This code is contributed by ipg2016107. |
<script> // Javascript program for the above approach // Function to find the missing numbers function getMissingNumbers(arr){
// Traverse the array arr[]
for (let num of arr)
// Update
arr[(Math.abs(num)-1)] = -(Math.abs(arr[(Math.abs(num)-1)]))
// Traverse the array arr[]
for (pos in arr)
// If Num is not present
if (arr[pos] > 0)
document.write(`${parseInt(pos) + 1} `)
} // Given Input let arr = [5, 5, 4, 4, 2] // Function Call getMissingNumbers(arr) // This code is contributed by _saurabh_jaiswal. </script> |
1 3
Time Complexity: O(N)
Auxiliary Space: O(1)