Find all numbers in range [1, N] that are not present in given Array

Last Updated : 24 Mar, 2023

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.

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++

// 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

// 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

Python3

# 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#

// 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.

Javascript

<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>

Output

1 3

Time Complexity: O(N)
Auxiliary Space: O(1)

Find all duplicate and missing numbers in given permutation array of 1 to N

aayushstar300

Improve

Article Tags :

Practice Tags :

Find all numbers in range [1, N] that are not present in given Array

C++

Java

Python3

C#

Javascript

Similar Reads

Thank You!

What kind of Experience do you want to share?