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Positive integers up to N that are not present in given Array

  • Difficulty Level : Hard
  • Last Updated : 20 Sep, 2021

Given an array a[] and an integer N, the task is to find all natural numbers from the range [1, N] that are nor present in the given array.

Examples:

Input: N = 5, a[] = {1, 2, 4, 5}
Output: 3
Explanation: 3 is the only integer from the range [1, 5] that is not present in the array.

Input: N = 10, a[] = {1, 3, 4, 6, 8, 10}
Output: 2 5 7 9

Naive Approach: The simplest approach to solve this problem is to traverse the range [1, N] and for each number from range, traverse the array and check if it is present in the array or not. 
Time Complexity: O(N * len), where len denotes the length of the array. 
Auxiliary Space: O(1)

Efficient Approach: The above approach can be optimized using HashSet. Traverse the given array and insert all array elements into the HashSet. Then, traverse the range [1, N] and for each element, check if it is present in the HashSet or not using contains() method, to compute search in O(1) complexity. 
Time Complexity: O(N) 
Auxiliary Space: O(N)
 

Alternate Approach: The given problem can be solved using BitSet in C++. Follow the steps below to solve the problem:

  1. Initialize a BitSet variable, bset with N as length.
  2. For each array element, set its bit to false, using bset.set(arr[i]-1, 0), where it sets the bit at position arr[i] – 1 to 0.
  3. Now, iterate from bset._Find_first() to bset.size() – 1 using a variable, say i.
  4. Print i + 1 and set bset._Find_next().

 

Below is the implementation of the above approach.

 

C++




// CPP program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find positive integers
// from 1 to N that are not present in the array
void findMissingNumbers(int arr[], int len)
{
    const int M = 15;
 
    // Declare bitset
    bitset<M> bset;
 
    // Iterate from 0 to M - 1
    for (int i = 0; i < M; i++) {
        bset.set(i);
    }
 
    // Iterate from 0 to len - 1
    for (int i = 0; i < len; i++) {
        bset.set(arr[i] - 1, 0);
    }
 
    // Iterate from bset._Find_first()
    // to bset.size() - 1
    for (int i = bset._Find_first();
         i < bset.size();
         i = bset._Find_next(i)) {
 
        if (i + 1 > len)
            break;
 
        cout << i + 1 << endl;
    }
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 2, 4, 6, 8, 9 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    findMissingNumbers(arr, n);
 
    return 0;
}

Java




// Java program for the above approach
import java.io.*;
import java.util.*;
class GFG
{
 
    // Function to find positive integers
    // from 1 to N that are not present in the array
    static void findMissingNumbers(int[] arr, int len)
    {
        int M = 15;
 
        // Declare bitset
        BitSet bset = new BitSet(M);
 
        // Iterate from 0 to M - 1
        for (int i = 0; i < M; i++)
        {
            bset.set(i);
        }
 
        // Iterate from 0 to len - 1
        for (int i = 0; i < len; i++)
        {
            bset.set(arr[i] - 1, false);
        }
 
        // Iterate from bset._Find_first()
        // to bset.size() - 1
        for (int i = bset.nextSetBit(0); i >= 0;
             i = bset.nextSetBit(i + 1))
        {
            if (i + 1 > len)
                break;
            System.out.println(i + 1);
        }
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int[] arr = new int[] { 1, 2, 4, 6, 8, 9 };
        int n = arr.length;
        findMissingNumbers(arr, n);
    }
}
 
// This code is contributed by Dharanendra L V

Python3




# Python 3 program for the above approach
 
#  Function to find positive integers
# from 1 to N that are not present in the array
def findMissingNumbers(arr, n):
 
    M = 15
 
    # Declare bitset
    bset = [0]*M
 
    # Iterate from 0 to M - 1
    for i in range(M):
        bset[i] = i
 
    # Iterate from 0 to n - 1
    for i in range(n):
        bset[arr[i] - 1] = 0
 
    bset = [i for i in bset if i != 0]
 
    # Iterate from bset._Find_first()
    # to bset.size() - 1
 
    for i in range(len(bset)):
 
        if (bset[i] + 1 > n):
            break
 
        print(bset[i] + 1)
 
# Driver Code
if __name__ == "__main__":
 
    arr = [1, 2, 4, 6, 8, 9]
    n = len(arr)
 
    findMissingNumbers(arr, n)
 
    # This code is contributed by ukasp.

 
 

Output: 
3
5

 

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


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