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:
- Initialize a BitSet variable, bset with N as length.
- 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.
- Now, iterate from bset._Find_first() to bset.size() – 1 using a variable, say i.
- 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 arrayvoid 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 Codeint 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 approachimport 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 arraydef 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 Codeif __name__ == "__main__": arr = [1, 2, 4, 6, 8, 9] n = len(arr) findMissingNumbers(arr, n) # This code is contributed by ukasp. |
C#
// C# program for the above approachusing System;using System.Collections;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 int[] bset = new int[M]; // Iterate from 0 to M - 1 for (int i = 0; i < M; i++) { bset[i] = i; } // Iterate from 0 to len - 1 for (int i = 0; i < len; i++) { bset[arr[i] - 1] = 0; } ArrayList temp = new ArrayList(); foreach(int x in bset){ if(x != 0){ temp.Add(x); } } // Iterate from bset._Find_first() // to bset.size() - 1 for (int i = 0; i < temp.Count; i++) { if ((int)temp[i] + 1 > len){ break; } Console.WriteLine((int)temp[i] + 1); } } // Driver Code public static void Main() { int[] arr = new int[] { 1, 2, 4, 6, 8, 9 }; int n = arr.Length; findMissingNumbers(arr, n); }}// This code is contributed by Saurabh Jaiswal |
Javascript
<script>// JavaScript program for the above approach// Function to find positive integers// from 1 to N that are not present in the arrayfunction findMissingNumbers(arr, n){ let M = 15 // Declare bitset let bset = new Array(M).fill(0) // Iterate from 0 to M - 1 for(let i=0;i<M;i++) bset[i] = i // Iterate from 0 to n - 1 for(let i=0;i<n;i++) bset[arr[i] - 1] = 0 bset = bset.filter((i)=>i != 0) // Iterate from bset._Find_first() // to bset.size() - 1 for(let i = 0; i < bset.length; i++) { if (bset[i] + 1 > n) break document.write(bset[i] + 1,"</br>") }}// Driver Codelet arr = [1, 2, 4, 6, 8, 9]let n = arr.lengthfindMissingNumbers(arr, n)// This code is contributed by shinjanpatra</script> |
3 5
Time Complexity: O(N)
Auxiliary Space: O(N)
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