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++
// 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> |
1 3
Time Complexity: O(N)
Auxiliary Space: O(1)
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