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

• Difficulty Level : Hard
• Last Updated : 31 Jan, 2022

Given an array arr[] of size N consisting of the first N natural numbers, the task is to find all the repeating and missing numbers over the range [1, N] in the given array.

Examples:

Input: arr[] = {1, 1, 2, 3, 3, 5}
Output:
Missing Numbers: [4, 6]
Duplicate Numbers: [1, 3]
Explanation:
As 4 and 6 are not in arr[] Therefore they are missing and 1 is repeating two times and 3 is repeating two times so they are duplicate numbers.

Input: arr[] = {1, 2, 2, 2, 4, 5, 7}
Output:
Missing Numbers: [3, 6]
Duplicate Numbers: [2]

Approach: The given problem can be solved using the idea discussed in this article where only one element is repeating and the other is duplicate. Follow the steps below to solve the given problem:

• Initialize an array, say missing[] that stores the missing elements.
• Initialize a set, say duplicate that stores the duplicate elements.
• Traverse the given array arr[] using the variable i and perform the following steps:
• If the value of arr[i] != arr[arr[i] – 1] is true, then the current element is not equal to the place where it is supposed to be if all numbers were present from 1 to N. So swap arr[i] and arr[arr[i] – 1].
• Otherwise, it means the same element is present at arr[arr[i] – 1].
• Traverse the given array arr[] using the variable i and if the value of arr[i] is not the same as (i + 1) then the missing element is (i + 1) and the duplicate element is arr[i].
• After completing the above steps, print the elements stored in missing[] and duplicate[] as the result.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach` `#include ``using` `namespace` `std;` `// Function to find the duplicate and``// the missing elements over the range``// [1, N]``void` `findElements(``int` `arr[], ``int` `N)``{``    ``int` `i = 0;` `    ``// Stores the missing and duplicate``    ``// numbers in the array arr[]``    ``vector<``int``> missing;``    ``set<``int``> duplicate;` `    ``// Making an iterator for set``    ``set<``int``>::iterator it;` `    ``// Traverse the given array arr[]``    ``while` `(i != N) {``        ``cout << i << ``" # "` `;``        ``// Check if the current element``        ``// is not same as the element at``        ``// index arr[i] - 1, then swap``        ``if` `(arr[i] != arr[arr[i] - 1]) {``            ``swap(arr[i], arr[arr[i] - 1]);``        ``}` `        ``// Otherwise, increment the index``        ``else` `{``            ``i++;``        ``}``    ``}` `    ``// Traverse the array again``    ``for` `(i = 0; i < N; i++) {` `        ``// If the element is not at its``        ``// correct position``        ``if` `(arr[i] != i + 1) {` `            ``// Stores the missing and the``            ``// duplicate elements``            ``missing.push_back(i + 1);``            ``duplicate.insert(arr[i]);``        ``}``    ``}` `    ``// Print the Missing Number``    ``cout << ``"Missing Numbers: "``;``    ``for` `(``auto``& it : missing)``        ``cout << it << ``' '``;` `    ``// Print the Duplicate Number``    ``cout << ``"\nDuplicate Numbers: "``;``    ``for` `(``auto``& it : duplicate)``        ``cout << it << ``' '``;``}` `// Driver code``int` `main()``{``    ``int` `arr[] = { 1, 2, 2, 2, 4, 5, 7 };``    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr[0]);``    ``findElements(arr, N);` `    ``return` `0;``}`

## Java

 `// Java program for the above approach``import` `java.util.ArrayList;``import` `java.util.HashSet;` `class` `GFG {` `    ``// Function to find the duplicate and``    ``// the missing elements over the range``    ``// [1, N]``    ``static` `void` `findElements(``int` `arr[], ``int` `N) {``        ``int` `i = ``0``;` `        ``// Stores the missing and duplicate``        ``// numbers in the array arr[]``        ``ArrayList missing = ``new` `ArrayList();``        ``HashSet duplicate = ``new` `HashSet();` `        ``// Traverse the given array arr[]``        ``while` `(i != N) {``            ``// Check if the current element``            ``// is not same as the element at``            ``// index arr[i] - 1, then swap``            ``if` `(arr[i] != arr[arr[i] - ``1``]) {``                ``int` `temp = arr[i];``                ``arr[i] = arr[arr[i] - ``1``];``                ``arr[temp - ``1``] = temp;``            ``}` `            ``// Otherwise, increment the index``            ``else` `{``                ``i++;``            ``}``        ``}` `        ``// Traverse the array again``        ``for` `(i = ``0``; i < N; i++) {` `            ``// If the element is not at its``            ``// correct position``            ``if` `(arr[i] != i + ``1``) {` `                ``// Stores the missing and the``                ``// duplicate elements``                ``missing.add(i + ``1``);``                ``duplicate.add(arr[i]);``            ``}``        ``}` `        ``// Print the Missing Number``        ``System.out.print(``"Missing Numbers: "``);``        ``for` `(Integer itr : missing)``            ``System.out.print(itr + ``" "``);` `        ``// Print the Duplicate Number``        ``System.out.print(``"\nDuplicate Numbers: "``);``        ``for` `(Integer itr : duplicate)``            ``System.out.print(itr + ``" "``);``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String args[]) {``        ``int` `arr[] = { ``1``, ``2``, ``2``, ``2``, ``4``, ``5``, ``7` `};``        ``int` `N = arr.length;``        ``findElements(arr, N);``    ``}` `}` `// This code is contributed by gfgking.`

## Python3

 `# Python3 program for the above approach` `# Function to find the duplicate and``# the missing elements over the range``# [1, N]``def` `findElements(arr, N) :``    ``i ``=` `0``;` `    ``# Stores the missing and duplicate``    ``# numbers in the array arr[]``    ``missing ``=` `[];``    ``duplicate ``=` `set``();` `    ``# Traverse the given array arr[]``    ``while` `(i !``=` `N) :``        ` `        ``# Check if the current element``        ``# is not same as the element at``        ``# index arr[i] - 1, then swap``        ``if` `(arr[i] !``=` `arr[arr[i] ``-` `1``]) :``            ` `            ``t ``=` `arr[i]``            ``arr[i] ``=` `arr[arr[i] ``-` `1``]``            ``arr[t ``-` `1``]  ``=` `t``            ` `        ``# Otherwise, increment the index``        ``else` `:``            ``i ``+``=` `1``;` `    ``# Traverse the array again``    ``for` `i ``in` `range``(N) :``        ` `        ``# If the element is not at its``        ``# correct position``        ``if` `(arr[i] !``=` `i ``+` `1``) :` `            ``# Stores the missing and the``            ``# duplicate elements``            ``missing.append(i ``+` `1``);``            ``duplicate.add(arr[i]);` `    ``# Print the Missing Number``    ``print``(``"Missing Numbers: "``,end``=``"");``    ` `    ``for` `it ``in` `missing:``        ``print``(it,end``=``" "``);` `    ``# Print the Duplicate Number``    ``print``(``"\nDuplicate Numbers: "``,end``=``"");``    ` `    ``for` `it ``in` `list``(duplicate) :``        ``print``(it, end``=``' '``);` `# Driver code``if` `__name__ ``=``=` `"__main__"` `:` `    ``arr ``=` `[ ``1``, ``2``, ``2``, ``2``, ``4``, ``5``, ``7` `];``    ``N ``=` `len``(arr);``    ``findElements(arr, N);` `    ``# This code is contributed by AnkThon`

## C#

 `// C# program for the above approach``using` `System;``using` `System.Collections.Generic;``class` `GFG {` `  ``// Function to find the duplicate and``  ``// the missing elements over the range``  ``// [1, N]``  ``static` `void` `findElements(``int``[] arr, ``int` `N)``  ``{``    ``int` `i = 0;` `    ``// Stores the missing and duplicate``    ``// numbers in the array arr[]``    ``List<``int``> missing = ``new` `List<``int``>();``    ``HashSet<``int``> duplicate = ``new` `HashSet<``int``>();` `    ``// Traverse the given array arr[]``    ``while` `(i != N) {``      ``// Check if the current element``      ``// is not same as the element at``      ``// index arr[i] - 1, then swap``      ``if` `(arr[i] != arr[arr[i] - 1]) {``        ``int` `temp = arr[i];``        ``arr[i] = arr[arr[i] - 1];``        ``arr[temp - 1] = temp;``      ``}` `      ``// Otherwise, increment the index``      ``else` `{``        ``i++;``      ``}``    ``}` `    ``// Traverse the array again``    ``for` `(i = 0; i < N; i++) {` `      ``// If the element is not at its``      ``// correct position``      ``if` `(arr[i] != i + 1) {` `        ``// Stores the missing and the``        ``// duplicate elements``        ``missing.Add(i + 1);``        ``duplicate.Add(arr[i]);``      ``}``    ``}` `    ``// Print the Missing Number``    ``Console.Write(``"Missing Numbers: "``);``    ``foreach``(``int` `itr ``in` `missing)``      ``Console.Write(itr + ``" "``);` `    ``// Print the Duplicate Number``    ``Console.Write(``"\nDuplicate Numbers: "``);``    ``foreach``(``int` `itr ``in` `duplicate)``      ``Console.Write(itr + ``" "``);``  ``}` `  ``// Driver code``  ``public` `static` `void` `Main()``  ``{``    ``int``[] arr = { 1, 2, 2, 2, 4, 5, 7 };``    ``int` `N = arr.Length;``    ``findElements(arr, N);``  ``}``}` `// This code is contributed by ukasp.`

## Javascript

 ``
Output:
```Missing Numbers: 3 6
Duplicate Numbers: 2```

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

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