Given an array** arr[]** of size **N**, the task is to count the maximum number of elements that can be removed from the given array without changing the** MEX **of the original array.

The

MEXis the smallest positive integer that is not present in the array.

**Examples:**

Input:arr[] = {2, 3, 5, 1, 6}Output:2Explanation:

The smallest positive integer which is not present in the array is 4.

Hence, MEX of the given array is 4.

Therefore, 5 and 6 can be removed without changing the MEX of the array.

Input:arr[] = {12, 4, 6, 1, 7, 2}Output:4Explanation:

The smallest positive integer which is not present in the array is 3.

Hence, MEX of the given array is 3.

Therefore, 4, 6, 7 and 12 can be removed without changing the MEX of the array.

**Naive Approach:** The simplest approach is to sort the array and then traverse the array from** i = 0 **while **arr[i]** equals to **i + 1**. After that, print the answer as **(N – i)** which is the maximum number of elements that can be removed from the given array without changing its **MEX**.

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

**Efficient Approach:** To optimize the above approach, the idea is to use Hashing. Observe that, the maximum number of elements that can be removed is the number of elements that are greater than the **MEX**. Follow the steps below to solve the problem:

- Initialize an array
**hash[]**of length**N+1**where**hash[i]**will be**1**if element**i**is present in the given array otherwise**hash[i] = 0.** - Initialize a variable
**mex**with**N + 1**to store**MEX**of the given array. - Traverse array
**hash[]**over the range**[1, N]**, and if for any index**hash[i]**equals to**0**, update**mex**as**mex = i**and break out of the loop. - Print
**N – (mex – 1)**as the maximum number of elements that can be removed from the given array.

Below is the implementation of the above approach:

## C++

`// C++ program for the above approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to find the maximum number` `// of elements that can be removed` `void` `countRemovableElem(` ` ` `int` `arr[], ` `int` `N)` `{` ` ` `// Initialize hash[] with 0s` ` ` `int` `hash[N + 1] = { 0 };` ` ` `// Initialize MEX` ` ` `int` `mex = N + 1;` ` ` `// Set hash[i] = 1, if i is` ` ` `// present in arr[]` ` ` `for` `(` `int` `i = 0; i < N; i++) {` ` ` `if` `(arr[i] <= N)` ` ` `hash[arr[i]] = 1;` ` ` `}` ` ` `// Find MEX from the hash` ` ` `for` `(` `int` `i = 1; i <= N; i++) {` ` ` `if` `(hash[i] == 0) {` ` ` `mex = i;` ` ` `break` `;` ` ` `}` ` ` `}` ` ` `// Print the maximum numbers` ` ` `// that can be removed` ` ` `cout << N - (mex - 1);` `}` `// Driver Code` `int` `main()` `{` ` ` `// Given array` ` ` `int` `arr[] = { 2, 3, 5, 1, 6 };` ` ` `// Size of the array` ` ` `int` `N = ` `sizeof` `(arr) / ` `sizeof` `(arr[0]);` ` ` `// Function Call` ` ` `countRemovableElem(arr, N);` ` ` `return` `0;` `}` |

## Java

`// Java program for the above approach` `import` `java.io.*;` `import` `java.util.*;` `class` `GFG{` `// Function to find the maximum number` `// of elements that can be removed` `static` `void` `countRemovableElem(` `int` `[] arr, ` `int` `N)` `{` ` ` ` ` `// Initialize hash[] with 0s` ` ` `int` `[] hash = ` `new` `int` `[N + ` `1` `];` ` ` `Arrays.fill(hash, ` `0` `);` ` ` `// Initialize MEX` ` ` `int` `mex = N + ` `1` `;` ` ` `// Set hash[i] = 1, if i is` ` ` `// present in arr[]` ` ` `for` `(` `int` `i = ` `0` `; i < N; i++)` ` ` `{` ` ` `if` `(arr[i] <= N)` ` ` `hash[arr[i]] = ` `1` `;` ` ` `}` ` ` `// Find MEX from the hash` ` ` `for` `(` `int` `i = ` `1` `; i <= N; i++)` ` ` `{` ` ` `if` `(hash[i] == ` `0` `)` ` ` `{` ` ` `mex = i;` ` ` `break` `;` ` ` `}` ` ` `}` ` ` `// Print the maximum numbers` ` ` `// that can be removed` ` ` `System.out.println(N - (mex - ` `1` `));` `}` `// Driver Code` `public` `static` `void` `main(String[] args)` `{` ` ` ` ` `// Given array` ` ` `int` `[] arr = { ` `2` `, ` `3` `, ` `5` `, ` `1` `, ` `6` `};` ` ` `// Size of the array` ` ` `int` `N = arr.length;` ` ` `// Function Call` ` ` `countRemovableElem(arr, N);` `}` `}` `// This code is contributed by akhilsaini` |

## Python3

`# Pyhton3 program for the above approach` `# Function to find the maximum number` `# of elements that can be removed` `def` `countRemovableElem(arr, N):` ` ` `# Initialize hash[] with 0s` ` ` `hash` `=` `[` `0` `] ` `*` `(N ` `+` `1` `)` ` ` `# Initialize MEX` ` ` `mex ` `=` `N ` `+` `1` ` ` `# Set hash[i] = 1, if i is` ` ` `# present in arr[]` ` ` `for` `i ` `in` `range` `(` `0` `, N):` ` ` `if` `(arr[i] <` `=` `N):` ` ` `hash` `[arr[i]] ` `=` `1` ` ` `# Find MEX from the hash` ` ` `for` `i ` `in` `range` `(` `1` `, N ` `+` `1` `):` ` ` `if` `(` `hash` `[i] ` `=` `=` `0` `):` ` ` `mex ` `=` `i` ` ` `break` ` ` `# Print the maximum numbers` ` ` `# that can be removed` ` ` `print` `(N ` `-` `(mex ` `-` `1` `))` `# Driver Code` `if` `__name__ ` `=` `=` `'__main__'` `:` ` ` ` ` `# Given array` ` ` `arr ` `=` `[ ` `2` `, ` `3` `, ` `5` `, ` `1` `, ` `6` `]` ` ` `# Size of the array` ` ` `N ` `=` `len` `(arr)` ` ` `# Function Call` ` ` `countRemovableElem(arr, N)` `# This code is contributed by akhilsaini` |

## C#

`// C# program for the above approach` `using` `System;` `using` `System.Collections.Generic;` `class` `GFG{` `// Function to find the maximum number` `// of elements that can be removed` `static` `void` `countRemovableElem(` `int` `[] arr, ` `int` `N)` `{` ` ` ` ` `// Initialize hash[] with 0s` ` ` `int` `[] hash = ` `new` `int` `[N + 1];` ` ` `Array.Fill(hash, 0);` ` ` `// Initialize MEX` ` ` `int` `mex = N + 1;` ` ` `// Set hash[i] = 1, if i is` ` ` `// present in arr[]` ` ` `for` `(` `int` `i = 0; i < N; i++)` ` ` `{` ` ` `if` `(arr[i] <= N)` ` ` `hash[arr[i]] = 1;` ` ` `}` ` ` `// Find MEX from the hash` ` ` `for` `(` `int` `i = 1; i <= N; i++)` ` ` `{` ` ` `if` `(hash[i] == 0)` ` ` `{` ` ` `mex = i;` ` ` `break` `;` ` ` `}` ` ` `}` ` ` `// Print the maximum numbers` ` ` `// that can be removed` ` ` `Console.WriteLine(N - (mex - 1));` `}` `// Driver Code` `public` `static` `void` `Main()` `{` ` ` ` ` `// Given array` ` ` `int` `[] arr = { 2, 3, 5, 1, 6 };` ` ` `// Size of the array` ` ` `int` `N = arr.Length;` ` ` `// Function Call` ` ` `countRemovableElem(arr, N);` `}` `}` `// This code is contributed by akhilsaini` |

## Javascript

`<script>` `// Javascript program to implement` `// the above approach` `// Function to find the maximum number` `// of elements that can be removed` `function` `countRemovableElem(arr, N)` `{` ` ` ` ` `// Initialize hash[] with 0s` ` ` `let hash = [];` ` ` `for` `(let i = 0; i < N; i++)` ` ` `{` ` ` `hash[i] = 0;` ` ` `}` ` ` ` ` `// Initialize MEX` ` ` `let mex = N + 1;` ` ` ` ` `// Set hash[i] = 1, if i is` ` ` `// present in arr[]` ` ` `for` `(let i = 0; i < N; i++)` ` ` `{` ` ` `if` `(arr[i] <= N)` ` ` `hash[arr[i]] = 1;` ` ` `}` ` ` ` ` `// Find MEX from the hash` ` ` `for` `(let i = 1; i <= N; i++)` ` ` `{` ` ` `if` `(hash[i] == 0)` ` ` `{` ` ` `mex = i;` ` ` `break` `;` ` ` `}` ` ` `}` ` ` ` ` `// Prlet the maximum numbers` ` ` `// that can be removed` ` ` `document.write(N - (mex - 1));` `}` ` ` `// Driver Code` ` ` ` ` `// Given array` ` ` `let arr = [2, 3, 5, 1, 6 ];` ` ` ` ` `// Size of the array` ` ` `let N = arr.length;` ` ` ` ` `// Function Call` ` ` `countRemovableElem(arr, N);` ` ` `</script>` |

**Output:**

2

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

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