# Elements to be added so that all elements of a range are present in array

• Difficulty Level : Basic
• Last Updated : 21 Jul, 2022

Given an array of size N. Let A and B be the minimum and maximum in the array respectively. Task is to find how many number should be added to the given array such that all the element in the range [A, B] occur at-least once in the array.
Examples:

```Input : arr[] = {4, 5, 3, 8, 6}
Output : 1
Only 7 to be added in the list.

Input : arr[] = {2, 1, 3}
Output : 0```
Recommended Practice

Method 1 (Sorting):

1. Sort the array.
2. Compare arr[i] == arr[i+1]-1 or not. If not, update count = arr[i+1]-arr[i]-1.
3. Return count.

Implementation:

## C++

 `// C++ program for above implementation``#include ``using` `namespace` `std;` `// Function to count numbers to be added``int` `countNum(``int` `arr[], ``int` `n)``{``    ``int` `count = 0;` `    ``// Sort the array``    ``sort(arr, arr + n);` `    ``// Check if elements are consecutive``    ``//  or not. If not, update count``    ``for` `(``int` `i = 0; i < n - 1; i++)``        ``if` `(arr[i] != arr[i+1] &&``            ``arr[i] != arr[i + 1] - 1)``            ``count += arr[i + 1] - arr[i] - 1;` `    ``return` `count;``}` `// Drivers code``int` `main()``{``    ``int` `arr[] = { 3, 5, 8, 6 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);``    ``cout << countNum(arr, n) << endl;``    ``return` `0;``}`

## Java

 `// java program for above implementation``import` `java.io.*;``import` `java.util.*;` `public` `class` `GFG {``    ` `    ``// Function to count numbers to be added``    ``static` `int` `countNum(``int` `[]arr, ``int` `n)``    ``{``        ``int` `count = ``0``;``    ` `        ``// Sort the array``        ``Arrays.sort(arr);``    ` `        ``// Check if elements are consecutive``        ``// or not. If not, update count``        ``for` `(``int` `i = ``0``; i < n - ``1``; i++)``            ``if` `(arr[i] != arr[i+``1``] &&``                ``arr[i] != arr[i + ``1``] - ``1``)``                ``count += arr[i + ``1``] - arr[i] - ``1``;``    ` `        ``return` `count;``    ``}``    ` `    ``// Drivers code``    ``static` `public` `void` `main (String[] args)``    ``{``        ` `        ``int` `[]arr = { ``3``, ``5``, ``8``, ``6` `};``        ``int` `n = arr.length;``        ` `        ``System.out.println(countNum(arr, n));``    ``}``}` `// This code is contributed by vt_m.`

## Python3

 `# python program for above implementation` `# Function to count numbers to be added``def` `countNum(arr, n):``    ` `    ``count ``=` `0` `    ``# Sort the array``    ``arr.sort()` `    ``# Check if elements are consecutive``    ``# or not. If not, update count``    ``for` `i ``in` `range``(``0``, n``-``1``):``        ``if` `(arr[i] !``=` `arr[i``+``1``] ``and``            ``arr[i] !``=` `arr[i ``+` `1``] ``-` `1``):``            ``count ``+``=` `arr[i ``+` `1``] ``-` `arr[i] ``-` `1``;` `    ``return` `count` `# Drivers code``arr ``=` `[ ``3``, ``5``, ``8``, ``6` `]``n ``=` `len``(arr)``print``(countNum(arr, n))` `# This code is contributed by Sam007`

## C#

 `// C# program for above implementation``using` `System;` `public` `class` `GFG {``    ` `    ``// Function to count numbers to be added``    ``static` `int` `countNum(``int` `[]arr, ``int` `n)``    ``{``        ``int` `count = 0;``    ` `        ``// Sort the array``        ``Array.Sort(arr);``    ` `        ``// Check if elements are consecutive``        ``// or not. If not, update count``        ``for` `(``int` `i = 0; i < n - 1; i++)``            ``if` `(arr[i] != arr[i+1] &&``                ``arr[i] != arr[i + 1] - 1)``                ``count += arr[i + 1] - arr[i] - 1;``    ` `        ``return` `count;``    ``}``    ` `    ``// Drivers code``    ``static` `public` `void` `Main ()``    ``{``        ` `        ``int` `[]arr = { 3, 5, 8, 6 };``        ``int` `n = arr.Length;``        ` `        ``Console.WriteLine(countNum(arr, n));``    ``}``}` `// This code is contributed by vt_m.`

## PHP

 ``

## Javascript

 ``

Output

`2`

Time Complexity: O(n log n)
Auxiliary Space: O(1)

Method 2 (Use Hashing):

1. Maintain a hash of array elements.
2. Store minimum and maximum element.
3. Traverse from minimum to maximum element in hash
And count if element is not in hash.
4. Return count.

Implementation:

## C++

 `// C++ program for above implementation``#include ``using` `namespace` `std;` `// Function to count numbers to be added``int` `countNum(``int` `arr[], ``int` `n)``{``    ``unordered_set<``int``> s;``    ``int` `count = 0, maxm = INT_MIN, minm = INT_MAX;` `    ``// Make a hash of elements``    ``// and store minimum and maximum element``    ``for` `(``int` `i = 0; i < n; i++) {``        ``s.insert(arr[i]);``        ``if` `(arr[i] < minm)``            ``minm = arr[i];``        ``if` `(arr[i] > maxm)``            ``maxm = arr[i];``    ``}` `    ``// Traverse all elements from minimum``    ``// to maximum and count if it is not``    ``// in the hash``    ``for` `(``int` `i = minm; i <= maxm; i++)``        ``if` `(s.find(arr[i]) == s.end())``            ``count++;``    ``return` `count;``}` `// Drivers code``int` `main()``{``    ``int` `arr[] = { 3, 5, 8, 6 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);``    ``cout << countNum(arr, n) << endl;``    ``return` `0;``}`

## Java

 `// Java implementation of the approach``import` `java.util.HashSet;` `class` `GFG``{` `// Function to count numbers to be added``static` `int` `countNum(``int` `arr[], ``int` `n)``{``    ``HashSet s = ``new` `HashSet<>();``    ``int` `count = ``0``,``        ``maxm = Integer.MIN_VALUE,``        ``minm = Integer.MAX_VALUE;` `    ``// Make a hash of elements``    ``// and store minimum and maximum element``    ``for` `(``int` `i = ``0``; i < n; i++)``    ``{``        ``s.add(arr[i]);``        ``if` `(arr[i] < minm)``            ``minm = arr[i];``        ``if` `(arr[i] > maxm)``            ``maxm = arr[i];``    ``}` `    ``// Traverse all elements from minimum``    ``// to maximum and count if it is not``    ``// in the hash``    ``for` `(``int` `i = minm; i <= maxm; i++)``        ``if` `(!s.contains(i))``            ``count++;``    ``return` `count;``}` `// Drivers code``public` `static` `void` `main(String[] args)``{``    ``int` `arr[] = { ``3``, ``5``, ``8``, ``6` `};``    ``int` `n = arr.length;``    ``System.out.println(countNum(arr, n));``}``}` `// This code is contributed by Rajput-Ji`

## Python3

 `# Function to count numbers to be added``def` `countNum(arr, n):` `    ``s ``=` `dict``()``    ``count, maxm, minm ``=` `0``, ``-``10``*``*``9``, ``10``*``*``9` `    ``# Make a hash of elements and store``    ``# minimum and maximum element``    ``for` `i ``in` `range``(n):``        ``s[arr[i]] ``=` `1``        ``if` `(arr[i] < minm):``            ``minm ``=` `arr[i]``        ``if` `(arr[i] > maxm):``            ``maxm ``=` `arr[i]``    ` `    ``# Traverse all elements from minimum``    ``# to maximum and count if it is not``    ``# in the hash``    ``for` `i ``in` `range``(minm, maxm ``+` `1``):``        ``if` `i ``not` `in` `s.keys():``            ``count ``+``=` `1``    ``return` `count` `# Driver code``arr ``=` `[``3``, ``5``, ``8``, ``6` `]``n ``=` `len``(arr)``print``(countNum(arr, n))``    ` `# This code is contributed by mohit kumar`

## C#

 `// C# implementation of the approach``using` `System;``using` `System.Collections.Generic;` `class` `GFG``{` `// Function to count numbers to be added``static` `int` `countNum(``int` `[]arr, ``int` `n)``{``    ``HashSet<``int``> s = ``new` `HashSet<``int``>();``    ``int` `count = 0,``        ``maxm = ``int``.MinValue,``        ``minm = ``int``.MaxValue;` `    ``// Make a hash of elements``    ``// and store minimum and maximum element``    ``for` `(``int` `i = 0; i < n; i++)``    ``{``        ``s.Add(arr[i]);``        ``if` `(arr[i] < minm)``            ``minm = arr[i];``        ``if` `(arr[i] > maxm)``            ``maxm = arr[i];``    ``}` `    ``// Traverse all elements from minimum``    ``// to maximum and count if it is not``    ``// in the hash``    ``for` `(``int` `i = minm; i <= maxm; i++)``        ``if` `(!s.Contains(i))``            ``count++;``    ``return` `count;``}` `// Drivers code``public` `static` `void` `Main(String[] args)``{``    ``int` `[]arr = { 3, 5, 8, 6 };``    ``int` `n = arr.Length;``    ``Console.WriteLine(countNum(arr, n));``}``}` `// This code is contributed by Rajput-Ji`

## Javascript

 ``

Output

`5`

Time Complexity: O(n + max – min + 1)
Auxiliary Space: O(1)

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