# Find maximum number of elements such that their absolute difference is less than or equal to 1

Given an array of n elements, find the maximum number of elements to select from the array such that the absolute difference between any two of the chosen elements is less than or equal to 1.
Examples

```Input : arr[] = {1, 2, 3}
Output : 2
We can either take 1, 2 or 2, 3.
Both will have the count 2 so maximum count is 2

Input : arr[] = {2, 2, 3, 4, 5}
Output : 3
The sequence with maximum count is 2, 2, 3.```

The absolute difference of 0 or 1 means that the numbers chosen can be of type x and x+1. Therefore, the idea is to store frequencies of array elements. So, the task now reduces to find the maximum sum of any two consecutive elements.

Below is the implementation of the above approach:

## C++

 `// CPP program to find maximum number of` `// elements such that their absolute` `// difference is less than or equal to 1` `#include ` `using` `namespace` `std;`   `// function to return maximum number of elements` `int` `maxCount(``int` `n,``int` `a[])` `{` `    ``// Counting frequencies of elements` `    ``map<``int``,``int``> freq;`   `    ``for``(``int` `i=0;i:: iterator it=freq.begin();`   `    ``while``(it!=freq.end())` `    ``{` `        ``key = it->first;`   `        ``// increment the iterator` `        ``++it;`   `        ``if``(freq[key+1]!=0)` `            ``ans=max(ans,freq[key]+freq[key+1]);`   `    ``}`   `    ``return` `ans;` `}`   `// Driver Code` `int` `main(){` `    ``int` `n = 5;` `    ``int` `arr[] = {2, 2, 3, 4, 5};`   `    ``// function call to print required answer` `    ``cout<

## Java

 `// Java program to find the maximum number ` `// of elements such that their absolute ` `// difference is less than or equal to 1 ` `import` `java.util.HashMap;` `import` `java.util.Map;` `import` `java.lang.Math;`   `class` `GfG` `{`   `    ``// function to return the maximum number of elements ` `    ``static` `int` `maxCount(``int` `n,``int` `a[]) ` `    ``{ ` `        ``// Counting frequencies of elements ` `        ``HashMap freq = ``new` `HashMap<>(); ` `    `  `        ``for``(``int` `i = ``0``; i < n; ++i)` `        ``{ ` `            ``if``(freq.containsKey(a[i])) ` `                ``freq.put(a[i], freq.get(a[i]) + ``1``); ` `            ``else` `                ``freq.put(a[i], ``1``); ` `        ``} ` `    `  `        ``// Finding max sum of adjacent indices ` `        ``int` `ans = ``0``; ` `    `  `        ``for` `(Integer key : freq.keySet()) ` `        ``{ ` `            ``if``(freq.containsKey(key+``1``)) ` `                ``ans = Math.max(ans, freq.get(key) + freq.get(key+``1``)); ` `        ``} ` `    `  `        ``return` `ans; ` `    ``} `   `    ``// Driver code` `    ``public` `static` `void` `main(String []args)` `    ``{` `        `  `        ``int` `n = ``5``; ` `        ``int` `arr[] = {``2``, ``2``, ``3``, ``4``, ``5``}; ` `    `  `        ``// function call to print required answer ` `        ``System.out.println(maxCount(n,arr));` `    ``}` `}`   `// This code is contributed by Rituraj Jain `

## Python3

 `# Python program to find maximum number of ` `# elements such that their absolute` `# difference is less than or equal to 1`   `def` `maxCount(a):`   `    ``# Counting frequencies of elements` `    ``freq ``=` `{}` `    ``for` `i ``in` `range``(n):` `        ``if` `(a[i] ``in` `freq): ` `            ``freq[a[i]] ``+``=` `1` `        ``else``: ` `            ``freq[a[i]] ``=` `1` `        `  `    `  `    ``# Finding max sum of adjacent indices    ` `    ``ans ``=` `0` `    ``for` `key, value ``in` `freq.items(): ` `        ``if` `(key``+``1` `in` `freq) :    ` `            ``ans ``=` `max``(ans, freq[key] ``+` `freq[key ``+` `1``]) ` `    `  `    ``return` `ans` `    `  `# Driver Code ` `n ``=` `5` `arr ``=` `[``2``, ``2``, ``3``, ``4``, ``5``]`   `print``(maxCount(arr))`

## C#

 `// C# program to find the maximum number ` `// of elements such that their absolute ` `// difference is less than or equal to 1 ` `using` `System;` `using` `System.Collections.Generic;`   `class` `GfG ` `{ `   `    ``// function to return the maximum number of elements ` `    ``static` `int` `maxCount(``int` `n,``int` `[]a) ` `    ``{ ` `        ``// Counting frequencies of elements ` `        ``Dictionary<``int``,``int``> mp = ``new` `Dictionary<``int``,``int``>();` `        `  `        ``// Increase the frequency of elements` `        ``for` `(``int` `i = 0 ; i < n; i++)` `        ``{` `            ``if``(mp.ContainsKey(a[i]))` `            ``{` `                ``var` `val = mp[a[i]];` `                ``mp.Remove(a[i]);` `                ``mp.Add(a[i], val + 1); ` `            ``}` `            ``else` `            ``{` `                ``mp.Add(a[i], 1);` `            ``}` `        ``} ` `    `  `        ``// Finding max sum of adjacent indices ` `        ``int` `ans = 0; ` `    `  `        ``foreach``(KeyValuePair<``int``, ``int``> e ``in` `mp)` `        ``{ ` `            ``if``(mp.ContainsKey(e.Key+1)) ` `                ``ans = Math.Max(ans, mp[e.Key] + mp[e.Key+1]); ` `        ``} ` `    `  `        ``return` `ans; ` `    ``} `   `    ``// Driver code ` `    ``public` `static` `void` `Main(String []args) ` `    ``{ ` `        `  `        ``int` `n = 5; ` `        ``int` `[]arr = {2, 2, 3, 4, 5}; ` `    `  `        ``// function call to print required answer ` `        ``Console.WriteLine(maxCount(n,arr)); ` `    ``} ` `} `   `/* This code is contributed by PrinciRaj1992 */`

## Javascript

 ``

Output

`3`

Complexity Analysis:

• Time Complexity: O(n * log(n))
• Auxiliary Space: O(n)

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