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 <bits/stdc++.h>
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<n;++i){
        if(freq[a[i]])
            freq[a[i]] += 1;
        else
            freq[a[i]] = 1;
    }

    // Finding max sum of adjacent indices
    int ans = 0, key;

    map<int,int>:: 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<<maxCount(n,arr);

    return 0;
}


// This code is contributed by Sanjit_Prasad

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<Integer, Integer> 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

<script>

// JavaScript program to find maximum number of
// elements such that their absolute
// difference is less than or equal to 1

// function to return maximum number of elements
function maxCount(n,a)
{
    // Counting frequencies of elements
    var freq = new Map();

    for(var i=0;i<n;++i){
        if(freq.has(a[i]))
            freq.set(a[i], freq.get(a[i])+1)
        else
            freq.set(a[i], 1)
    }

    // Finding max sum of adjacent indices
    var ans = 0, key;

    freq.forEach((value, key) => {

        if(freq.has(key+1))
            ans=Math.max(ans,freq.get(key)+freq.get(key+1));
    });

    return ans;
}

// Driver Code

var n = 5;
var arr =  [2, 2, 3, 4, 5];

// function call to print required answer
document.write( maxCount(n,arr));

</script>

Output

3

Complexity Analysis:

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