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

• Difficulty Level : Easy
• Last Updated : 25 May, 2021

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 elementsint maxCount(int n,int a[]){    // Counting frequencies of elements    map 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 Codeint 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 1import 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 Coden = 5arr = [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 1using 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 mp = new Dictionary();                 // 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 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

Time Complexity: O(n * log(n))

Auxiliary Space: O(n)

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