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 elementsint 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 Codeint 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 1def 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 = 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 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 elementsfunction 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 Codevar n = 5;var arr = [2, 2, 3, 4, 5];// function call to print required answerdocument.write( maxCount(n,arr));</script> |
Output
3
Complexity Analysis:
- Time Complexity: O(n * log(n))
- Auxiliary Space: O(n)
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