Maximize count of unique array elements by incrementing array elements by K

Last Updated : 01 Feb, 2021

Given an array arr[] consisting of N integers and an integer K, the task is to find the maximum number of unique elements possible by increasing any array element by K only once.

Examples:

Input: arr[] = {0, 2, 4, 3, 4}, K = 1
Output: 5
Explanation:
Increase arr[2] ( = 4) by K ( = 1). Therefore, new array is {0, 2, 4, 3, 5} which has 5 unique elements.

Input: arr[] = {2, 3, 2, 4, 5, 5, 7, 4}, K = 2
Output: 7
Explanation:
Increase 4 by 2 = 6.
Increase element 7 by 2 = 9
Increase 5 by 2 = 7.
The new array is {2, 3, 2, 4, 5, 7, 9, 6} which contains 7 unique elements.

Approach: The idea to solve this problem is to store the frequency of elements of the array in a Map and change the array elements accordingly to get unique elements in the array after incrementing any values by K. Follow the steps below to solve the problem:

Traverse the array and store the frequency of every array element in a Map.
Traverse the Map and perform the following steps:
- If the count of the current element is 1, then don’t change the array element.
- If the count of the current element is more than 1, then increase the current element by K and increase the count of (current_element + K) by 1 in the Map.
After completing the above steps, print the size of the Map as the maximum number of unique elements in the array.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the maximum unique
// elements in array after incrementing
// any element by K
void maxDifferent(int arr[], int N, int K)
{
    // Stores the count of element
    // in array
    map<int, int> M;
 
    // Traverse the array
    for (int i = 0; i < N; i++) {
 
        // Increase the counter of
        // the array element by 1
        M[arr[i]]++;
    }
 
    // Traverse the map
    for (auto it = M.begin();
         it != M.end(); it++) {
 
        // Extract the current element
        int current_element = it->first;
 
        // Number of times the current
        // element is present in array
        int count = it->second;
 
        // If element is present only
        // once, then do not change it
        if (count == 1)
            continue;
 
        // If the count > 1 then change
        // one of the same current
        // elements to (current_element + K)
        // and increase its count by 1
        M[current_element + K]++;
    }
 
    // The size of the map is the
    // required answer
    cout << M.size();
}
 
// Driver Code
int main()
{
    int arr[] = { 2, 3, 2, 4, 5, 5, 7, 4 };
    int K = 2;
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call
    maxDifferent(arr, N, K);
 
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.util.*;
 
class GFG{
 
// Function to find the maximum unique
// elements in array after incrementing
// any element by K
static void maxDifferent(int arr[], int N, int K)
{
     
    // Stores the count of element
    // in array
    HashMap<Integer, 
            Integer> M = new HashMap<Integer, 
                                     Integer>();
 
    // Traverse the array
    for(int i = 0; i < N; i++) 
    {
         
        // Increase the counter of
        // the array element by 1
        Integer count = M.get(arr[i]);
         
        if (count == null)
        {
            M.put(arr[i], 1);
        }
        else
        {
            M.put(arr[i], count + 1);
        }
    }
 
    // Iterator itr = M.entrySet().iterator();
    Iterator<Map.Entry<Integer, 
                       Integer>> itr = M.entrySet().iterator();
 
    int[] ar1 = new int[N];
 
    // Traverse the map
    while (itr.hasNext()) 
    {
        Map.Entry<Integer, Integer> Element = itr.next();
 
        // Extract the current element
        int current_element = (int)Element.getKey();
 
        // Number of times the current
        // element is present in array
        int count = (int)Element.getValue();
 
        // If element is present only
        // once, then do not change it
        if (count == 1)
            continue;
 
        // If the count > 1 then change
        // one of the same current
        // elements to (current_element + K)
        // and increase its count by 1
        ar1[current_element + K]++;
    }
 
    for(int i = 0; i < N; i++) 
    {
        if (ar1[i] >= 0)
        {
            Integer count = M.get(ar1[i]);
             
              if (count == null)
              {
                  M.put(ar1[i], 1);
            }
            else
            {
                M.put(ar1[i], count + 1);
            }
        }
    }
 
    // The size of the map is the
    // required answer
    System.out.println(M.size());
}
 
// Driver Code
public static void main(String[] args)
{
    int arr[] = { 2, 3, 2, 4, 5, 5, 7, 4 };
    int K = 2;
    int N = arr.length;
     
    // Function Call
    maxDifferent(arr, N, K);
}
}
 
// This code is contributed by Dharanendra L V

Python3

# Python3 program for the above approach
 
# Function to find the maximum unique
# elements in array after incrementing
# any element by K
def maxDifferent(arr, N, K):
     
    # Stores the count of element
    # in array
    M = {}
 
    # Traverse the array
    for i in range(N):
 
        # Increase the counter of
        # the array element by 1
        M[arr[i]] = M.get(arr[i], 0) + 1
 
    # Traverse the map
    for it in list(M.keys()):
 
        # Extract the current element
        current_element = it
 
        # Number of times the current
        # element is present in array
        count = M[it]
 
        # If element is present only
        # once, then do not change it
        if (count == 1):
            continue
 
        # If the count > 1 then change
        # one of the same current
        # elements to (current_element + K)
        # and increase its count by 1
        M[current_element + K] = M.get(current_element, 0) + 1
 
    # The size of the map is the
    # required answer
    print(len(M))
 
# Driver Code
if __name__ == '__main__':
    arr=[2, 3, 2, 4, 5, 5, 7, 4]
    K = 2
    N = len(arr)
 
    # Function Call
    maxDifferent(arr, N, K)
 
    # This code is contributed by mohit kumar 29

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
 
public class GFG
{
 
// Function to find the maximum unique
// elements in array after incrementing
// any element by K
static void maxDifferent(int []arr, int N, int K)
{
     
    // Stores the count of element
    // in array
    Dictionary<int, 
            int> M = new Dictionary<int, 
                                     int>();
 
    // Traverse the array
    for(int i = 0; i < N; i++) 
    {
         
        // Increase the counter of
        // the array element by 1
        int count = M.ContainsKey(arr[i]) ? M[arr[i]] : 0;
         
        if (count == 0)
        {
            M.Add(arr[i], 1);
        }
        else
        {
            M[arr[i]] = count + 1;
        }
    }
 
    int[] ar1 = new int[N];
 
    // Traverse the map
    foreach(KeyValuePair<int, int> Element in M) 
    {
 
        // Extract the current element
        int current_element = (int)Element.Key;
 
        // Number of times the current
        // element is present in array
        int count = (int)Element.Value;
 
        // If element is present only
        // once, then do not change it
        if (count == 1)
            continue;
 
        // If the count > 1 then change
        // one of the same current
        // elements to (current_element + K)
        // and increase its count by 1
        ar1[current_element + K]++;
    }
 
    for(int i = 0; i < N; i++) 
    {
        if (ar1[i] >= 0)
        {
            int count = M.ContainsKey(ar1[i]) ? M[ar1[i]] : 0;            
              if (count == 0)
              {
                  M.Add(ar1[i], 1);
            }
            else
            {
                M[ar1[i]] = count + 1;
            }
        }
    }
 
    // The size of the map is the
    // required answer
    Console.WriteLine(M.Count);
}
 
// Driver Code
public static void Main(String[] args)
{
    int []arr = { 2, 3, 2, 4, 5, 5, 7, 4 };
    int K = 2;
    int N = arr.Length;
     
    // Function Call
    maxDifferent(arr, N, K);
}
}
 
// This code contributed by shikhasingrajput

Output:

Time Complexity: O(N)
Auxiliary Space: O(N)

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