Maximize the count of distinct elements in Array after at most K changes

Last Updated : 18 May, 2022

Given an array arr[], the task is to find the maximum number of distinct numbers in arr after at most K changes. In each change pick any element X from arr and change it to Y such that L <= Y <= R.

Examples:

Input: arr[] = {1, 2, 1, 4, 6, 4, 4}, L = 1, R = 5 and K = 2
Output: 6
Explanation:

Following are the operations performed on the given array elements:

1. Changing arr[2] to 3 modifies array to {1, 2, 3, 4, 6, 4, 4}
2. Changing arr[3] to 5 modifies array to {1, 2, 3, 5, 6, 4, 4}

As K = 2, no more changes can be done
Therefore, Distinct elements are {1, 2, 3, 5, 6, 4} and the number of maximum distinct elements is 6.

Input: arr[] = {1, 2, 1, 4, 6, 4, 4}, L = 1, R = 5 and K = 1
Output: 5
Explanation:

Following are the operations performed on the given array elements:

1. Changing arr[2] to 3 modifies array to {1, 2, 3, 4, 6, 4, 4}

As K = 1, no more changes can be done
Therefore, Distinct elements are {1, 2, 3, 6, 4} and the number of maximum distinct elements is 5.

Approach: This problem can be solved by using an unordered map. Store frequency of all the elements in the map and then traverse from L to R, see which number is not already present in the map we can use that number to replace any duplicate element in arr.

At first, store the frequency of all the elements in the map.
Total distinct elements will be equal to the size of the map.
The number of extra elements will be equal to the (size of array - size of map).
Then iterate from L to R and check how many elements inside this range can be used to replace the extra elements present in the array.
Update the map for each change in any extra element to any other value.
At last, the size of the map will be containing all the distinct elements.
Return the size of the map as the answer.

C++

// C++ program for above approach
#include <bits/stdc++.h>
using namespace std;

// Function to calculate
// maximum distinct elements possible
// after at most K changes
int maxDistinctElements(int* arr, int K,
                        int L, int R, int n)
{
    // Map to store frequency of all the elements
    unordered_map<int, int> frequency;

    // Count frequency of each element
    for (int x = 0; x < n; x++) {
        frequency[arr[x]] += 1;
    }

    // To store number of extra elements
    // that needs to be changed
    int extra = (n - frequency.size());

    // Traverse from L to R
    // and see which number is not
    // present in map, use that number
    // to change extra duplicate element
    for (int i = L;
         i <= R and K != 0 and extra != 0;
         i++) {
        if (!frequency[i]) {
            frequency[i] = 1;
            K--;
            extra--;
        }
    }

    // Total distinct element will be equal
    // to the size of updated frequency map.
    int ans = frequency.size();

    // Return answer
    return ans;
}

// Driver Code
int main()
{
    // Test case 1
    int N = 7, L = 1, R = 5, K = 2;
    int arr[7] = { 1, 2, 1, 4, 6, 4, 4 };
    cout << maxDistinctElements(arr, K, L, R, N)
         << endl;

    // Test case 2
    K = 1;
    cout << maxDistinctElements(arr, K, L, R, N)
         << endl;
}

Java

// Java program for above approach
import java.util.*;

class GFG {

    // Function to calculate
    // maximum distinct elements possible
    // after at most K changes
    static int maxDistinctElements(int[] arr, int K, int L, int R, int n)
    {
      
        // Map to store frequency of all the elements
        HashMap<Integer, Integer> frequency = new HashMap<Integer, Integer>();

        // Count frequency of each element
        for (int x = 0; x < n; x++) {
            if (frequency.containsKey(arr[x])) {
                frequency.put(arr[x], frequency.get(arr[x]) + 1);
            } else {
                frequency.put(arr[x], 1);
            }
        }

        // To store number of extra elements
        // that needs to be changed
        int extra = (n - frequency.size());

        // Traverse from L to R
        // and see which number is not
        // present in map, use that number
        // to change extra duplicate element
        for (int i = L; i <= R && K != 0 && extra != 0; i++) {
            if (!frequency.containsKey(i)) {
                frequency.put(i, 1);
                K--;
                extra--;
            }
        }

        // Total distinct element will be equal
        // to the size of updated frequency map.
        int ans = frequency.size();

        // Return answer
        return ans;
    }

    // Driver Code
    public static void main(String[] args) {
        // Test case 1
        int N = 7, L = 1, R = 5, K = 2;
        int arr[] = { 1, 2, 1, 4, 6, 4, 4 };
        System.out.print(maxDistinctElements(arr, K, L, R, N) + "\n");

        // Test case 2
        K = 1;
        System.out.print(maxDistinctElements(arr, K, L, R, N) + "\n");
    }
}

// This code is contributed by 29AjayKumar

Python3

# Python 3 program for above approach

# Function to calculate
# maximum distinct elements possible
# after at most K changes
def maxDistinctElements(arr, K, L, R, n):
  
    # Map to store frequency of all the elements
    frequency = {}

    # Count frequency of each element
    for x in range(n):
        if arr[x] in frequency:
            frequency[arr[x]] += 1
        else:
            frequency[arr[x]] = 1
        
    # To store number of extra elements
    # that needs to be changed
    extra = (n - len(frequency))

    # Traverse from L to R
    # and see which number is not
    # present in map, use that number
    # to change extra duplicate element
    i = L
    while(i <= R and K != 0 and extra != 0):
        if (i not in frequency):
            frequency[i] = 1
            K -= 1
            extra -= 1
        else:
            frequency[i] = 1
            
        i += 1

    # Total distinct element will be equal
    # to the size of updated frequency map.
    ans = len(frequency)

    # Return answer
    return ans

# Driver Code
if __name__ == '__main__':
  
    # Test case 1
    N = 7
    L = 1
    R = 5
    K = 2
    arr = [1, 2, 1, 4, 6, 4, 4]
    print(maxDistinctElements(arr, K, L, R, N))

    # Test case 2
    K = 1
    print(maxDistinctElements(arr, K, L, R, N))
    
    # This code is contributed by SURENDRA_GANGWAR.

// C# program for above approach
using System;
using System.Collections.Generic;
class GFG {
    // Function to calculate
    // maximum distinct elements possible
    // after at most K changes
    static int maxDistinctElements(int[] arr, int K, int L,
                                   int R, int n)
    {
        // Map to store frequency of all the elements
        Dictionary<int, int> frequency
            = new Dictionary<int, int>();

        // Count frequency of each element
        for (int x = 0; x < n; x++) {
            if (frequency.ContainsKey(arr[x]))
                frequency[arr[x]] += 1;
            else
                frequency[arr[x]] = 1;
        }

        // To store number of extra elements
        // that needs to be changed
        int extra = (n - frequency.Count);

        // Traverse from L to R
        // and see which number is not
        // present in map, use that number
        // to change extra duplicate element
        for (int i = L; i <= R && K != 0 && extra != 0;
             i++) {
            if (!frequency.ContainsKey(i)) {
                frequency[i] = 1;
                K--;
                extra--;
            }
        }

        // Total distinct element will be equal
        // to the size of updated frequency map.
        int ans = frequency.Count;

        // Return answer
        return ans;
    }

    // Driver Code
    public static void Main()
    {
        // Test case 1
        int N = 7, L = 1, R = 5, K = 2;
        int[] arr = { 1, 2, 1, 4, 6, 4, 4 };
        Console.WriteLine(
            maxDistinctElements(arr, K, L, R, N));

        // Test case 2
        K = 1;
        Console.WriteLine(
            maxDistinctElements(arr, K, L, R, N));
    }
}

// This code is contributed by decode2207.

JavaScript

<script>
        // JavaScript Program to implement
        // the above approach

        // Function to calculate
        // maximum distinct elements possible
        // after at most K changes
        function maxDistinctElements(arr, K, L, R, n) {
            // Map to store frequency of all the elements
            let frequency = new Map();

            // Count frequency of each element
            for (let x = 0; x < n; x++) {
                if (frequency.has(arr[x])) {
                    frequency.set(frequency.get(arr[x]), frequency.get(arr[x]) + 1);
                }
                else {
                    frequency.set(arr[x], 1)
                }
            }

            // To store number of extra elements
            // that needs to be changed
            let extra = (n - frequency.size);

            // Traverse from L to R
            // and see which number is not
            // present in map, use that number
            // to change extra duplicate element

            for (let i = L;
                i <= R && K != 0 && extra != 0;
                i++) {
                if (!frequency.has(i)) {
                    frequency.set(i, 1);
                    K--;
                    extra--;
                }
            }

            // Total distinct element will be equal
            // to the size of updated frequency map.
            let ans = frequency.size;

            // Return answer
            return ans;
        }

        // Driver Code

        // Test case 1
        let N = 7, L = 1, R = 5, K = 2;
        let arr = [1, 2, 1, 4, 6, 4, 4];
        document.write(maxDistinctElements(arr, K, L, R, N) + "<br>");

        // Test case 2
        K = 1;
        document.write(maxDistinctElements(arr, K, L, R, N) + "<br>");


// This code is contributed by Potta Lokesh
    </script>

Output:

6
5

Time Complexity: O(N).

Auxiliary Space: O(N).

Count of Unique elements after inserting average of MEX and Array Max K times

adityamutharia

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Maximize the count of distinct elements in Array after at most K changes

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