Maximise occurrence of an element after K replacements within Array elements

Last Updated : 23 Feb, 2023

Given an array arr[] having N integers, and an integer K, the task is to find an array such that it contains a single elements maximum number of times possible after K replacements within array elements.

Examples:

Input: N = 7, arr[] = {1, 2, 1, 5, 1, 6, 7}, K = 3
Output: {1, 1, 1, 1, 1, 1, 7}
Explanation: We can replace element on index 1, 3, and 5 with 1, i.e. A[1] = A[0], A[3] = A[0] and A[5] = A[0]

Input: N = 6, arr[] = {2, 2, 2, 5, 1, 6 }, K = 3
Output: {2, 2, 2, 2, 2, 2 }

Approach: The approach to solve this problem is based on following idea:

If we dont do any replacement, and if there are duplicates present in the array, then there will be an element present already occurring maximum number of times in the array.

Now when K replacements are allowed, we can simply try to change replace elements not equal to maximum occurring element with maximum occurring element K times, to get the desired array.

Following is the algorithm to implement above discussed approach:

Create a hashmap to store the frequency of every distinct element.
Find the maximum occurring element in the array.
Then assign every element value of maximum occurring element.
Make sure to take care of frequency before assigning the element and after assigning the element.
For that first decrease the frequency of the current element and after assigning it the maximum occurring value increase its value in the hashmap.

Below is the implementation of the above approach:

C++

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

// Function to maximizeTheElements
vector<int> maximizeTheElements(
    int N, vector<int> arr, int K)
{

    // Map to store the frequency
    // of the elements
    unordered_map<int, int> mp;

    int max_freq = 0, max_element = -1;
    for (auto x : arr) {
        mp[x]++;

        // Getting max_element
        if (mp[x] > max_freq) {
            max_freq = mp[x];
            max_element = x;
        }
    }

    for (int i = 0; i < N && K > 0; i++) {

        // If the element is not equal
        // to the max_element
        if (arr[i] != max_element) {

            // Decrease its frequency from the map
            mp[arr[i]] -= 1;

            // Assign the max frequency element
            arr[i] = max_element;

            // Increase its frequency in the map
            mp[arr[i]] += 1;

            // Decrease the operation by 1
            K -= 1;
        }
    }

    // Return the modified array
    return arr;
}

// Driver Function
int main()
{

    int N = 7;
    vector<int> arr = { 1, 2, 1, 5, 1, 6, 7 };
    int K = 3;

    // Function call
    vector<int> res = maximizeTheElements(N, arr, K);

    for (auto x : res) {
        cout << x << " ";
    }
    cout << endl;

    return 0;
}

Java

// Java code for the above discussed approach
import java.util.*;

class GFG {

  // Function to maximizeTheElements
  static int[] maximizeTheElements(
    int N, int arr[], int K)
  {

    // Map to store the frequency
    // of the elements
    HashMap<Integer, Integer> m = new HashMap<Integer, Integer>();
    int max_freq = 0, max_element = -1;
    for(int i = 0; i  < arr.length; i++){
      if(m.containsKey(arr[i])){
        m.put(arr[i], m.get(arr[i]) + 1);
      }
      else{
        m.put(arr[i],1);
      }

      // Getting max_element
      if (m.get(arr[i]) > max_freq) {
        max_freq = m.get(arr[i]);
        max_element = arr[i];
      }
    }

    for (int i = 0; i < N; i++) {
      if(K <= 0)break;
      // If the element is not equal
      // to the max_element
      if (arr[i] != max_element) {

        // Decrease its frequency from the map
        m.put(arr[i], m.get(arr[i]) - 1);

        // Assign the max frequency element
        arr[i] = max_element;

        // Increase its frequency in the map
        m.put(arr[i], m.get(arr[i]) + 1);

        // Decrease the operation by 1
        K -= 1;
      }
    }

    // Return the modified array
    return arr;
  }

  // Driver Function
  public static void main (String[] args) {
    int N = 7;
    int  arr[] = { 1, 2, 1, 5, 1, 6, 7 };
    int K = 3;

    // Function call
    int res[] = maximizeTheElements(N, arr, K);

    for (int x : res) {
      System.out.print(x + " ");
    }

  }
}

// This code is contributed by hrithikgarg03188.

Python3

# Python3 code for the above discussed approach

# Function to maximizeTheElements
def maximizeTheElements(N, arr, K):

        # Map to store the frequency
        # of the elements
    mp = {}

    max_freq, max_element = 0, -1
    for x in arr:
        mp[x] = mp[x] + 1 if x in mp else 1

        # Getting max_element
        if (mp[x] > max_freq):
            max_freq = mp[x]
            max_element = x

    for i in range(0, N):
        if K <= 0:
            break
            # If the element is not equal
            # to the max_element
        if (arr[i] != max_element):

                        # Decrease its frequency from the map
            mp[arr[i]] -= 1

            # Assign the max frequency element
            arr[i] = max_element

            # Increase its frequency in the map
            mp[arr[i]] += 1

            # Decrease the operation by 1
            K -= 1

        # Return the modified array
    return arr

# Driver Function
if __name__ == "__main__":

    N = 7
    arr = [1, 2, 1, 5, 1, 6, 7]
    K = 3

    # Function call
    res = maximizeTheElements(N, arr, K)

    for x in res:
        print(x, end=" ")

    # This code is contributed by rakeshsahni

// C# code for the above discussed approach
using System;
using System.Collections.Generic;

public class GFG 
{

  // Function to maximizeTheElements
  static int[] maximizeTheElements(int N, int[] arr,
                                   int K)
  {

    // Map to store the frequency
    // of the elements
    var m = new Dictionary<int, int>();
    int max_freq = 0, max_element = -1;
    for (int i = 0; i < arr.Length; i++) {
      if (m.ContainsKey(arr[i])) {
        m[arr[i]] += 1;
      }
      else {
        m[arr[i]] = 1;
      }

      // Getting max_element
      if (m[arr[i]] > max_freq) {
        max_freq = m[arr[i]];
        max_element = arr[i];
      }
    }

    for (int i = 0; i < N; i++) {
      if (K <= 0)
        break;

      // If the element is not equal
      // to the max_element
      if (arr[i] != max_element) {

        // Decrease its frequency from the map
        m[arr[i]] -= 1;

        // Assign the max frequency element
        arr[i] = max_element;

        // Increase its frequency in the map
        m[arr[i]] += 1;

        // Decrease the operation by 1
        K -= 1;
      }
    }

    // Return the modified array
    return arr;
  }

  // Driver Code
  public static void Main(string[] args)
  {
    int N = 7;
    int[] arr = { 1, 2, 1, 5, 1, 6, 7 };
    int K = 3;

    // Function call
    int[] res = maximizeTheElements(N, arr, K);

    for (int i = 0; i < res.Length; i++) {
      Console.Write(res[i] + " ");
    }
  }
}

// This code is contributed by phasing17

JavaScript

 <script>
        // JavaScript code for the above approach

        // Function to maximizeTheElements
        function maximizeTheElements(
            N, arr, K) {

            // Map to store the frequency
            // of the elements
            let mp = new Map();

            let max_freq = 0, max_element = -1;
            for (let x of arr) {
                if (!mp.has(x)) {
                    mp.set(x, 1)
                }
                else {
                    mp.set(x, mp.get(x) + 1)
                }

                // Getting max_element
                if (mp.get(x) > max_freq) {
                    max_freq = mp.get(x);
                    max_element = x;
                }
            }

            for (let i = 0; i < N && K > 0; i++) {

                // If the element is not equal
                // to the max_element
                if (arr[i] != max_element) {

                    // Decrease its frequency from the map
                    mp.set(arr[i], mp.get(arr[i]) - 1)

                    // Assign the max frequency element
                    arr[i] = max_element;

                    // Increase its frequency in the map
                    mp.set(arr[i], mp.get(arr[i]) + 1)

                    // Decrease the operation by 1
                    K -= 1;
                }
            }

            // Return the modified array
            return arr;
        }

        // Driver Function
        let N = 7;
        let arr = [1, 2, 1, 5, 1, 6, 7];
        let K = 3;

        // Function call
        let res = maximizeTheElements(N, arr, K);

        for (let x of res) {
            document.write(x + " ");
        }

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

Output

1 1 1 1 1 1 7

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

Remove all occurrences of any element for maximum array sum

stryker27

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