Minimum delete operations to make all elements of array same

Given an array of n elements such that elements may repeat. We can delete any number of elements from the array. The task is to find a minimum number of elements to be deleted from the array to make it equal.
Examples:

Input: arr[] = {4, 3, 4, 4, 2, 4}
Output: 2
After deleting 2 and 3 from array, array becomes 
arr[] = {4, 4, 4, 4} 

Input: arr[] = {1, 2, 3, 4, 5}
Output: 4
We can delete any four elements from array.

In this problem, we need to minimize the delete operations. The approach is simple, we count the frequency of each element in an array, then find the frequency of the most frequent elements in count array. Let this frequency be max_freq. To get the minimum number of elements to be deleted from the array calculate n – max_freq where n is a number of elements in the given array.

C++

// C++ program to find minimum
// number of deletes required
// to make all elements same.
#include <bits/stdc++.h>
using namespace std;

// Function to get minimum number of elements to be deleted
// from array to make array elements equal
int minDelete(int arr[], int n)
{
    // Create an hash map and store frequencies of all
    // array elements in it using element as key and
    // frequency as value
    unordered_map<int, int> freq;
    for (int i = 0; i < n; i++)
        freq[arr[i]]++;

    // Find maximum frequency among all frequencies.
    int max_freq = INT_MIN;
    for (auto itr = freq.begin(); itr != freq.end(); itr++)
        max_freq = max(max_freq, itr->second);

    // To minimize delete operations, we remove all
    // elements but the most frequent element.
    return n - max_freq;
}

// Driver program to run the case
int main()
{
    int arr[] = { 4, 3, 4, 4, 2, 4 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << minDelete(arr, n);
    return 0;
}

Java

// Java program to find minimum number
// of deletes required to make all
// elements same.
import java.util.*;

class GFG{

// Function to get minimum number of
// elements to be deleted from array
// to make array elements equal
static int minDelete(int arr[], int n)
{

    // Create an hash map and store
    // frequencies of all array elements
    // in it using element as key and
    // frequency as value
    HashMap<Integer, Integer> freq = new HashMap<>();
    for(int i = 0; i < n; i++)
        freq.put(arr[i], freq.getOrDefault(arr[i], 0) + 1);

    // Find maximum frequency among all frequencies.
    int max_freq = Integer.MIN_VALUE;
    for(Map.Entry<Integer,
                  Integer> entry : freq.entrySet())
        max_freq = Math.max(max_freq,
                            entry.getValue());

    // To minimize delete operations,
    // we remove all elements but the
    // most frequent element.
    return n - max_freq ;
}

// Driver code
public static void main(String[] args)
{
    int arr[] = { 4, 3, 4, 4, 2, 4 };
    int n = arr.length;

    System.out.print(minDelete(arr, n));
}
}

// This code is contributed by amal kumar choubey and corrected by Leela Kotte

Python3

# Python3 program to find minimum
# number of deletes required to
# make all elements same.

# Function to get minimum number
# of elements to be deleted from
# array to make array elements equal
def minDelete(arr, n):

    # Create an dictionary and store
    # frequencies of all array
    # elements in it using
    # element as key and
    # frequency as value
    freq = {}
    for i in range(n):
        if arr[i] in freq:
            freq[arr[i]] += 1
        else:
            freq[arr[i]] = 1;


    # Find maximum frequency
    # among all frequencies.
    max_freq = 0;
    for i, j in freq.items():
        max_freq = max(max_freq, j);

    # To minimize delete operations,
    # we remove all elements but the
    # most frequent element.
    return n - max_freq;

# Driver code
arr = [ 4, 3, 4, 4, 2, 4 ];
n = len(arr)

print(minDelete(arr, n));

# This code is contributed by grand_master

C#

// C# program to find minimum number
// of deletes required to make all
// elements same.
using System;
using System.Collections.Generic;
class GFG {

    // Function to get minimum number of
    // elements to be deleted from array
    // to make array elements equal
    static int minDelete(int[] arr, int n)
    {

        // Create an hash map and store
        // frequencies of all array elements
        // in it using element as key and
        // frequency as value
        Dictionary<int, int> freq
            = new Dictionary<int, int>();
        for (int i = 0; i < n; i++)
            if (freq.ContainsKey(arr[i]))
            {
                freq[arr[i]] = freq[arr[i]] + 1;
            }
            else
            {
                freq.Add(arr[i], 1);
            }

        // Find maximum frequency among all frequencies.
        int max_freq = int.MinValue;
        foreach(KeyValuePair<int, int> entry in freq)
            max_freq = Math.Max(max_freq, entry.Value);

        // To minimize delete operations,
        // we remove all elements but the
        // most frequent element.
        return n - max_freq + 1;
    }

    // Driver code
    public static void Main(String[] args)
    {
        int[] arr = {4, 3, 4, 4, 2, 4};
        int n = arr.Length;
          Console.Write(minDelete(arr, n));
    }
}

// This code is contributed by Amit Katiyar

Javascript

<script>
// Javascript program to find minimum number
// of deletes required to make all
// elements same.

// Function to get minimum number of
// elements to be deleted from array
// to make array elements equal
function minDelete(arr, n)
{

    // Create an hash map and store
    // frequencies of all array elements
    // in it using element as key and
    // frequency as value
    let freq = new Map();
    for(let i = 0; i < n; i++){
        if(freq.has(arr[i])){
            freq.set(arr[i], freq.get(arr[i]) + 1)
        }else{
            freq.set(arr[i], 1)
        }
    }

    // Find maximum frequency among all frequencies.
    let max_freq = Number.MIN_SAFE_INTEGER;
    for(let entry of freq)
        max_freq = Math.max(max_freq, entry[1]);

    // To minimize delete operations,
    // we remove all elements but the
    // most frequent element.
    return n - max_freq ;
}

// Driver code
    let arr = [4, 3, 4, 4, 2, 4 ];
    let n = arr.length;

    document.write(minDelete(arr, n));

// This code is contributed by _saurabh_jaiswal.
</script>

Output:

2

Time complexity: O(n)

Auxiliary Space: O(n)

Another Approach Using binary search :

• .First , we will sort the array for binary search function . Then we can find frequency of all array elements using lower_bound and upper bound .
• Then our answer will be n – max_frequency .
• Then return final answer

Below is the implementation of the above approach:

C++

// C++ implementation of the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to find minimum operation required to
// make all array elements equal
int minDelete(int arr[], int n)
{
   int max_freq = 1;
   sort(arr,arr+n);//sort array for binary search

    // Iterating the whole array
   for(int i = 0 ; i < n ;i++)
   {
     //index of first and last occ of arr[i]
     int first_index = lower_bound(arr,arr+n,arr[i])- arr;
     int last_index = upper_bound(arr,arr+n,arr[i])- arr-1;
     i = last_index; // assign i to last_index to avoid
                     // same element multiple time

     int fre = last_index-first_index+1;//finding frequency

      // Finding maximum frequency from all array elements
      max_freq = max( max_freq , fre );
   }
   return n-max_freq;// return answer
}

// Drive code
int main()
{
    int arr[] = { 4, 3, 4, 4, 2, 4 };
    int n = sizeof(arr) / sizeof(arr[0]);

    // Function call
    cout << minDelete(arr, n);

    return 0;
}

// This Approach is contributed by nikhilsainiofficial546

Java

import java.util.Arrays;

public class Main {
    // Function to find minimum operation required to make all
   // array elements equal
    public static int minDelete(int[] arr) {
        int n = arr.length;
        int max_freq = 1;
        Arrays.sort(arr); // sort array for binary search

        // Iterating the whole array
        for (int i = 0; i < n; i++) {
            // index of first and last occurrence of arr[i]
            int first_index = Arrays.binarySearch(arr, arr[i]);
            int last_index = first_index;
            while (last_index < n - 1 && arr[last_index + 1] == arr[i]) {
                last_index++;
            }
            i = last_index; // assign i to last_index to avoid same element multiple times

            int fre = last_index - first_index + 1; // finding frequency

            // Finding maximum frequency from all array elements
            max_freq = Math.max(max_freq, fre);
        }
        return n - max_freq; // return answer
    }

    public static void main(String[] args) {
        int[] arr = {4, 3, 4, 4, 2, 4};
        System.out.println(minDelete(arr));
    }
}

Python3

# Function to find minimum operation required to
# make all array elements equal
def minDelete(arr, n):
    max_freq = 1
    arr.sort() # sort array for binary search

    # Iterating the whole array
    i = 0
    while i < n:

        # index of first occ of arr[i]
        first_index = arr.index(arr[i])

        # find the last occ of arr[i]
        j = n - 1
        while j >= 0 and arr[j] != arr[i]:
            j -= 1
        last_index = j
        fre = last_index - first_index + 1 # finding frequency
        # Finding maximum frequency from all array elements
        max_freq = max(max_freq, fre)

        i = last_index + 1 # assign i to next index

    return n - max_freq # return answer

# Drive code
arr = [4, 3, 4, 4, 2, 4]
n = len(arr)

# Function call
print(minDelete(arr, n))

C#

using System;

class Program {

  // Function to find minimum operation required to
  // make all array elements equal
  static int minDelete(int[] arr, int n)
  {
    int max_freq = 1;
    Array.Sort(arr); // sort array for binary search

    // Iterating the whole array
    for (int i = 0; i < n; i++) {
      // index of first and last occ of arr[i]
      int first_index
        = Array.BinarySearch(arr, arr[i]);
      int last_index = Array.LastIndexOf(arr, arr[i]);
      i = last_index; // assign i to last_index to
      // avoid same element multiple
      // time

      int fre = last_index - first_index
        + 1; // finding frequency

      // Finding maximum frequency from all array
      // elements
      max_freq = Math.Max(max_freq, fre);
    }
    return n - max_freq; // return answer
  }

  static void Main(string[] args)
  {
    int[] arr = { 4, 3, 4, 4, 2, 4 };
    int n = arr.Length;

    // Function call
    Console.WriteLine(minDelete(arr, n));
  }
}

// This code is contributed by sarojmcy2e

Javascript

// Function to find minimum operation required to
// make all array elements equal
function minDelete(arr, n) {
  let max_freq = 1;
  arr.sort((a, b) => a - b); // sort array for binary search

  // Iterating the whole array
  for (let i = 0; i < n; i++) {
    // index of first and last occ of arr[i]
    let first_index = arr.indexOf(arr[i]);
    let last_index = arr.lastIndexOf(arr[i]);
    i = last_index; // assign i to last_index to avoid
    // same element multiple time

    let fre = last_index - first_index + 1; // finding frequency

    // Finding maximum frequency from all array elements
    max_freq = Math.max(max_freq, fre);
  }
  return n - max_freq; // return answer
}

// Drive code
let arr = [4, 3, 4, 4, 2, 4];
let n = arr.length;

// Function call
console.log(minDelete(arr, n));

Output

2

Time Complexity: O(N*Log₂N), where N is the size of the input array
Auxiliary Space: O(1)

Note: Here we can optimize the extra space to count the frequency of each element to O(1) but for this, we have to modify our original array. See this article.

This article is contributed by Aarti_Rathi and Shashank Mishra ( Gullu ). If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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