HackerRank Equalize the Array Problem Solution

In this post, we will solve HackerRank Equalize the Array Problem Solution.

Given an array of integers, determine the minimum number of elements to delete to leave only elements of equal value.
Example
arr = [1, 2, 2, 3]
Delete the 2 elements 1 and 3 leaving arr= [2,2]. If both twos plus either the 1 or the 3 are deleted, it takes 3 deletions to leave either [3] or [1]. The minimum number of deletions is 2.

Function Description

Complete the equalizeArray function in the editor below.

equalizeArray has the following parameter(s):

int arr[n]: an array of integers

Returns

int: the minimum number of deletions required

Input Format
The first line contains an integer n, the number of elements in arr.
The next line contains n space-separated integers arr[i].

Sample Input

STDIN       Function
-----       --------
5           arr[] size n = 5
3 3 2 1 3   arr = [3, 3, 2, 1, 3]

Sample Output

Explanation
Delete arr[2] = 2 and arr[3] = 1 to leave arr’ = [3, 3, 3]. This is minimal. The only other options are to delete 4 elements to get an array of either [1] or [2].

Table of Contents

Equalize the Array C Solution

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int main() 
{
    int n;
    scanf("%d", &n);
    int* count = (int*)malloc(100*sizeof(int));
    int i;
    for(i=0; i<100; i++){count[i]=0;}
    for(i=0; i<n; i++)
    {
        int t;
        scanf("%d",&t);
        count[t-1]++;
    }
    int max = 0;
    for(i=0; i<100; i++)
    {
        if(max<count[i])max=count[i];
    }
    printf("%d\n",n-max);
    
    return 0;
}

Equalize the Array C++ Solution

#include <iostream>
#include <cstring>
#include <climits>
#include <cmath>
#include <vector>
#include <cstdio>
#include <algorithm>
#include <float.h>
#include <map>
#include <unordered_map>
#include <queue>
#include <bitset>

#define ui unsigned int
#define ull unsigned long long
#define ll long long
#define iloop(i,j,n) for(ll i=j;i<n;i++)
#define dloop(i,j,n) for(ll i=j;i>=n;i--)
#define all(v) v.begin(),v.end()
#define sc(n) scanf("%d",&n)
#define sc2(i,j) scanf("%d %d",&i,&j)
#define sc3(i,j,k) scanf("%d %d %d",&i,&j,&k)
#define sc4(i,j,k,l) scanf("%d %d %d %d",&i,&j,&k,&l)
#define llsc(n) scanf("%lld",&n)
#define llsc2(i,j) scanf("%lld %lld",&i,&j)
#define llsc3(i,j,k) scanf("%lld %lld %lld",&i,&j,&k)
#define llsc4(i,j,k,l) scanf("%lld %lld %lld %lld",&i,&j,&k,&l)
#define pll(n) printf("%lld\n",n);
#define pi(n) printf("%d\n",n);
#define pb(i) push_back(i)
#define dbl double
#define mp(i,j) make_pair(i,j)
#define set(a,i) memset(a,i,sizeof(a))
#define vi vector <int> 
#define vll vector <ll>
#define vpi vector < pair <int,int> >
#define vpll vector < pair <ll,ll> >
#define f first
#define s second;

using namespace std;

const int mod = 1000000007;

int main()
{
	int n;sc(n);
	int a[n];
	int mapping[101];
	set(mapping,0);
	iloop(i,0,n)
	{
		sc(a[i]);
		mapping[a[i]]++;
	}
	int ct=0;
	iloop(i,0,101)
	{
		if(ct < mapping[i])
			ct = mapping[i];
	}
	cout << (n-ct) << endl;
	return 0;
}

Equalize the Array C Sharp Solution

using System;
using System.Collections.Generic;
using System.IO;
using System.Linq;

class Solution {
    static void Main(String[] args) {
        Console.ReadLine();
        int[] arr = Array.ConvertAll(Console.ReadLine().Split(' '), int.Parse);
        
        Console.WriteLine(EqualizeArray(arr));
    }
    
    static int EqualizeArray(int[] arr) {
        Dictionary<int, int> freq = new Dictionary<int, int>();
        
        foreach(var item in arr) {
            if (freq.ContainsKey(item))
                freq[item]++;
            else 
                freq.Add(item, 1);
        }
        
        int maxFreq = freq.Values.Max();
        
        return arr.Length - maxFreq;
    }
}

Equalize the Array Java Solution

import java.io.*;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import java.util.stream.Stream;

import static java.util.stream.Collectors.toList;

class Result {

    /*
     * Complete the 'equalizeArray' function below.
     *
     * The function is expected to return an INTEGER.
     * The function accepts INTEGER_ARRAY arr as parameter.
     */

    public static int equalizeArray(List<Integer> arr) {

        List<Integer> values = List.copyOf(arr);


        HashMap<Integer, Integer> objectFrequencyMap = new HashMap<>();
        arr.stream().distinct().collect(toList()).forEach(uniqueObject -> {
            objectFrequencyMap.put(uniqueObject, (int) values.stream().filter(f -> f.equals(uniqueObject)).count());
        });

        objectFrequencyMap.entrySet()
                .stream()
                .sorted(Map.Entry.comparingByValue())
                .forEach(System.out::println);

        Integer maxFrequency = objectFrequencyMap.values().stream().sorted().collect(toList()).get(objectFrequencyMap.size() - 1);


        return arr.size() - maxFrequency;


    }

}

public class Solution {
    public static void main(String[] args) throws IOException {
        BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(System.in));
        BufferedWriter bufferedWriter = new BufferedWriter(new FileWriter(System.getenv("OUTPUT_PATH")));

        int n = Integer.parseInt(bufferedReader.readLine().trim());

        List<Integer> arr = Stream.of(bufferedReader.readLine().replaceAll("\\s+$", "").split(" "))
                .map(Integer::parseInt)
                .collect(toList());

        int result = Result.equalizeArray(arr);

        bufferedWriter.write(String.valueOf(result));
        bufferedWriter.newLine();

        bufferedReader.close();
        bufferedWriter.close();
    }
}

Equalize the Array Javascript Solution

function processData(input) {
  // Read input
  input = input.split('\n');
  let line = 0,
    readLine = () => input[line++], // Read a string
    readNumber = () => Number(readLine()), // Read a single integer
    readNumberList = () => readLine().split(' ').map(Number), // Read multiple integers
    log = (x) => console.log(x);

  const n = readNumber();
  const arr = readNumberList();
    
  let c = Array(Math.max(...arr)+1).fill(0);
    
    for (let i=0; i<n; i++){
        c[arr[i]]++;
    }
    
    let max = Math.max(...c);
    console.log(n-max);
} 

process.stdin.resume();
process.stdin.setEncoding("ascii");
_input = "";
process.stdin.on("data", function (input) {
    _input += input;
});

process.stdin.on("end", function () {
   processData(_input);
});

Equalize the Array Python Solution

s = int(input())
A = [int(i) for i in input().split(' ')]

if A.count(A[0]) == s: print(0)
else: print(s - A.count(max(set(A), key=A.count)))

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