Elemento menos frecuente en una array

Dada una array, encuentre el elemento menos frecuente en ella. Si hay varios elementos que aparecen la menor cantidad de veces, imprima cualquiera de ellos.
Ejemplos: 
 

Input : arr[] = {1, 3, 2, 1, 2, 2, 3, 1}
Output : 3
3 appears minimum number of times in given
array.

Input : arr[] = {10, 20, 30}
Output : 10 or 20 or 30

Una solución simple es ejecutar dos bucles. El bucle exterior recoge todos los elementos uno por uno. El bucle interno encuentra la frecuencia del elemento elegido y la compara con el mínimo hasta el momento. La complejidad temporal de esta solución es O(n 2 )
Una mejor solución es clasificar. Primero ordenamos la array, luego recorremos linealmente la array.
 

C++

// CPP program to find the least frequent element
// in an array.
#include <bits/stdc++.h>
using namespace std;
 
int leastFrequent(int arr[], int n)
{
    // Sort the array
    sort(arr, arr + n);
 
    // find the min frequency using linear traversal
    int min_count = n+1, res = -1, curr_count = 1;
    for (int i = 1; i < n; i++) {
        if (arr[i] == arr[i - 1])
            curr_count++;
        else {
            if (curr_count < min_count) {
                min_count = curr_count;
                res = arr[i - 1];
            }
            curr_count = 1;
        }
    }
  
    // If last element is least frequent
    if (curr_count < min_count)
    {
        min_count = curr_count;
        res = arr[n - 1];
    }
 
    return res;
}
 
// driver program
int main()
{
    int arr[] = {1, 3, 2, 1, 2, 2, 3, 1};
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << leastFrequent(arr, n);
    return 0;
}

Java

// Java program to find the least frequent element
// in an array.
import java.io.*;
import java.util.*;
 
class GFG {
     
    static int leastFrequent(int arr[], int n)
    {
         
        // Sort the array
        Arrays.sort(arr);
     
        // find the min frequency using
        // linear traversal
        int min_count = n+1, res = -1;
        int curr_count = 1;
         
        for (int i = 1; i < n; i++) {
            if (arr[i] == arr[i - 1])
                curr_count++;
            else {
                if (curr_count < min_count) {
                    min_count = curr_count;
                    res = arr[i - 1];
                }
                 
                curr_count = 1;
            }
        }
     
        // If last element is least frequent
        if (curr_count < min_count)
        {
            min_count = curr_count;
            res = arr[n - 1];
        }
     
        return res;
    }
     
    // driver program
    public static void main(String args[])
    {
        int arr[] = {1, 3, 2, 1, 2, 2, 3, 1};
        int n = arr.length;
        System.out.print(leastFrequent(arr, n));
         
    }
}
 
/*This code is contributed by Nikita Tiwari.*/

Python3

# Python 3 program to find the least
# frequent element in an array.
 
 
def leastFrequent(arr, n) :
 
    # Sort the array
    arr.sort()
  
    # find the min frequency using
    # linear traversal
    min_count = n + 1
    res = -1
    curr_count = 1
    for i in range(1, n) :
        if (arr[i] == arr[i - 1]) :
            curr_count = curr_count + 1
        else :
            if (curr_count < min_count) :
                min_count = curr_count
                res = arr[i - 1]
             
            curr_count = 1
             
   
    # If last element is least frequent
    if (curr_count < min_count) :
        min_count = curr_count
        res = arr[n - 1]
     
    return res
     
  
# Driver program
arr = [1, 3, 2, 1, 2, 2, 3, 1]
n = len(arr)
print(leastFrequent(arr, n))
 
 
# This code is contributed
# by Nikita Tiwari.

C#

// C# program to find the least
// frequent element in an array.
using System;
 
class GFG {
     
    static int leastFrequent(int[] arr, int n)
    {
        // Sort the array
        Array.Sort(arr);
     
        // find the min frequency
        // using linear traversal
        int min_count = n + 1, res = -1;
        int curr_count = 1;
         
        for (int i = 1; i < n; i++)
        {
            if (arr[i] == arr[i - 1])
                curr_count++;
            else
            {
                if (curr_count < min_count)
                {
                    min_count = curr_count;
                    res = arr[i - 1];
                }
                 
                curr_count = 1;
            }
        }
     
        // If last element is least frequent
        if (curr_count < min_count)
        {
            min_count = curr_count;
            res = arr[n - 1];
        }
     
        return res;
    }
     
    // Driver code
    static public void Main ()
    {
        int[] arr = {1, 3, 2, 1, 2, 2, 3, 1};
        int n = arr.Length;
         
        // Function calling
        Console.Write(leastFrequent(arr, n));
    }
}
 
// This code is contributed by Shrikant13

PHP

<?php
// PHP program to find the
// least frequent element
// in an array.
 
function leastFrequent($arr, $n)
{
     
    // Sort the array
    sort($arr);
    sort($arr , $n);
 
    // find the min frequency
    // using linear traversal
    $min_count = $n + 1;
    $res = -1;
    $curr_count = 1;
    for($i = 1; $i < $n; $i++)
    {
        if ($arr[$i] == $arr[$i - 1])
            $curr_count++;
        else
        {
            if ($curr_count < $min_count)
            {
                $min_count = $curr_count;
                $res = $arr[$i - 1];
            }
            $curr_count = 1;
        }
    }
 
    // If last element is
    // least frequent
    if ($curr_count < $min_count)
    {
        $min_count = $curr_count;
        $res = $arr[$n - 1];
    }
 
    return $res;
}
 
// Driver Code
{
    $arr = array(1, 3, 2, 1, 2, 2, 3, 1);
    $n = sizeof($arr) / sizeof($arr[0]);
    echo leastFrequent($arr, $n);
    return 0;
}
 
// This code is contributed by nitin mittal
?>

Javascript

<script>
 
// JavaScript program to find the least frequent element
// in an array.
 
    function leastFrequent(arr, n)
    {
           
        // Sort the array
        arr.sort();
       
        // find the min frequency using
        // linear traversal
        let min_count = n+1, res = -1;
        let curr_count = 1;
           
        for (let i = 1; i < n; i++) {
            if (arr[i] == arr[i - 1])
                curr_count++;
            else {
                if (curr_count < min_count) {
                    min_count = curr_count;
                    res = arr[i - 1];
                }
                   
                curr_count = 1;
            }
        }
       
        // If last element is least frequent
        if (curr_count < min_count)
        {
            min_count = curr_count;
            res = arr[n - 1];
        }
       
        return res;
    }
 
// Driver code
 
        let arr = [1, 3, 2, 1, 2, 2, 3, 1];
        let n = arr.length;
        document.write(leastFrequent(arr, n));
                                
</script>
Producción: 

3

 

Complejidad de tiempo: O(n Log n) 
Espacio auxiliar: O(1)
Una solución eficiente es usar hashing. Creamos una tabla hash y almacenamos elementos y su frecuencia cuenta como pares de valores clave. Finalmente recorremos la tabla hash e imprimimos la clave con el valor mínimo.
 

C++

// CPP program to find the least frequent element
// in an array.
#include <bits/stdc++.h>
using namespace std;
 
int leastFrequent(int arr[], int n)
{
    // Insert all elements in hash.
    unordered_map<int, int> hash;
    for (int i = 0; i < n; i++)
        hash[arr[i]]++;
 
    // find the min frequency
    int min_count = n+1, res = -1;
    for (auto i : hash) {
        if (min_count >= i.second) {
            res = i.first;
            min_count = i.second;
        }
    }
 
    return res;
}
 
// driver program
int main()
{
    int arr[] = {1, 3, 2, 1, 2, 2, 3, 1};
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << leastFrequent(arr, n);
    return 0;
}

Java

//Java program to find the least frequent element
//in an array
import java.util.HashMap;
import java.util.Map;
import java.util.Map.Entry;
 
class GFG {
     
    static int leastFrequent(int arr[],int n)
    {
         
        // Insert all elements in hash.
        Map<Integer,Integer> count =
                   new HashMap<Integer,Integer>();
                    
        for(int i = 0; i < n; i++)
        {
            int key = arr[i];
            if(count.containsKey(key))
            {
                int freq = count.get(key);
                freq++;
                count.put(key,freq);
            }
            else
                count.put(key,1);
        }
         
        // find min frequency.
        int min_count = n+1, res = -1;
        for(Entry<Integer,Integer> val : count.entrySet())
        {
            if (min_count >= val.getValue())
            {
                res = val.getKey();
                min_count = val.getValue();
            }
        }
         
        return res;
    }
     
    // driver program
    public static void main (String[] args) {
         
        int arr[] = {1, 3, 2, 1, 2, 2, 3, 1};
        int n = arr.length;
         
        System.out.println(leastFrequent(arr,n));
    }
}
 
// This code is contributed by Akash Singh.

Python3

# Python3 program to find the most
# frequent element in an array.
import math as mt
 
def leastFrequent(arr, n):
 
    # Insert all elements in Hash.
    Hash = dict()
    for i in range(n):
        if arr[i] in Hash.keys():
            Hash[arr[i]] += 1
        else:
            Hash[arr[i]] = 1
 
    # find the max frequency
    min_count = n + 1
    res = -1
    for i in Hash:
        if (min_count >= Hash[i]):
            res = i
            min_count = Hash[i]
         
    return res
 
# Driver Code
arr = [1, 3, 2, 1, 2, 2, 3, 1]
n = len(arr)
print(leastFrequent(arr, n))
 
# This code is contributed by
# mohit kumar 29

C#

// C# program to find the
// least frequent element
// in an array.
using System;
using System.Collections.Generic;
 
class GFG
{
    static int leastFrequent(int []arr,
                             int n)
    {
        // Insert all elements in hash.
        Dictionary<int, int> count =
                        new Dictionary<int,
                                       int>();
        for (int i = 0; i < n; i++)
        {
            int key = arr[i];
            if(count.ContainsKey(key))
            {
                int freq = count[key];
                freq++;
                count[key] = freq;
            }
            else
                count.Add(key, 1);
        }
         
        // find the min frequency
        int min_count = n + 1, res = -1;
        foreach (KeyValuePair<int,
                    int> pair in count)
        {
            if (min_count >= pair.Value)
            {
                res = pair.Key;
                min_count = pair.Value;
            }
        }
        return res;
    }
     
    // Driver Code
    static void Main()
    {
        int []arr = new int[]{1, 3, 2, 1,
                              2, 2, 3, 1};
        int n = arr.Length;
        Console.Write(leastFrequent(arr, n));
    }
}
 
// This code is contributed by
// Manish Shaw(manishshaw1)

Javascript

<script>
 
// JavaScript program to find the least frequent element
// in an array.
 
function leastFrequent(arr, n)
{
    // Insert all elements in hash.
    var hash = new Map();
    for (var i = 0; i < n; i++)
    {
        if(hash.has(arr[i]))
            hash.set(arr[i], hash.get(arr[i])+1)
        else
            hash.set(arr[i], 1);
    }
 
    // find the min frequency
    var min_count = n+1, res = -1;
 
    hash.forEach((value, key) => {
         
        if (min_count >= value) {
            res = key;
            min_count = value;
        }
    });
 
    return res;
}
 
// driver program
var arr = [1, 3, 2, 1, 2, 2, 3, 1];
var n = arr.length;
document.write( leastFrequent(arr, n));
 
</script>
Producción: 

3

 

Tiempo Complejidad : O(n) 
Espacio Auxiliar : O(n)
 

Publicación traducida automáticamente

Artículo escrito por kartik y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA

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