Encuentra la frecuencia del valor más pequeño en una array

Dada una array A de N elementos. Encuentre la frecuencia del valor más pequeño en la array.
Ejemplos: 
 

Input : N = 5, arr[] = {3, 2, 3, 4, 4} 
Output : 1
The smallest element in the array is 2 
and it occurs only once.

Input : N = 6, arr[] = {4, 3, 5, 3, 3, 6}
Output : 3
The smallest element in the array is 3 
and it occurs 3 times.

Enfoque simple : un método simple es encontrar primero el elemento mínimo en la array en el primer recorrido. Luego recorra la array nuevamente y encuentre el número de ocurrencias del elemento mínimo.
Enfoque eficiente : el enfoque eficiente es hacer esto en un solo recorrido. 
Supongamos que el primer elemento es el mínimo actual, por lo que la frecuencia del mínimo actual sería 1. Ahora iteremos a través de la array (1 a N), nos encontramos con 2 casos: 
 

• El elemento actual es más pequeño que nuestro mínimo actual: cambiamos el mínimo actual para que sea igual al elemento actual y dado que es la primera vez que nos encontramos con este elemento, hacemos que su frecuencia sea 1.
• El elemento actual es igual al mínimo actual: Incrementamos la frecuencia del mínimo actual.

A continuación se muestra la implementación del enfoque anterior:
 

// C++ program to find the frequency of
// minimum element in the array
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the frequency of the
// smallest value in the array
int frequencyOfSmallest(int n, int arr[])
{
    int mn = arr[0], freq = 1;
 
    for (int i = 1; i < n; i++) {
 
        // If current element is smaller
        // than minimum
        if (arr[i] < mn) {
            mn = arr[i];
            freq = 1;
        }
        // If current element is equal
        // to smallest
        else if (arr[i] == mn)
            freq++;
    }
 
    return freq;
}
 
// Driver Code
int main()
{
    int N = 5;
    int arr[] = { 3, 2, 3, 4, 4 };
 
    cout << frequencyOfSmallest(N, arr);
 
    return 0;
}

// Java program to find the frequency of
// minimum element in the array
import java.io.*;
 
class GFG
{
     
// Function to find the frequency of the
// smallest value in the array
static int frequencyOfSmallest(int n, int arr[])
{
    int mn = arr[0], freq = 1;
 
    for (int i = 1; i < n; i++)
    {
 
        // If current element is smaller
        // than minimum
        if (arr[i] < mn)
        {
            mn = arr[i];
            freq = 1;
        }
         
        // If current element is equal
        // to smallest
        else if (arr[i] == mn)
            freq++;
    }
    return freq;
}
 
    // Driver Code
    public static void main (String[] args)
    {
        int N = 5;
        int arr[] = { 3, 2, 3, 4, 4 };
        System.out.println (frequencyOfSmallest(N, arr));
    }
}
 
// This code is contributed by Tushil.

# Python 3 program to find the frequency of
# minimum element in the array
 
 
# Function to find the frequency of the
# smallest value in the array
def frequencyOfSmallest(n,arr):
    mn = arr[0]
    freq = 1
 
    for i in range(1,n):
        # If current element is smaller
        # than minimum
        if (arr[i] < mn):
            mn = arr[i]
            freq = 1
       
        # If current element is equal
        # to smallest
        elif(arr[i] == mn):
            freq += 1
 
    return freq
 
# Driver Code
if __name__ == '__main__':
    N = 5
    arr = [3, 2, 3, 4, 4]
 
    print(frequencyOfSmallest(N, arr))
 
# This code is contributed by
# Surendra_Gangwar

// C# program to find the frequency of
// minimum element in the array
using System;
 
class GFG
{
         
    // Function to find the frequency of the
    // smallest value in the array
    static int frequencyOfSmallest(int n, int []arr)
    {
        int mn = arr[0], freq = 1;
     
        for (int i = 1; i < n; i++)
        {
     
            // If current element is smaller
            // than minimum
            if (arr[i] < mn)
            {
                mn = arr[i];
                freq = 1;
            }
             
            // If current element is equal
            // to smallest
            else if (arr[i] == mn)
                freq++;
        }
        return freq;
    }
     
    // Driver Code
    public static void Main()
    {
        int N = 5;
        int []arr = { 3, 2, 3, 4, 4 };
             
        Console.WriteLine(frequencyOfSmallest(N, arr));
    }
}
 
// This code is contributed by Ryuga

<?php
// PHP program to find the frequency of
// minimum element in the array
 
// Function to find the frequency of the
// smallest value in the array
function frequencyOfSmallest($n, $arr)
{
    $mn = $arr[0];
    $freq = 1;
 
    for ($i = 1; $i < $n; $i++)
    {
 
        // If current element is smaller
        // than minimum
        if ($arr[$i] < $mn)
        {
            $mn = $arr[$i];
            $freq = 1;
        }
        // If current element is equal
        // to smallest
        else if ($arr[$i] == $mn)
            $freq++;
    }
 
    return $freq;
}
 
// Driver Code
$N = 5;
$arr = array( 3, 2, 3, 4, 4 );
 
print(frequencyOfSmallest($N, $arr));
 
// This code is contributed by mits
?>

<script>
    // Javascript program to find the frequency of
    // minimum element in the array
    // required answer
     
    // Function to find the frequency of the
    // smallest value in the array
    function frequencyOfSmallest(n, arr)
    {
        let mn = arr[0], freq = 1;
       
        for (let i = 1; i < n; i++)
        {
       
            // If current element is smaller
            // than minimum
            if (arr[i] < mn)
            {
                mn = arr[i];
                freq = 1;
            }
               
            // If current element is equal
            // to smallest
            else if (arr[i] == mn)
                freq++;
        }
        return freq;
    }
     
    let N = 5;
    let arr = [ 3, 2, 3, 4, 4 ];
 
    document.write(frequencyOfSmallest(N, arr));
 
</script>

1

Complejidad de tiempo: O(N)