Find frequency of smallest value in an array

Given an array A of N elements. Find the frequency of the smallest value in the array.

Examples: 

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.

Simple Approach: A simple method is to first find the minimum element in the array in first traversal. Then traverse the array again and find the number of occurrences of the minimum element.

Efficient Approach: The efficient approach is to do this in a single traversal. 
Let us assume the first element to be the current minimum, so the frequency of the current minimum would be 1. Now let's iterate through the array (1 to N), we come across 2 cases: 

• Current element is smaller than our current minimum: We change the current minimum to be equal to current element and since this is the first time we are coming across this element, we make its frequency 1.
• Current element is equal to the current minimum: We increment the frequency of current minimum.

Below is the implementation of the above approach:

// 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

Time Complexity: O(N), where N is the size of the given array.
Auxiliary Space: O(1), no extra space is required, so it is a constant.