查找数组中最小值的频率

给定一个由N个元素组成的数组A。在数组中找到最小值的频率。

例子：

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.

简单方法：一个简单的方法是在第一次遍历中首先找到数组中的最小元素。然后再次遍历数组，找到最小元素的出现次数。

高效的方法：高效的方法是一次遍历。
让我们假设第一个元素是当前最小值，因此当前最小值的频率将是1。现在让我们遍历数组(从1到N)，我们遇到2种情况：

• 当前元素小于我们的当前最小值：我们将当前最小值更改为等于当前元素，并且由于这是我们第一次遇到此元素，因此将其设为频率1。
• 当前元素等于当前最小值：我们增加当前最小值的频率。

下面是上述方法的实现：

C++

// C++ program to find the frequency of
// minimum element in the array
  
#include 
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

// 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.

Python3

# 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#

// 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

输出：

1

时间复杂度： O(N)