竞争性编程的频率测量技术

测量数组中元素的频率是一项非常方便的技能，并且需要很多竞争性编码问题。在很多问题中，我们需要测量各种元素（如数字、字母、符号等）的频率，作为我们问题的一部分。

朴素的方法

例子：

Input :  arr[] = {10, 20, 20, 10, 10, 20, 5, 20}
Output : 10 3
         20 4
         5  1

Input : arr[] = {10, 20, 20}
Output : 10 2
         20 1

我们运行两个循环。对于每个项目计数它发生的次数。为避免重复打印，请跟踪处理的项目。

C++

// C++ program to count frequencies of array items
#include 
using namespace std;
  
void countFreq(int arr[], int n)
{
    // Mark all array elements as not visited
    vector visited(n, false);
  
    // Traverse through array elements and
    // count frequencies
    for (int i = 0; i < n; i++) {
  
        // Skip this element if already processed
        if (visited[i] == true)
            continue;
  
        // Count frequency
        int count = 1;
        for (int j = i + 1; j < n; j++) {
            if (arr[i] == arr[j]) {
                visited[j] = true;
                count++;
            }
        }
        cout << arr[i] << " " << count << endl;
    }
}
  
int main()
{
    int arr[] = { 10, 20, 20, 10, 10, 20, 5, 20 };
    int n = sizeof(arr) / sizeof(arr[0]);
    countFreq(arr, n);
    return 0;
}

Java

// Java program to count frequencies
// of array items
import java.util.*;
 
class GFG
{
static void countFreq(int arr[], int n)
{
    // Mark all array elements as not visited
    boolean []visited = new boolean[n];
 
    // Traverse through array elements and
    // count frequencies
    for (int i = 0; i < n; i++)
    {
 
        // Skip this element if already processed
        if (visited[i] == true)
            continue;
 
        // Count frequency
        int count = 1;
        for (int j = i + 1; j < n; j++)
        {
            if (arr[i] == arr[j])
            {
                visited[j] = true;
                count++;
            }
        }
        System.out.println(arr[i] + " " + count);
    }
}
 
// Driver Code
public static void main(String[] args)
{
    int arr[] = { 10, 20, 20, 10, 10, 20, 5, 20 };
    int n = arr.length;
    countFreq(arr, n);
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python program to count frequencies
# of array items
def countFreq(arr, n):
     
    # mark all array elements as not visited
    visited = [False for i in range(n)]
     
    # Traverse through array elements
    # and count frequencies
    for i in range(n):
         
        # Skip this element if already processed
        if visited[i] == True:
            continue
             
        # count frequency
        count = 1
         
        for j in range(i + 1, n):
            if arr[i] == arr[j]:
                visited[j] = True
                count += 1
        print(arr[i], count)
     
# Driver code
a = [10, 20, 20, 10, 10, 20, 5, 20]
 
n = len(a)
 
countFreq(a, n)
 
# This code is contributed
# by Mohit kumar 29

C#

// C# program to count frequencies
// of array items
using System;
     
class GFG
{
static void countFreq(int []arr, int n)
{
    // Mark all array elements as not visited
    Boolean []visited = new Boolean[n];
 
    // Traverse through array elements and
    // count frequencies
    for (int i = 0; i < n; i++)
    {
 
        // Skip this element if already processed
        if (visited[i] == true)
            continue;
 
        // Count frequency
        int count = 1;
        for (int j = i + 1; j < n; j++)
        {
            if (arr[i] == arr[j])
            {
                visited[j] = true;
                count++;
            }
        }
        Console.WriteLine(arr[i] + " " + count);
    }
}
 
// Driver Code
public static void Main(String[] args)
{
    int []arr = { 10, 20, 20, 10,
                  10, 20, 5, 20 };
    int n = arr.Length;
    countFreq(arr, n);
}
}
 
// This code is contributed by Rajput-Ji

Javascript

    // Javascript program to count frequencies of array items
     
    function countFreq(arr, n)
    {
        // Mark all array elements as not visited
        let visited = new Array(n);
        visited.fill(false);
 
        // Traverse through array elements and
        // count frequencies
        for (let i = 0; i < n; i++) {
 
            // Skip this element if already processed
            if (visited[i] == true)
                continue;
 
            // Count frequency
            let count = 1;
            for (let j = i + 1; j < n; j++) {
                if (arr[i] == arr[j]) {
                    visited[j] = true;
                    count++;
                }
            }
            document.write(arr[i] + " " + count + "</br>");
        }
    }
 
    let arr = [ 10, 20, 20, 10, 10, 20, 5, 20 ];
    let n = arr.length;
    countFreq(arr, n);
     
    // This code is contributed by mukesh07.

输出：

10 3
20 4
5 1

优化方法：

当元素受值限制时测量频率
如果我们的输入数组的值很小，我们可以使用数组元素作为计数数组中的索引并增加计数。在下面的示例中，元素最多为 10。

Input :  arr[] = {5, 5, 6, 6, 5, 6, 1, 2, 3, 10, 10}
         limit = 10
Output : 1 1
         2 1
         3 1
         5 3
         6 3
         10 2

C++

// C++ program to count frequencies of array items
// having small values.
#include 
using namespace std;
  
void countFreq(int arr[], int n, int limit)
{
    // Create an array to store counts. The size
    // of array is limit+1 and all values are
    // initially 0
    vector count(limit+1, 0);
  
    // Traverse through array elements and
    // count frequencies (assuming that elements
    // are limited by limit)
    for (int i = 0; i < n; i++)
        count[arr[i]]++;
  
    for (int i = 0; i <= limit; i++)
       if (count[i] > 0)
        cout << i << " " << count[i] << endl;
     
}
  
int main()
{
    int arr[] = {5, 5, 6, 6, 5, 6, 1, 2, 3, 10, 10};
    int n = sizeof(arr) / sizeof(arr[0]);
    int limit = 10;
    countFreq(arr, n, limit);
    return 0;
}

Java

// Java program to count frequencies of array items
// having small values.
class GFG
{
     
static void countFreq(int arr[], int n, int limit)
{
    // Create an array to store counts. The size
    // of array is limit+1 and all values are
    // initially 0
    int []count = new int[limit + 1];
 
    // Traverse through array elements and
    // count frequencies (assuming that elements
    // are limited by limit)
    for (int i = 0; i < n; i++)
        count[arr[i]]++;
 
    for (int i = 0; i <= limit; i++)
    if (count[i] > 0)
            System.out.println(i + " " + count[i]);
}
 
// Driver Code
public static void main(String[] args)
{
    int arr[] = {5, 5, 6, 6, 5, 6, 1, 2, 3, 10, 10};
    int n = arr.length;
    int limit = 10;
    countFreq(arr, n, limit);
}
}
 
// This code is contributed by PrinciRaj1992

蟒蛇3

# Python3 program to count frequencies of
# array items having small values.
def countFreq(arr, n, limit):
     
    # Create an array to store counts.
    # The size of array is limit+1 and
    # all values are initially 0
    count = [0 for i in range(limit + 1)]
 
    # Traverse through array elements and
    # count frequencies (assuming that
    # elements are limited by limit)
    for i in range(n):
        count[arr[i]] += 1
 
    for i in range(limit + 1):
        if (count[i] > 0):
            print(i, count[i])
 
# Driver Code
arr = [ 5, 5, 6, 6, 5, 6,
        1, 2, 3, 10, 10 ]
n = len(arr)
limit = 10
 
countFreq(arr, n, limit)
 
# This code is contributed by avanitrachhadiya2155

C＃

// C# program to count frequencies of
// array items having small values.
using System;
     
class GFG
{
     
static void countFreq(int []arr,
                      int n, int limit)
{
    // Create an array to store counts.
    // The size of array is limit+1 and
    // all values are initially 0
    int []count = new int[limit + 1];
 
    // Traverse through array elements and
    // count frequencies (assuming that
    // elements are limited by limit)
    for (int i = 0; i < n; i++)
        count[arr[i]]++;
 
    for (int i = 0; i <= limit; i++)
    if (count[i] > 0)
            Console.WriteLine(i + " " +
                              count[i]);
}
 
// Driver Code
public static void Main(String[] args)
{
    int []arr = {5, 5, 6, 6, 5, 6,
                 1, 2, 3, 10, 10};
    int n = arr.Length;
    int limit = 10;
    countFreq(arr, n, limit);
}
}
 
// This code is contributed
// by Princi Singh

Javascript

 
// Javascript program to count frequencies
// of array items having small values.
function countFreq(arr, n, limit)
{
     
    // Create an array to store counts.
    // The size of array is limit+1 and
    // all values are initially 0
    let count = new Array(limit + 1);
    count.fill(0);
 
    // Traverse through array elements and
    // count frequencies (assuming that
    // elements are limited by limit)
    for(let i = 0; i < n; i++)
        count[arr[i]]++;
 
    for(let i = 0; i <= limit; i++)
        if (count[i] > 0)
            document.write(i + " " +
                           count[i] + "</br>");
}
 
// Driver code
let arr = [ 5, 5, 6, 6, 5, 6,
            1, 2, 3, 10, 10 ];
let n = arr.length;
let limit = 10;
 
countFreq(arr, n, limit);
 
// This code is contributed by rameshtravel07  
 

输出：

1 1
2 1
3 1
5 3
6 3
10 2

C++

// C++ program to count frequencies of array items
#include 
using namespace std;
 
const int limit = 255;
 
void countFreq(string str)
{
    // Create an array to store counts. The size
    // of array is limit+1 and all values are
    // initially 0
    vector count(limit+1, 0);
  
    // Traverse through string characters and
    // count frequencies
    for (int i = 0; i < str.length(); i++)
        count[str[i]]++;
  
    for (int i = 0; i <= limit; i++)
       if (count[i] > 0)
        cout << (char)i << " " << count[i] << endl;
     
}
  
int main()
{
    string str = "GeeksforGeeks";
    countFreq(str);
    return 0;
}

Java

// Java program to count frequencies of array items
class GFG
{
static int limit = 255;
 
static void countFreq(String str)
{
    // Create an array to store counts. The size
    // of array is limit+1 and all values are
    // initially 0
    int []count= new int[limit + 1];
 
    // Traverse through string characters and
    // count frequencies
    for (int i = 0; i < str.length(); i++)
        count[str.charAt(i)]++;
 
    for (int i = 0; i <= limit; i++)
    if (count[i] > 0)
        System.out.println((char)i + " " + count[i]);
}
 
// Driver Code
public static void main(String[] args)
{
    String str = "GeeksforGeeks";
    countFreq(str);
}
}
 
// This code is contributed by PrinciRaj1992

蟒蛇3

# Python3 program to count frequencies of array items
limit = 255
def countFreq(Str) :
 
    # Create an array to store counts. The size
    # of array is limit+1 and all values are
    # initially 0
    count = [0] * (limit + 1)
   
    # Traverse through string characters and
    # count frequencies
    for i in range(len(Str)) :
        count[ord(Str[i])] += 1
   
    for i in range(limit + 1) :
       if (count[i] > 0) :
        print(chr(i), count[i])
  
Str = "GeeksforGeeks"
countFreq(Str)
 
# This code is contributed by divyeshrabadiya07

C＃

// C# program to count frequencies
// of array items
using System;
     
class GFG
{
static int limit = 25;
 
static void countFreq(String str)
{
    // Create an array to store counts.
    // The size of array is limit+1 and
    // all values are initially 0
    int []count = new int[limit + 1];
 
    // Traverse through string characters
    // and count frequencies
    for (int i = 0; i < str.Length; i++)
        count[str[i] - 'A']++;
 
    for (int i = 0; i <= limit; i++)
    if (count[i] > 0)
        Console.WriteLine((char)(i + 'A') +
                          " " + count[i]);
}
 
// Driver Code
public static void Main(String[] args)
{
    String str = "GEEKSFORGEEKS";
    countFreq(str);
}
}
 
// This code is contributed by PrinciRaj1992

Javascript

 
// Javascript program to count frequencies of array items
     
    let limit = 255;
     
function countFreq(str)
{
        // Create an array to store counts. The size
    // of array is limit+1 and all values are
    // initially 0
    let count= new Array(limit + 1);
    for(let i=0;i<count.length;i++)
    {
        count[i]=0;
    }
  
    // Traverse through string characters and
    // count frequencies
    for (let i = 0; i < str.length; i++)
        count[str[i].charCodeAt(0)]++;
  
    for (let i = 0; i <= limit; i++)
    {    if (count[i] > 0)
            document.write(String.fromCharCode(i) + " " + count[i]+"<br>");
    }
}
 
// Driver Code
let str = "GeeksforGeeks";
countFreq(str);
     
     
    // This code is contributed by unknown2108

输出：

G 2
e 4
f 1
k 2
o 1
r 1
s 2

当元件在有限范围内时测量频率
例如，考虑一个只包含大写字母的字符串。字符串的元素限制在从“A”到“Z”的范围内。这个想法是减去最小元素（在本例中为’A’）以获得元素的索引。

C++

// C++ program to count frequencies of array items
#include 
using namespace std;
 
const int limit = 25;
 
void countFreq(string str)
{
    // Create an array to store counts. The size
    // of array is limit+1 and all values are
    // initially 0
    vector count(limit+1, 0);
  
    // Traverse through string characters and
    // count frequencies
    for (int i = 0; i < str.length(); i++)
        count[str[i] - 'A']++;
  
    for (int i = 0; i <= limit; i++)
       if (count[i] > 0)
        cout << (char)(i + 'A') << " " << count[i] << endl;
     
}
  
int main()
{
    string str = "GEEKSFORGEEKS";
    countFreq(str);
    return 0;
}

Java

// Java program to count frequencies
// of array items
import java.util.*;
 
class GFG
{
static int limit = 25;
 
static void countFreq(String str)
{
    // Create an array to store counts.
    // The size of array is limit+1 and
    // all values are initially 0
    int []count = new int[limit + 1];
 
    // Traverse through string characters
    // and count frequencies
    for (int i = 0; i < str.length(); i++)
        count[str.charAt(i) - 'A']++;
 
    for (int i = 0; i <= limit; i++)
    if (count[i] > 0)
        System.out.println((char)(i + 'A') +
                           " " + count[i]);
}
 
// Driver Code
public static void main(String[] args)
{
    String str = "GEEKSFORGEEKS";
    countFreq(str);
}
}
 
// This code is contributed by PrinciRaj1992

蟒蛇3

# Python3 program to count frequencies of array items
limit = 25;
def countFreq(str):
     
    # Create an array to store counts. The size
    # of array is limit+1 and all values are
    # initially 0
    count = [0 for i in range(limit + 1)]
   
    # Traverse through string characters and
    # count frequencies
    for i in range(len(str)):  
        count[ord(str[i]) - ord('A')] += 1
   
    for i in range(limit + 1):
       if (count[i] > 0):
           print(chr(i + ord('A')), count[i])
 
# Driver code
if __name__=='__main__':
     
    str = "GEEKSFORGEEKS";
    countFreq(str);
     
# This code is contributed by Pratham76

C＃

// C# program to count frequencies
// of array items
using System;
     
class GFG
{
static int limit = 25;
  
static void countFreq(String str)
{
    // Create an array to store counts.
    // The size of array is limit+1 and
    // all values are initially 0
    int []count = new int[limit + 1];
  
    // Traverse through string characters
    // and count frequencies
    for (int i = 0; i < str.Length; i++)
        count[str[i] - 'A']++;
  
    for (int i = 0; i <= limit; i++)
    if (count[i] > 0)
        Console.WriteLine((char)(i + 'A') +
                           " " + count[i]);
}
  
// Driver Code
public static void Main(String[] args)
{
    String str = "GEEKSFORGEEKS";
    countFreq(str);
}
}
  
// This code contributed by PrinciRaj1992

Javascript

 
// JavaScript program to count frequencies
// of array items
 
let limit = 25;
function countFreq(str)
{
    // Create an array to store counts.
    // The size of array is limit+1 and
    // all values are initially 0
    let count = new Array(limit + 1);
    for(let i=0;i<limit+1;i++)
    {
        count[i]=0;
    }
  
    // Traverse through string characters
    // and count frequencies
    for (let i = 0; i < str.length; i++)
        count[str[i].charCodeAt(0) - 'A'.charCodeAt(0)]++;
  
    for (let i = 0; i <= limit; i++)
    if (count[i] > 0)
        document.write(String.fromCharCode(i + 'A'.charCodeAt(0)) +
                           " " + count[i]+"<br>");
}
 
// Driver Code
let str = "GEEKSFORGEEKS";
countFreq(str);
 
// This code is contributed by rag2127
 

输出：

E 4
F 1
G 2
K 2
O 1
R 1
S 2

如果没有范围和限制，测量频率
这个想法是使用散列（C++ 中的 unordered_map 和Java的HashMap ）来获取频率。

C++

// C++ program to count frequencies of array items
#include 
using namespace std;
 
void countFreq(int arr[], int n)
{
    unordered_map mp;
 
    // Traverse through array elements and
    // count frequencies
    for (int i = 0; i < n; i++)
        mp[arr[i]]++;
 
    // Traverse through map and print frequencies
    for (auto x : mp)
        cout << x.first << " " << x.second << endl;
}
 
int main()
{
    int arr[] = { 10, 20, 20, 10, 10, 20, 5, 20 };
    int n = sizeof(arr) / sizeof(arr[0]);
    countFreq(arr, n);
    return 0;
}

Java

// Java program to count frequencies of array items
import java.util.*;
class GFG
{
static void countFreq(int arr[], int n)
{
    HashMapmp = new HashMap();
                                      
    // Traverse through array elements and
    // count frequencies
    for (int i = 0 ; i < n; i++)
    {
        if(mp.containsKey(arr[i]))
        {
            mp.put(arr[i], mp.get(arr[i]) + 1);
        }
        else
        {
            mp.put(arr[i], 1);
        }
    }
     
    // Traverse through map and print frequencies
    for (Map.Entry entry : mp.entrySet())
        System.out.println(entry.getKey() + " " +
                           entry.getValue());
}
 
// Driver Code
public static void main(String[] args)
{
    int arr[] = { 10, 20, 20, 10, 10, 20, 5, 20 };
    int n = arr.length;
    countFreq(arr, n);   
}
}
 
// This code is contributed by 29AjayKumar

蟒蛇3

# Python3 program to count frequencies of array items
def countFreq(arr, n):
 
    mp = {}
  
    # Traverse through array elements and
    # count frequencies
    for i in range(n):
        if arr[i] in mp:
            mp[arr[i]] += 1
        else:
            mp[arr[i]] = 1
  
    # Traverse through map and print frequencies
    for x in sorted(mp):
        print(x, mp[x])
 
# Driver Code
arr = [ 10, 20, 20, 10, 10, 20, 5, 20 ]
n = len(arr)
 
countFreq(arr, n)
 
# This code is contributed by divyesh072019

C＃

// C# program to count frequencies of array items
using System;
using System.Collections.Generic;
 
class GFG
{
static void countFreq(int []arr, int n)
{
    Dictionary mp = new Dictionary();
                                     
    // Traverse through array elements and
    // count frequencies
    for (int i = 0 ; i < n; i++)
    {
        if(mp.ContainsKey(arr[i]))
        {
            var val = mp[arr[i]];
            mp.Remove(arr[i]);
            mp.Add(arr[i], val + 1);
        }
        else
        {
            mp.Add(arr[i], 1);
        }
    }
     
    // Traverse through map and print frequencies
    foreach(KeyValuePair entry in mp)
        Console.WriteLine(entry.Key + " " +
                          entry.Value);
}
 
// Driver Code
public static void Main(String[] args)
{
    int []arr = { 10, 20, 20, 10,
                  10, 20, 5, 20 };
    int n = arr.Length;
    countFreq(arr, n);
}
}
 
// This code is contributed by Princi Singh

Javascript

// Javascript program to count frequencies of array items
function  countFreq(arr, n)
{
    let mp = new Map();
                                       
    // Traverse through array elements and
    // count frequencies
    for (let i = 0 ; i < n; i++)
    {
        if(mp.has(arr[i]))
        {
            mp.set(arr[i], mp.get(arr[i]) + 1);
        }
        else
        {
            mp.set(arr[i], 1);
        }
    }
      
    // Traverse through map and print frequencies
    for (let [key, value] of mp.entries())
        document.write(key + " " +
                           value+"<br>");
}
 
// Driver Code
let arr=[10, 20, 20, 10, 10, 20, 5, 20];
let n = arr.length;
countFreq(arr, n);  
 
// This code is contributed by ab2127

输出：

5 1
10 3
20 4

时间复杂度： O(n)
辅助空间： O(n)

在上述有效的解决方案中，如何以与输入中出现的相同顺序打印元素？

C++

// C++ program to count frequencies of array items
#include 
using namespace std;
 
void countFreq(int arr[], int n)
{
    unordered_map mp;
 
    // Traverse through array elements and
    // count frequencies
    for (int i = 0; i < n; i++)
        mp[arr[i]]++;
 
    // To print elements according to first
    // occurrence, traverse array one more time
    // print frequencies of elements and mark
    // frequencies as -1 so that same element
    // is not printed multiple times.
    for (int i = 0; i < n; i++) {
      if (mp[arr[i]] != -1)
      {
          cout << arr[i] << " " << mp[arr[i]] << endl;
          mp[arr[i]] = -1;
      }
    }
}
 
int main()
{
    int arr[] = { 10, 20, 20, 10, 10, 20, 5, 20 };
    int n = sizeof(arr) / sizeof(arr[0]);
    countFreq(arr, n);
    return 0;
}

Java

// Java program to count frequencies of array items
import java.util.*;
 
class GFG
{
static void countFreq(int arr[], int n)
{
    HashMap mp = new HashMap();
 
    // Traverse through array elements and
    // count frequencies
    for (int i = 0 ; i < n; i++)
    {
        if(mp.containsKey(arr[i]))
        {
            mp.put(arr[i], mp.get(arr[i]) + 1);
        }
        else
        {
            mp.put(arr[i], 1);
        }
    }
     
    // To print elements according to first
    // occurrence, traverse array one more time
    // print frequencies of elements and mark
    // frequencies as -1 so that same element
    // is not printed multiple times.
    for (int i = 0; i < n; i++)
    {
        if (mp.get(arr[i]) != -1)
        {
            System.out.println(arr[i] + " " +
                               mp.get(arr[i]));
            mp. put(arr[i], -1);
        }
    }
}
 
// Driver Code
public static void main(String[] args)
{
    int arr[] = { 10, 20, 20, 10, 10, 20, 5, 20 };
    int n = arr.length;
    countFreq(arr, n);
}
}
 
// This code is contributed by Princi Singh

蟒蛇3

# Python3 program to count frequencies of array items
def countFreq(arr, n):
    mp = {}
     
    # Traverse through array elements and
    # count frequencies
    for i in range(n):
        if arr[i] not in mp:
            mp[arr[i]] = 1
        else:
            mp[arr[i]] += 1
 
    # To print elements according to first
    # occurrence, traverse array one more time
    # print frequencies of elements and mark
    # frequencies as -1 so that same element
    # is not printed multiple times.
    for i in range(n):
        if(mp[arr[i]] != -1):
            print(arr[i], mp[arr[i]])
            mp[arr[i]] = -1
 
# Driver code
arr = [10, 20, 20, 10, 10, 20, 5, 20 ]
n = len(arr)
countFreq(arr, n)
 
# This code is contributed by rag2127

C＃

// C# program to count frequencies of array items
using System;
using System.Collections.Generic;
     
class GFG
{
static void countFreq(int []arr, int n)
{
    Dictionary mp = new Dictionary();
 
    // Traverse through array elements and
    // count frequencies
    for (int i = 0 ; i < n; i++)
    {
        if(mp.ContainsKey(arr[i]))
        {
            mp[arr[i]] = mp[arr[i]] + 1;
        }
        else
        {
            mp.Add(arr[i], 1);
        }
    }
     
    // To print elements according to first
    // occurrence, traverse array one more time
    // print frequencies of elements and mark
    // frequencies as -1 so that same element
    // is not printed multiple times.
    for (int i = 0; i < n; i++)
    {
        if (mp[arr[i]] != -1)
        {
            Console.WriteLine(arr[i] + " " +
                              mp[arr[i]]);
            mp[arr[i]] = - 1;
        }
    }
}
 
// Driver Code
public static void Main(String[] args)
{
    int []arr = { 10, 20, 20, 10,
                  10, 20, 5, 20 };
    int n = arr.Length;
    countFreq(arr, n);
}
}
 
// This code is contributed by Rajput-Ji

Javascript

 
// Javascript program to count
// frequencies of array items
function countFreq(arr, n)
{
    let mp = new Map();
  
    // Traverse through array elements and
    // count frequencies
    for(let i = 0 ; i < n; i++)
    {
        if (mp.has(arr[i]))
        {
            mp.set(arr[i], mp.get(arr[i]) + 1);
        }
        else
        {
            mp.set(arr[i], 1);
        }
    }
      
    // To print elements according to first
    // occurrence, traverse array one more time
    // print frequencies of elements and mark
    // frequencies as -1 so that same element
    // is not printed multiple times.
    for(let i = 0; i < n; i++)
    {
        if (mp.get(arr[i]) != -1)
        {
            document.write(arr[i] + " " +
                    mp.get(arr[i]) + "<br>");
            mp.set(arr[i], -1);
        }
    }
}
 
// Driver Code
let arr = [ 10, 20, 20, 10, 10, 20, 5, 20 ];
let n = arr.length;   
 
countFreq(arr, n);
 
// This code is contributed by patel2127
 

输出：

10 3
20 4
5 1

时间复杂度： O(n)
辅助空间： O(n)

在Java，我们可以使用 LinkedHashMap 以相同的顺序获取元素。因此我们不需要额外的循环。
很多问题都基于频率测量，如果我们知道如何计算给定数组中各种元素的频率，这将是一个问题。例如，尝试以下基于频率测量的问题：

1. 字谜
2. 按频率对数组的元素进行排序
3. 单号