由相同元素组成的子序列的计数

给定一个由N 个整数组成的数组A[] ，任务是找到在整个子序列中只包含一个重复的不同数字的子序列的总数。

例子：

Input: A[] = {1, 2, 1, 5, 2} 
Output: 7 
Explanation: 
Subsequences {1}, {2}, {1}, {5}, {2}, {1, 1} and {2, 2} satisfy the required conditions.

Input: A[] = {5, 4, 4, 5, 10, 4} 
Output: 11 
Explanation: 
Subsequences {5}, {4}, {4}, {5}, {10}, {4}, {5, 5}, {4, 4}, {4, 4}, {4, 4} and {4, 4, 4} satisfy the required conditions. 

方法：
请按照以下步骤解决问题：

• 遍历数组并计算 HashMap 中每个元素的频率。
• 遍历HashMap。对于每个元素，通过等式计算可能的所需子序列的数量：

 Number of subsequences possible by arr[i] = 2^freq[arr[i]] – 1 

• 计算给定数组中可能的总子序列。

下面是上述方法的实现：

C++

// C++ program to implement
// the above approach
#include 
using namespace std;
 
// Function to count subsequences in
// array containing same element
void CountSubSequence(int A[], int N)
{
    // Stores the count
    // of subsequences
    int result = 0;
 
    // Stores the frequency
    // of array elements
    map mp;
 
    for (int i = 0; i < N; i++) {
 
        // Update frequency of A[i]
        mp[A[i]]++;
    }
 
    for (auto it : mp) {
 
        // Calculate number of subsequences
        result
            = result + pow(2, it.second) - 1;
    }
 
    // Print the result
    cout << result << endl;
}
 
// Driver code
int main()
{
    int A[] = { 5, 4, 4, 5, 10, 4 };
 
    int N = sizeof(A) / sizeof(A[0]);
 
    CountSubSequence(A, N);
 
    return 0;
}

Java

// Java program to implement
// the above approach
import java.util.*;
 
class GFG{
 
// Function to count subsequences in
// array containing same element
static void CountSubSequence(int A[], int N)
{
     
    // Stores the count
    // of subsequences
    int result = 0;
 
    // Stores the frequency
    // of array elements
    Map mp = new HashMap();
 
    for(int i = 0; i < N; i++)
    {
         
        // Update frequency of A[i]
        mp.put(A[i], mp.getOrDefault(A[i], 0) + 1);
    }
 
    for(Integer it : mp.values())
    {
         
        // Calculate number of subsequences
        result = result + (int)Math.pow(2, it) - 1;
    }
     
    // Print the result
    System.out.println(result);
}
 
// Driver code
public static void main(String[] args)
{
    int A[] = { 5, 4, 4, 5, 10, 4 };
    int N = A.length;
     
    CountSubSequence(A, N);
}
}
 
// This code is contributed by offbeat

Python3

# Python3 program to implement 
# the above approach 
 
# Function to count subsequences in 
# array containing same element 
def CountSubSequence(A, N):
     
    # Stores the frequency 
    # of array elements 
    mp = {}
     
    for element in A:
        if element in mp:
            mp[element] += 1
        else:
            mp[element] = 1
             
    result = 0
     
    for key, value in mp.items():
         
        # Calculate number of subsequences 
        result += pow(2, value) - 1
         
    # Print the result     
    print(result)
 
# Driver code
A = [ 5, 4, 4, 5, 10, 4 ]
N = len(A)
 
CountSubSequence(A, N)
 
# This code is contributed by jojo9911

C#

// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to count subsequences in
// array containing same element
public static void CountSubSequence(int []A, int N)
{
     
    // Stores the count
    // of subsequences
    int result = 0;
 
    // Stores the frequency
    // of array elements
    var mp = new Dictionary();
 
    for(int i = 0; i < N; i++)
    {
         
        // Update frequency of A[i]
        if(mp.ContainsKey(A[i]))
            mp[A[i]] += 1;
        else
            mp.Add(A[i], 1);
    }
 
    foreach(var it in mp)
    {
         
        // Calculate number of subsequences
        result = result +
                 (int)Math.Pow(2, it.Value) - 1;
    }
     
    // Print the result
    Console.Write(result);
}
 
// Driver code
public static void Main()
{
    int []A = { 5, 4, 4, 5, 10, 4 };
    int N = A.Length;
     
    CountSubSequence(A, N);
}
}
 
// This code is contributed by grand_master

Javascript

 
// Javascript program to implement
// the above approach
 
// Function to count subsequences in
// array containing same element
function CountSubSequence(A, N)
{
    // Stores the count
    // of subsequences
    var result = 0;
 
    // Stores the frequency
    // of array elements
    var mp = new Map(); 
 
    for (var i = 0; i < N; i++) {
 
        // Update frequency of A[i]
        if(mp.has(A[i]))
            mp.set(A[i], mp.get(A[i])+1)
        else
            mp.set(A[i], 1)
    }
 
    mp.forEach((value, key) => {
         
 
        // Calculate number of subsequences
        result
            = result + Math.pow(2, value) - 1;
    });
 
    // Print the result
    document.write( result );
}
 
// Driver code
var A = [5, 4, 4, 5, 10, 4];
var N = A.length;
CountSubSequence(A, N);
 
// This code is contributed by itsok.

输出：

11

时间复杂度： O(NlogN)
辅助空间： O(N)