由相同元素组成的子序列数

给定一个由N个整数组成的数组A [] ，任务是查找子序列的总数，该子序列仅包含在整个子序列中重复的一个不同的数。

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，可通过以下公式计算所需的子序列数：

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

• 计算给定数组中可能存在的所有子序列。

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

输出：

11

时间复杂度： O(NlogN)
辅助空间： O(N)