给定一个大小为n的arr 。问题是要计算具有最大数量的不同元素的所有子序列。
例子:
Input : arr[] = {4, 7, 6, 7}
Output : 2
The indexes for the subsequences are:
{0, 1, 2} - Subsequence is {4, 7, 6} and
{0, 2, 3} - Subsequence is {4, 6, 7}
Input : arr[] = {9, 6, 4, 4, 5, 9, 6, 1, 2}
Output : 8
天真的方法:考虑所有具有不同元素的子序列,并计算具有最大不同元素的子序列。
高效的方法:创建一个哈希表来存储数组中每个元素的频率。取所有频率的乘积。
该解决方案基于以下事实:当所有元素都不同时,总有1个子序列。如果元素重复,则每次出现重复元素都会构成一系列不同元素的子序列。
C++
// C++ implementation to count subsequences having
// maximum distinct elements
#include
using namespace std;
typedef unsigned long long int ull;
// function to count subsequences having
// maximum distinct elements
ull countSubseq(int arr[], int n)
{
// unordered_map 'um' implemented as
// hash table
unordered_map um;
ull count = 1;
// count frequency of each element
for (int i = 0; i < n; i++)
um[arr[i]]++;
// traverse 'um'
for (auto itr = um.begin(); itr != um.end(); itr++)
// multiply frequency of each element
// and accumulate it in 'count'
count *= (itr->second);
// required number of subsequences
return count;
}
// Driver program to test above
int main()
{
int arr[] = { 4, 7, 6, 7 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << "Count = "
<< countSubseq(arr, n);
return 0;
}
Java
// Java implementation to count subsequences having
// maximum distinct elements
import java.util.HashMap;
class geeks
{
// function to count subsequences having
// maximum distinct elements
public static long countSubseq(int[] arr, int n)
{
// unordered_map 'um' implemented as
// hash table
HashMap um = new HashMap<>();
long count = 1;
// count frequency of each element
for (int i = 0; i < n; i++)
{
if (um.get(arr[i]) != null)
{
int a = um.get(arr[i]);
um.put(arr[i], ++a);
}
else
um.put(arr[i], 1);
}
// traverse 'um'
for (HashMap.Entry entry : um.entrySet())
{
// multiply frequency of each element
// and accumulate it in 'count'
count *= entry.getValue();
}
// required number of subsequences
return count;
}
// Driver Code
public static void main(String[] args)
{
int[] arr = { 4, 7, 6, 7 };
int n = arr.length;
System.out.println("Count = " + countSubseq(arr, n));
}
}
// This code is contributed by
// sanjeev2552
Python3
# Python 3 implementation to count subsequences
# having maximum distinct elements
# function to count subsequences having
# maximum distinct elements
def countSubseq(arr, n):
# unordered_map 'um' implemented
# as hash table
# take range equal to maximum
# value of arr
um = {i:0 for i in range(8)}
count = 1
# count frequency of each element
for i in range(n):
um[arr[i]] += 1
# traverse 'um'
for key, values in um.items():
# multiply frequency of each element
# and accumulate it in 'count'
if(values > 0):
count *= values
# required number of subsequences
return count
# Driver Code
if __name__ == '__main__':
arr = [4, 7, 6, 7]
n = len(arr)
print("Count =", countSubseq(arr, n))
# This code is contributed by
# Surendra_Gangwar
C#
// C# implementation to count subsequences
// having maximum distinct elements
using System;
using System.Collections.Generic;
class GFG
{
// function to count subsequences having
// maximum distinct elements
public static long countSubseq(int[] arr,
int n)
{
// unordered_map 'um' implemented as
// hash table
Dictionary um = new Dictionary();
long count = 1;
// count frequency of each element
for (int i = 0; i < n; i++)
{
if (um.ContainsKey(arr[i]))
{
int a = um[arr[i]];
um.Remove(arr[i]);
um.Add(arr[i], ++a);
}
else
um.Add(arr[i], 1);
}
// traverse 'um'
foreach(KeyValuePair entry in um)
{
// multiply frequency of each element
// and accumulate it in 'count'
count *= entry.Value;
}
// required number of subsequences
return count;
}
// Driver Code
public static void Main(String[] args)
{
int[] arr = { 4, 7, 6, 7 };
int n = arr.Length;
Console.WriteLine("Count = " +
countSubseq(arr, n));
}
}
// This code is contributed by Princi Singh
输出:
Count = 2
时间复杂度: O(n)。
辅助空间: O(n)。