给定N个数字和整数K的数组。任务是打印长度为K的唯一子序列的数量。
例子：
Input : a[] = {1, 2, 3, 4}, k = 3
Output : 4.
Unique Subsequences are:
{1, 2, 3}, {1, 2, 4}, {1, 3, 4}, {2, 3, 4}
Input: a[] = {1, 1, 1, 2, 2, 2 }, k = 3
Output : 4
Unique Subsequences are
{1, 1, 1}, {1, 1, 2}, {1, 2, 2}, {2, 2, 2}
方法：有一个众所周知的公式，可以从N个唯一对象中选择多少个固定长度K的子序列。但是这里的问题有几个不同之处。其中之一是子序列的顺序很重要，必须保留原始序列中的顺序。对于这样的问题，可能没有现成的组合公式，因为结果取决于原始数组的顺序。
主要思想是按子序列的长度循环处理。在每个循环步骤中，从末尾移动到开始，并使用上一步中较短的唯一组合的数量对唯一组合进行计数。更严格地说，在每个步骤j上，我们都保留一个长度为N的数组，位置p中的每个元素表示在位置i的元素右侧(包括i本身)我们发现了多少个长度为j的唯一子序列。
下面是上述方法的实现。
C++
#include
using namespace std;
// Function which returns the numbe of
// unique subsequences of length K
int solution(vector& A, int k)
{
// seiz of the vector
// which does is constant
const int N = A.size();
// bases cases
if (N < k || N < 1 || k < 1)
return 0;
if (N == k)
return 1;
// Prepare arrays for recursion
vector v1(N, 0);
vector v2(N, 0);
vector v3(N, 0);
// initiate separately for k = 1
// intiate the last element
v2[N - 1] = 1;
v3[A[N - 1] - 1] = 1;
// initiate all other elements of k = 1
for (int i = N - 2; i >= 0; i--) {
// initialize the front element
// to vector v2
v2[i] = v2[i + 1];
// if element v[a[i]-1] is 0
// then increment it in vector v2
if (v3[A[i] - 1] == 0) {
v2[i]++;
v3[A[i] - 1] = 1;
}
}
// iterate for all possible values of K
for (int j = 1; j < k; j++) {
// fill the vectors with 0
fill(v3.begin(), v3.end(), 0);
// fill(v1.begin(), v1.end(), 0)
// the last must be 0 as from last no unique
// subarray can be formed
v1[N - 1] = 0;
// Iterate for all index from which unique
// subsequences can be formed
for (int i = N - 2; i >= 0; i--) {
// add the number of subsequence formed
// from the next index
v1[i] = v1[i + 1];
// start with combinations on the
// next index
v1[i] = v1[i] + v2[i + 1];
// Remove the elements which have
// already been counted
v1[i] = v1[i] - v3[A[i] - 1];
// Update the number used
v3[A[i] - 1] = v2[i + 1];
}
// prepare the next iteration
// by filling v2 in v1
v2 = v1;
}
// last answer is stored in v2
return v2[0];
}
// Function to push the vector into an array
// and print all the unique subarrays
void solve(int a[], int n, int k)
{
vector v;
// fill the vector with a[]
v.assign(a, a + n);
// Function call to print the count
// of unique susequences of size K
cout << solution(v, k);
}
// Driver Code
int main()
{
int a[] = { 1, 2, 3, 4 };
int n = sizeof(a) / sizeof(a[0]);
int k = 3;
solve(a, n, k);
return 0;
} Java
import java.util.*;
class GFG{
// Function which returns the numbe of
// unique subsequences of length K
static int solution(int[] A, int N, int k)
{
// Bases cases
if (N < k || N < 1 || k < 1)
return 0;
if (N == k)
return 1;
// Prepare arrays for recursion
int[] v1 = new int[N];
int[] v2 = new int[N];
int[] v3 = new int[N];
// Initiate separately for k = 1
// intiate the last element
v2[N - 1] = 1;
v3[A[N - 1] - 1] = 1;
// Initiate all other elements of k = 1
for(int i = N - 2; i >= 0; i--)
{
// Initialize the front element
// to vector v2
v2[i] = v2[i + 1];
// If element v[a[i]-1] is 0
// then increment it in vector v2
if (v3[A[i] - 1] == 0)
{
v2[i]++;
v3[A[i] - 1] = 1;
}
}
// Iterate for all possible values of K
for(int j = 1; j < k; j++)
{
// Fill the vectors with 0
Arrays.fill(v3, 0);
// Fill(v1.begin(), v1.end(), 0)
// the last must be 0 as from last
// no unique subarray can be formed
v1[N - 1] = 0;
// Iterate for all index from which
// unique subsequences can be formed
for(int i = N - 2; i >= 0; i--)
{
// Add the number of subsequence
// formed from the next index
v1[i] = v1[i + 1];
// Start with combinations on the
// next index
v1[i] = v1[i] + v2[i + 1];
// Remove the elements which have
// already been counted
v1[i] = v1[i] - v3[A[i] - 1];
// Update the number used
v3[A[i] - 1] = v2[i + 1];
}
}
// Last answer is stored in v2
return v2[0];
}
// Driver Code
public static void main(String[] args)
{
int a[] = { 1, 2, 3, 4 };
int n = a.length;
int k = 3;
System.out.print(solution(a, n, k));
}
}
// This code is contributed by amal kumar choubeyPython3
# Function which returns the numbe of
# unique subsequences of length K
def solution( A, k):
# seiz of the vector
# which does is constant
N = len(A)
# bases cases
if (N < k or N < 1 or k < 1):
return 0
if (N == k):
return 1
# Prepare arrays for recursion
v1 = [0]*(N)
v2 = [0]*N
v3 = [0]*N
# initiate separately for k = 1
# intiate the last element
v2[N - 1] = 1
v3[A[N - 1] - 1] = 1
# initiate all other elements of k = 1
for i in range(N - 2,-1,-1):
# initialize the front element
# to vector v2
v2[i] = v2[i + 1]
# if element v[a[i]-1] is 0
# then increment it in vector v2
if (v3[A[i] - 1] == 0):
v2[i] += 1
v3[A[i] - 1] = 1
# iterate for all possible values of K
for j in range( 1, k) :
# fill the vectors with 0
v3 = [0]*N
# fill(v1.begin(), v1.end(), 0)
# the last must be 0 as from last no unique
# subarray can be formed
v1[N - 1] = 0
# Iterate for all index from which unique
# subsequences can be formed
for i in range( N - 2, -1, -1) :
# add the number of subsequence formed
# from the next index
v1[i] = v1[i + 1]
# start with combinations on the
# next index
v1[i] = v1[i] + v2[i + 1]
# Remove the elements which have
# already been counted
v1[i] = v1[i] - v3[A[i] - 1]
# Update the number used
v3[A[i] - 1] = v2[i + 1]
# prepare the next iteration
# by filling v2 in v1
for i in range(len(v1)):
v2[i] = v1[i]
# last answer is stored in v2
return v2[0]
# Function to push the vector into an array
# and print all the unique subarrays
def solve(a, n, k):
# fill the vector with a[]
v = a
# Function call to print the count
# of unique susequences of size K
print( solution(v, k))
# Driver Code
if __name__ == "__main__":
a = [ 1, 2, 3, 4 ]
n = len(a)
k = 3
solve(a, n, k)
# This code is contributed by chitranayalC#
using System;
class GFG{
// Function which returns the numbe of
// unique subsequences of length K
static int solution(int[] A, int N, int k)
{
// Bases cases
if (N < k || N < 1 || k < 1)
return 0;
if (N == k)
return 1;
// Prepare arrays for recursion
int[] v1 = new int[N];
int[] v2 = new int[N];
int[] v3 = new int[N];
// Initiate separately for k = 1
// intiate the last element
v2[N - 1] = 1;
v3[A[N - 1] - 1] = 1;
// Initiate all other elements of k = 1
for(int i = N - 2; i >= 0; i--)
{
// Initialize the front element
// to vector v2
v2[i] = v2[i + 1];
// If element v[a[i]-1] is 0
// then increment it in vector v2
if (v3[A[i] - 1] == 0)
{
v2[i]++;
v3[A[i] - 1] = 1;
}
}
// Iterate for all possible values of K
for(int j = 1; j < k; j++)
{
// Fill the vectors with 0
for(int i = 0; i < v3.GetLength(0); i++)
v3[i] = 0;
// Fill(v1.begin(), v1.end(), 0)
// the last must be 0 as from last
// no unique subarray can be formed
v1[N - 1] = 0;
// Iterate for all index from which
// unique subsequences can be formed
for(int i = N - 2; i >= 0; i--)
{
// Add the number of subsequence
// formed from the next index
v1[i] = v1[i + 1];
// Start with combinations on the
// next index
v1[i] = v1[i] + v2[i + 1];
// Remove the elements which have
// already been counted
v1[i] = v1[i] - v3[A[i] - 1];
// Update the number used
v3[A[i] - 1] = v2[i + 1];
}
}
// Last answer is stored in v2
return v2[0];
}
// Driver Code
public static void Main(String[] args)
{
int []a = { 1, 2, 3, 4 };
int n = a.Length;
int k = 3;
Console.Write(solution(a, n, k));
}
}
// This code is contributed by Rohit_ranjan输出：
4