# Count unique subsequences of length K

Given an array of N numbers and an integer K. The task is to print the number of unique subsequences possible of length K.

Examples:

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}

**Approach:** There is a well-known formula how many subsequences of fixed length K can be chosen from N unique objects. But the problem here has several differences. One among them is the order in subsequences is important and must be preserved as in the original sequence. For such a problem there can be no ready combinatorics formula because the results depend on the order of the original array.

The main idea is to deal recurrently by the length of the subsequence. On each recurrent step, move from the end to the beginning and count the unique combinations using the count of shorter unique combinations from the previous step. More strictly on every step j we keep an array of length N and every element in the place p means how many unique subsequences with length j we found to the right of the element in place i, including i itself.

Below is the implementation of the above approach.

`#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Function which returns the numbe of ` `// unique subsequences of length K ` `int` `solution(vector<` `int` `>& 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<` `int` `> v1(N, 0); ` ` ` `vector<` `int` `> v2(N, 0); ` ` ` `vector<` `int` `> 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 v1 ` ` ` `return` `v1[0]; ` `} ` ` ` `// Function to push the vector into an array ` `// and print all the unique subarrays ` `void` `solve(` `int` `a[], ` `int` `n, ` `int` `k) ` `{ ` ` ` `vector<` `int` `> 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; ` `} ` |

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**Output:**

4

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