Given an array arr[] of N elements and a positive integer K such that K ≤ N. The task is to find the number of subsequences of maximum length K i.e. subsequences of length 0, 1, 2, …, K – 1, K that have all distinct elements.
Examples:
Input: arr[] = {2, 2, 3, 3, 5}, K = 2
Output: 14
All the valid subsequences are {}, {2}, {2}, {3}, {3}, {5},
{2, 3}, {2, 3}, {2, 3}, {2, 3}, {2, 5}, {2, 5}, {3, 5} and {3, 5}.Input: arr[] = {1, 2, 3, 4, 4}, K = 4
Output: 24
Approach:
- Sort the array a[] if it is not already sorted and in a vector arr[] store the frequencies of each element of the original array. For example, if a[] = {2, 2, 3, 3, 5} then arr[] = {2, 2, 1} because 2 is present twice, 3 is present twice and 5 only once.
- Say m is the length of the vector arr[]. So m will be the number of distinct elements. There can be subsequences of maximum length m without repetition. If m < k then there is no subsequence of length k. So, declare n = minimum(m, k).
- Now apply dynamic programming. Create a 2-d array dp[n + 1][m + 1] such that dp[i][j] will store the number of subsequences of length i whose first element begins after j-th element from arr[]. For example, dp[1][1] = 3 because it means number
of subsequences of length 1 whose first element starts after 1-st element of arr[] which are {3}, {3}, {5}.- Initialize first row of dp[][] to 1.
- Run two loops top to bottom and right to left inside of the former loop.
- If j > m – i that means there cannot be any such sequences due to lack of elements. So dp[i][j] = 0.
- Else, dp[i][j] = dp[i][j + 1] + arr[j] * dp[i – 1][j + 1] as the number will be number of already existing subsequences of length i plus the number of subsequences of length i – 1 multiplied by arr[j] due to repetition.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Returns number of subsequences// of maximum length k and// contains no repeated elementint countSubSeq(int a[], int n, int k){ // Sort the array a[] sort(a, a + n); vector<int> arr; // Store the frequencies of all the // distinct element in the vector arr for (int i = 0; i < n;) { int count = 1, x = a[i]; i++; while (i < n && a[i] == x) { count++; i++; } arr.push_back(count); } int m = arr.size(); n = min(m, k); // count is the number // of such subsequences int count = 1; // Create a 2-d array dp[n+1][m+1] to // store the intermediate result int dp[n + 1][m + 1]; // Initialize the first row to 1 for (int i = 0; i <= m; i++) dp[0][i] = 1; // Update the dp[][] array based // on the recurrence relation for (int i = 1; i <= n; i++) { for (int j = m; j >= 0; j--) { if (j > m - i) dp[i][j] = 0; else { dp[i][j] = dp[i][j + 1] + arr[j] * dp[i - 1][j + 1]; } } count = count + dp[i][0]; } // Return the number of subsequences return count;}// Driver codeint main(){ int a[] = { 2, 2, 3, 3, 5 }; int n = sizeof(a) / sizeof(int); int k = 3; cout << countSubSeq(a, n, k); return 0;} |
Java
// Java implementation of the approachimport java.util.*;class GFG {// Returns number of subsequences// of maximum length k and// contains no repeated elementstatic int countSubSeq(int a[], int n, int k){ // Sort the array a[] Arrays.sort(a); List<Integer> arr = new LinkedList<>(); // Store the frequencies of all the // distinct element in the vector arr for (int i = 0; i < n;) { int count = 1, x = a[i]; i++; while (i < n && a[i] == x) { count++; i++; } arr.add(count); } int m = arr.size(); n = Math.min(m, k); // count is the number // of such subsequences int count = 1; // Create a 2-d array dp[n+1][m+1] to // store the intermediate result int [][]dp = new int[n + 1][m + 1]; // Initialize the first row to 1 for (int i = 0; i <= m; i++) dp[0][i] = 1; // Update the dp[][] array based // on the recurrence relation for (int i = 1; i <= n; i++) { for (int j = m; j >= 0; j--) { if (j > m - i) dp[i][j] = 0; else { dp[i][j] = dp[i][j + 1] + arr.get(j) * dp[i - 1][j + 1]; } } count = count + dp[i][0]; } // Return the number of subsequences return count;}// Driver codepublic static void main(String[] args) { int a[] = { 2, 2, 3, 3, 5 }; int n = a.length; int k = 3; System.out.println(countSubSeq(a, n, k));}}// This code is contributed by PrinciRaj1992 |
Python 3
# Python 3 implementation of the approach# Returns number of subsequences# of maximum length k and# contains no repeated elementdef countSubSeq(a, n, k): # Sort the array a[] a.sort(reverse = False) arr = [] # Store the frequencies of all the # distinct element in the vector arr i = 0 while(i < n): count = 1 x = a[i] i += 1 while (i < n and a[i] == x): count += 1 i += 1 arr.append(count) m = len(arr) n = min(m, k) # count is the number # of such subsequences count = 1 # Create a 2-d array dp[n+1][m+1] to # store the intermediate result dp = [[0 for i in range(m + 1)] for j in range(n + 1)] # Initialize the first row to 1 for i in range(m + 1): dp[0][i] = 1 # Update the dp[][] array based # on the recurrence relation for i in range(1, n + 1, 1): j = m while(j >= 0): if (j > m - i): dp[i][j] = 0 else: dp[i][j] = dp[i][j + 1] + \ arr[j] * dp[i - 1][j + 1] j -= 1 count = count + dp[i][0] # Return the number of subsequences return count# Driver codeif __name__ == '__main__': a = [2, 2, 3, 3, 5] n = len(a) k = 3 print(countSubSeq(a, n, k))# This code is contributed by Surendra_Gangwar |
C#
// C# implementation of the approachusing System;using System.Collections.Generic; class GFG {// Returns number of subsequences// of maximum length k and// contains no repeated elementstatic int countSubSeq(int []a, int n, int k){ // Sort the array a[] Array.Sort(a); List<int> arr = new List<int>(); int count, x; // Store the frequencies of all the // distinct element in the vector arr for (int i = 0; i < n;) { count = 1; x = a[i]; i++; while (i < n && a[i] == x) { count++; i++; } arr.Add(count); } int m = arr.Count; n = Math.Min(m, k); // count is the number // of such subsequences count = 1; // Create a 2-d array dp[n+1][m+1] to // store the intermediate result int [,]dp = new int[n + 1, m + 1]; // Initialize the first row to 1 for (int i = 0; i <= m; i++) dp[0, i] = 1; // Update the dp[][] array based // on the recurrence relation for (int i = 1; i <= n; i++) { for (int j = m; j >= 0; j--) { if (j > m - i) dp[i, j] = 0; else { dp[i, j] = dp[i, j + 1] + arr[j] * dp[i - 1, j + 1]; } } count = count + dp[i, 0]; } // Return the number of subsequences return count;}// Driver codepublic static void Main(String[] args) { int []a = { 2, 2, 3, 3, 5 }; int n = a.Length; int k = 3; Console.WriteLine(countSubSeq(a, n, k));}}// This code is contributed by 29AjayKumar |
Javascript
<script>// Javascript implementation of the approach// Returns number of subsequences// of maximum length k and// contains no repeated elementfunction countSubSeq(a, n, k){ // Sort the array a[] a.sort(); var arr = []; // Store the frequencies of all the // distinct element in the vector arr for (var i = 0; i < n;) { var count = 1, x = a[i]; i++; while (i < n && a[i] == x) { count++; i++; } arr.push(count); } var m = arr.length; n = Math.min(m, k); // count is the number // of such subsequences var count = 1; // Create a 2-d array dp[n+1][m+1] to // store the intermediate result var dp = Array.from(Array(n+1), ()=>Array(m+1)); // Initialize the first row to 1 for (var i = 0; i <= m; i++) dp[0][i] = 1; // Update the dp[][] array based // on the recurrence relation for (var i = 1; i <= n; i++) { for (var j = m; j >= 0; j--) { if (j > m - i) dp[i][j] = 0; else { dp[i][j] = dp[i][j + 1] + arr[j] * dp[i - 1][j + 1]; } } count = count + dp[i][0]; } // Return the number of subsequences return count;}// Driver codevar a = [2, 2, 3, 3, 5];var n = a.length;var k = 3;document.write( countSubSeq(a, n, k));</script> |
Output
18
Time Complexity: O(n*log(n)+n*m) where m is the size of the array and n=min(m,k).
Auxiliary Space: O(n*m)
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Returns number of subsequences// of maximum length k and// contains no repeated elementint countSubSeq(int a[], int n, int k){ // Sort the array a[] sort(a, a + n); vector<int> arr; // Store the frequencies of all the // distinct element in the vector arr for (int i = 0; i < n;) { int count = 1, x = a[i]; i++; while (i < n && a[i] == x) { count++; i++; } arr.push_back(count); } // count is the number // of such subsequences int m = arr.size(); n = min(m, k); int count = 1; // to store computations of subproblems vector<int> dp(m + 1, 1); // iterate over subproblems to // get the current solution from previous computations for (int i = 1; i <= n; i++) { // vector to store current values vector<int> curr(m + 1, 0); for (int j = m; j >= 0; j--) { if (j > m - i) curr[j] = 0; else { curr[j] = curr[j + 1] + arr[j] * dp[j + 1]; } } // assigning values ot iterate further dp = curr; // update count count = count + dp[0]; } // return final answer return count;}// Driver codeint main(){ int a[] = { 2, 2, 3, 3, 5 }; int n = sizeof(a) / sizeof(int); int k = 3; cout << countSubSeq(a, n, k); return 0;} |
Java
import java.util.Arrays;public class GFG { public static int countSubSeq(int[] a, int k) { // Sort the input array in ascending order Arrays.sort(a); int[] arr = new int[a.length]; // Count the occurrences of each unique element in the sorted array int m = 0; for (int i = 0; i < a.length; i++) { int count = 1; int x = a[i]; while (i + 1 < a.length && a[i + 1] == x) { count++; i++; } arr[m++] = count; } m = Math.min(m, k); int count = 1; int[] dp = new int[m + 1]; Arrays.fill(dp, 1); // Dynamic programming to calculate the count of subsequences for (int i = 1; i <= m; i++) { int[] curr = new int[m + 1]; for (int j = m; j >= 0; j--) { if (j > m - i) { curr[j] = 0; } else { curr[j] = curr[j + 1] + arr[j] * dp[j + 1]; } } dp = curr; count += dp[0]; } return count; } public static void main(String[] args) { int[] a = {2, 2, 3, 3, 5}; // Given input array int k = 3; // Maximum number of elements in a subsequence // Calculate and print the total count of subsequences with at most 'k' elements System.out.println(countSubSeq(a, k)); }} |
Python
def countSubSeq(a, k): """ Function to count the number of subsequences of an array 'a' with at most 'k' elements. Args: a (list): The input list containing the elements. k (int): The maximum number of elements in a subsequence. Returns: int: The total count of subsequences with at most 'k' elements. """ a.sort() # Sort the input list in ascending order arr = [] # Count the occurrences of each unique element in the sorted array i = 0 while i < len(a): count = 1 x = a[i] i += 1 while i < len(a) and a[i] == x: count += 1 i += 1 arr.append(count) m = len(arr) n = min(m, k) count = 1 dp = [1] * (m + 1) # Dynamic programming to calculate the count of subsequences for i in range(1, n + 1): curr = [0] * (m + 1) for j in range(m, -1, -1): if j > m - i: curr[j] = 0 else: curr[j] = curr[j + 1] + arr[j] * dp[j + 1] dp = curr count += dp[0] return count# Driver codea = [2, 2, 3, 3, 5] # Given input listk = 3 # Maximum number of elements in a subsequence# Calculate and print the total count of subsequences with at most 'k' elementsprint(countSubSeq(a, k))#user_dtewbxkn77n |
C#
using System;public class GFG { public static int CountSubSeq(int[] a, int k) { // Sort the input array in ascending order Array.Sort(a); int[] arr = new int[a.Length]; // Count the occurrences of each unique element in // the sorted array int m = 0; for (int i = 0; i < a.Length; i++) { int count = 1; int x = a[i]; while (i + 1 < a.Length && a[i + 1] == x) { count++; i++; } arr[m++] = count; } m = Math.Min(m, k); int totalCount = 1; // Renamed 'count' to 'totalCount' int[] dp = new int[m + 1]; Array.Fill(dp, 1); // Dynamic programming to calculate the count of // subsequences for (int i = 1; i <= m; i++) { int[] curr = new int[m + 1]; for (int j = m; j >= 0; j--) { if (j > m - i) { curr[j] = 0; } else { curr[j] = curr[j + 1] + arr[j] * dp[j + 1]; } } dp = curr; totalCount += dp[0]; // Renamed 'count' to 'totalCount' } return totalCount; // Renamed 'count' to // 'totalCount' } public static void Main() { int[] a = { 2, 2, 3, 3, 5 }; // Given input array int k = 3; // Maximum number of elements in a // subsequence // Calculate and print the total count of // subsequences with at most 'k' elements Console.WriteLine(CountSubSeq(a, k)); }} |
Javascript
// Returns number of subsequences// of maximum length k and// contains no repeated elementfunction countSubSeq(a, n, k) { // Sort the array a[] a.sort((x, y) => x - y); const arr = []; // Store the frequencies of all the // distinct element in the array arr let i = 0; while (i < n) { let count = 1; const x = a[i]; i++; while (i < n && a[i] === x) { count++; i++; } arr.push(count); } // count is the number // of such subsequences let m = arr.length; n = Math.min(m, k); let count = 1; // to store computations of subproblems const dp = new Array(m + 1).fill(1); // iterate over subproblems to // get the current solution from previous computations for (let i = 1; i <= n; i++) { // array to store current values const curr = new Array(m + 1).fill(0); for (let j = m; j >= 0; j--) { if (j > m - i) curr[j] = 0; else { curr[j] = curr[j + 1] + arr[j] * dp[j + 1]; } } // assigning values to iterate further dp.splice(0, dp.length, ...curr); // update count count = count + dp[0]; } // return final answer return count;}// Driver codeconst a = [2, 2, 3, 3, 5];const n = a.length;const k = 3;console.log(countSubSeq(a, n, k)); |
Output
18
Time Complexity: O(n*log(n)+n*m) where m is the size of the array and n=min(m,k).
Auxiliary Space: O(m)
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