Number of Longest Increasing Subsequences
Given an array arr[] of size N, the task is to count the number of longest increasing subsequences present in the given array.
Examples:
Input: arr[] = {2, 2, 2, 2, 2}
Output: 5
Explanation: The length of the longest increasing subsequence is 1, i.e. {2}. Therefore, count of longest increasing subsequences of length 1 is 5.Input: arr[] = {1, 3, 5, 4, 7}
Output: 2
Explanation: The length of the longest increasing subsequence is 4, and there are 2 longest increasing subsequences of length 4, i.e. {1, 3, 4, 7} and {1, 3, 5, 7}.
Naive Approach: The simplest approach is to generate all possible subsequences present in the given array arr[] and count the increasing subsequences of maximum length. Print the count after checking all subsequences.
Time Complexity: O(N*2N)
Auxiliary Space: O(1)
Efficient Approach: To optimize the above approach, the idea is to use Dynamic Programming as the above problem has overlapping subproblems that need to be calculated more than once, and to reduce that calculation use tabulation or memoization. Follow the steps below to solve the problem:
- Initialize two arrays dp_l[] and dp_c[] to store the length of the longest increasing subsequences and the count of the longest increasing subsequence at each index respectively.
- Iterate over the range [1, N – 1] using the variable i:
- Iterate over the range [0, i – 1] using the variable j:
- If arr[i] > arr[j] then check for the following cases:
~ If (dp_l[j]+1) greater than dp_l[i], then update dp_l[i] as dp_l[j] + 1 and dp_c[i] as dp_c[j]
~ Else if (dp_l[j] + 1) is the same as dp_l[i], then update dp_c[i] as dp_c[i] + dp_c[j].
- If arr[i] > arr[j] then check for the following cases:
- Iterate over the range [0, i – 1] using the variable j:
- Find the maximum element in the array dp_l[] and store it in a variable max_length that will give the length of LIS.
- Initialize a variable count with 0 to store the number of the longest increasing subsequence.
- Traverse the array dp_l[] and if at any index idx, dp_l[idx] is the same as max_length then increment the count by dp_c[idx].
- After the above steps, print the value of count, which is the number of longest increasing subsequences in the given array.
Below is the implementation of the above approach:
C++
// C++ program for the // above approach #include <bits/stdc++.h> using namespace std; //Function to count the number // of LIS in the array nums[] int findNumberOfLIS(vector< int > nums) { //Base Case if (nums.size() == 0) return 0; int n = nums.size(); //Initialize dp_l array with // 1s vector< int > dp_l(n, 1); //Initialize dp_c array with // 1s vector< int > dp_c(n, 1); for ( int i = 0; i < n; i++) { for ( int j = 0; j < i; j++) { //If current element is // smaller if (nums[i] <= nums[j]) continue ; if (dp_l[j] + 1 > dp_l[i]) { dp_l[i] = dp_l[j] + 1; dp_c[i] = dp_c[j]; } else if (dp_l[j] + 1 == dp_l[i]) dp_c[i] += dp_c[j]; } } //Store the maximum element // from dp_l int max_length = 0; for ( int i : dp_l) max_length = max(i,max_length); //Stores the count of LIS int count = 0; //Traverse dp_l and dp_c // simultaneously for ( int i = 0; i < n; i++) { //Update the count if (dp_l[i] == max_length) count += dp_c[i]; } //Return the count of LIS return count; } //Driver code int main() { //Given array arr[] vector< int > arr = {1, 3, 5, 4, 7}; //Function Call cout << findNumberOfLIS(arr) << endl; } // This code is contributed by Mohit Kumar 29 |
Java
// Java program for the // above approach import java.util.*; class GFG{ // Function to count the number // of LIS in the array nums[] static int findNumberOfLIS( int [] nums) { // Base Case if (nums.length == 0 ) return 0 ; int n = nums.length; // Initialize dp_l array with // 1s int [] dp_l = new int [n]; Arrays.fill(dp_l, 1 ); // Initialize dp_c array with // 1s int [] dp_c = new int [n]; Arrays.fill(dp_c, 1 ); for ( int i = 0 ; i < n; i++) { for ( int j = 0 ; j < i; j++) { // If current element is // smaller if (nums[i] <= nums[j]) continue ; if (dp_l[j] + 1 > dp_l[i]) { dp_l[i] = dp_l[j] + 1 ; dp_c[i] = dp_c[j]; } else if (dp_l[j] + 1 == dp_l[i]) dp_c[i] += dp_c[j]; } } // Store the maximum element // from dp_l int max_length = 0 ; for ( int i : dp_l) max_length = Math.max(i, max_length); // Stores the count of LIS int count = 0 ; // Traverse dp_l and dp_c // simultaneously for ( int i = 0 ; i < n; i++) { // Update the count if (dp_l[i] == max_length) count += dp_c[i]; } // Return the count of LIS return count; } // Driver code public static void main(String[] args) { // Given array arr[] int [] arr = { 1 , 3 , 5 , 4 , 7 }; // Function Call System.out.print(findNumberOfLIS(arr) + "\n" ); } } // This code is contributed by shikhasingrajput |
Python3
# Python3 program for the above approach # Function to count the number of LIS # in the array nums[] def findNumberOfLIS(nums): # Base Case if not nums: return 0 n = len (nums) # Initialize dp_l array with 1s dp_l = [ 1 ] * n # Initialize dp_c array with 1s dp_c = [ 1 ] * n for i, num in enumerate (nums): for j in range (i): # If current element is smaller if nums[i] < = nums[j]: continue # Otherwise if dp_l[j] + 1 > dp_l[i]: dp_l[i] = dp_l[j] + 1 dp_c[i] = dp_c[j] elif dp_l[j] + 1 = = dp_l[i]: dp_c[i] + = dp_c[j] # Store the maximum element from dp_l max_length = max (x for x in dp_l) # Stores the count of LIS count = 0 # Traverse dp_l and dp_c simultaneously for l, c in zip (dp_l, dp_c): # Update the count if l = = max_length: count + = c # Return the count of LIS return count # Driver Code # Given array arr[] arr = [ 1 , 3 , 5 , 4 , 7 ] # Function Call print (findNumberOfLIS(arr)) |
C#
// C# program to implement // the above approach using System; using System.Collections.Generic; class GFG { // Function to count the number // of LIS in the array nums[] static int findNumberOfLIS( int [] nums) { // Base Case if (nums.Length == 0) return 0; int n = nums.Length; // Initialize dp_l array with // 1s int [] dp_l = new int [n]; Array.Fill(dp_l, 1); // Initialize dp_c array with // 1s int [] dp_c = new int [n]; Array.Fill(dp_c, 1); for ( int i = 0; i < n; i++) { for ( int j = 0; j < i; j++) { // If current element is // smaller if (nums[i] <= nums[j]) continue ; if (dp_l[j] + 1 > dp_l[i]) { dp_l[i] = dp_l[j] + 1; dp_c[i] = dp_c[j]; } else if (dp_l[j] + 1 == dp_l[i]) dp_c[i] += dp_c[j]; } } // Store the maximum element // from dp_l int max_length = 0; foreach ( int i in dp_l) max_length = Math.Max(i, max_length); // Stores the count of LIS int count = 0; // Traverse dp_l and dp_c // simultaneously for ( int i = 0; i < n; i++) { // Update the count if (dp_l[i] == max_length) count += dp_c[i]; } // Return the count of LIS return count; } // Driver code static void Main() { // Given array arr[] int [] arr = { 1, 3, 5, 4, 7 }; // Function Call Console.WriteLine(findNumberOfLIS(arr)); } } // This code is contributed by divyeshrabadiya07 |
Javascript
<script> // Javascript program for the // above approach // Function to count the number // of LIS in the array nums[] function findNumberOfLIS(nums) { //Base Case if (nums.length == 0) return 0; var n = nums.length; // Initialize dp_l array with // 1s var dp_l = Array(n).fill(1); // Initialize dp_c array with // 1s var dp_c = Array(n).fill(1); for ( var i = 0; i < n; i++) { for ( var j = 0; j < i; j++) { // If current element is // smaller if (nums[i] <= nums[j]) continue ; if (dp_l[j] + 1 > dp_l[i]) { dp_l[i] = dp_l[j] + 1; dp_c[i] = dp_c[j]; } else if (dp_l[j] + 1 == dp_l[i]) dp_c[i] += dp_c[j]; } } // Store the maximum element // from dp_l var max_length = 0; dp_l.forEach(i => { max_length = Math.max(i,max_length); }); // Stores the count of LIS var count = 0; //Traverse dp_l and dp_c // simultaneously for ( var i = 0; i < n; i++) { // Update the count if (dp_l[i] == max_length) count += dp_c[i]; } // Return the count of LIS return count; } // Driver code // Given array arr[] var arr = [1, 3, 5, 4, 7]; // Function Call document.write( findNumberOfLIS(arr)); // This code is contributed by rutvik_56. </script> |
2
Time Complexity: O(N2)
Auxiliary Space: O(N)
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