Given an array arr[] of size N, the task is to find the length of the Longest Increasing Subsequence (LIS) i.e., the longest possible subsequence in which the elements of the subsequence are sorted in increasing order.
Examples:
Input: arr[] = {3, 10, 2, 1, 20}
Output: 3
Explanation: The longest increasing subsequence is 3, 10, 20Input: arr[] = {3, 2}
Output:1
Explanation: The longest increasing subsequences are {3} and {2}Input: arr[] = {50, 3, 10, 7, 40, 80}
Output: 4
Explanation: The longest increasing subsequence is {3, 7, 40, 80}
Longest Increasing Sequence using Recursion:
The problem can be solved based on the following idea:
Let L(i) be the length of the LIS ending at index i such that arr[i] is the last element of the LIS. Then, L(i) can be recursively written as:
- L(i) = 1 + max(L(j) ) where 0 < j < i and arr[j] < arr[i]; or
- L(i) = 1, if no such j exists.
Formally, the length of LIS ending at index i, is 1 greater than the maximum of lengths of all LIS ending at some index j such that arr[j] < arr[i] where j < i.
We can see that the above recurrence relation follows the optimal substructure property.
Below is the implementation of the recursive approach:
Python3
# A naive Python implementation of LIS problem# Global variable to store the maximumglobal maximum# To make use of recursive calls, this function must return# two things:# 1) Length of LIS ending with element arr[n-1]. We use# max_ending_here for this purpose# 2) Overall maximum as the LIS may end with an element# before arr[n-1] max_ref is used this purpose.# The value of LIS of full array of size n is stored in# *max_ref which is our final resultdef _lis(arr, n): # To allow the access of global variable global maximum # Base Case if n == 1: return 1 # maxEndingHere is the length of LIS ending with arr[n-1] maxEndingHere = 1 # Recursively get all LIS ending with # arr[0], arr[1]..arr[n-2] # If arr[i-1] is smaller than arr[n-1], and # max ending with arr[n-1] needs to be updated, # then update it for i in range(1, n): res = _lis(arr, i) if arr[i-1] < arr[n-1] and res+1 > maxEndingHere: maxEndingHere = res + 1 # Compare maxEndingHere with overall maximum. And # update the overall maximum if needed maximum = max(maximum, maxEndingHere) return maxEndingHeredef lis(arr): # To allow the access of global variable global maximum # Length of arr n = len(arr) # Maximum variable holds the result maximum = 1 # The function _lis() stores its result in maximum _lis(arr, n) return maximum# Driver program to test the above functionif __name__ == '__main__': arr = [10, 22, 9, 33, 21, 50, 41, 60] n = len(arr) # Function call print("Length of lis is", lis(arr))# This code is contributed by NIKHIL KUMAR SINGH |
Length of lis is 5
Complexity Analysis:
- Time Complexity: O(2n) The time complexity of this recursive approach is exponential as there is a case of overlapping subproblems as explained in the recursive tree diagram above.
- Auxiliary Space: O(1). No external space is used for storing values apart from the internal stack space.
Longest Increasing Subsequence using Memoization:
If noticed carefully, we can see that the above recursive solution also follows the overlapping subproblems property i.e., same substructure solved again and again in different recursion call paths. We can avoid this using the memoization approach.
We can see that each state can be uniquely identified using two parameters:
- Current index (denotes the last index of the LIS) and
- Previous index (denotes the ending index of the previous LIS behind which the arr[i] is being concatenated).
Below is the implementation of the above approach.
Python3
# A Naive Python recursive implementation# of LIS problemimport sys# To make use of recursive calls, this# function must return two things:# 1) Length of LIS ending with element arr[n-1].# We use max_ending_here for this purpose# 2) Overall maximum as the LIS may end with# an element before arr[n-1] max_ref is# used this purpose.# The value of LIS of full array of size n# is stored in *max_ref which is our final resultdef f(idx, prev_idx, n, a, dp): if (idx == n): return 0 if (dp[idx][prev_idx + 1] != -1): return dp[idx][prev_idx + 1] notTake = 0 + f(idx + 1, prev_idx, n, a, dp) take = -sys.maxsize - 1 if (prev_idx == -1 or a[idx] > a[prev_idx]): take = 1 + f(idx + 1, idx, n, a, dp) dp[idx][prev_idx + 1] = max(take, notTake) return dp[idx][prev_idx + 1]# Function to find length of longest increasing# subsequence.def longestSubsequence(n, a): dp = [[-1 for i in range(n + 1)]for j in range(n + 1)] return f(0, -1, n, a, dp)# Driver program to test above functionif __name__ == '__main__': a = [3, 10, 2, 1, 20] n = len(a) # Function call print("Length of lis is", longestSubsequence(n, a))# This code is contributed by shinjanpatra |
Length of lis is 3
Time Complexity: O(N2)
Auxiliary Space: O(N2)
Longest Increasing Subsequence using Dynamic Programming:
Because of the optimal substructure and overlapping subproblem property, we can also utilise Dynamic programming to solve the problem. Instead of memoization, we can use the nested loop to implement the recursive relation.
The outer loop will run from i = 1 to N and the inner loop will run from j = 0 to i and use the recurrence relation to solve the problem.
Below is the implementation of the above approach:
Python3
# Dynamic programming Python implementation# of LIS problem# lis returns length of the longest# increasing subsequence in arr of size ndef lis(arr): n = len(arr) # Declare the list (array) for LIS and # initialize LIS values for all indexes lis = [1]*n # Compute optimized LIS values in bottom up manner for i in range(1, n): for j in range(0, i): if arr[i] > arr[j] and lis[i] < lis[j] + 1: lis[i] = lis[j]+1 # Initialize maximum to 0 to get # the maximum of all LIS maximum = 0 # Pick maximum of all LIS values for i in range(n): maximum = max(maximum, lis[i]) return maximum# Driver program to test above functionif __name__ == '__main__': arr = [10, 22, 9, 33, 21, 50, 41, 60] print("Length of lis is", lis(arr))# This code is contributed by Nikhil Kumar Singh |
Length of lis is 5
Time Complexity: O(N2) As a nested loop is used.
Auxiliary Space: O(N) Use of any array to store LIS values at each index.
Please Refer Longest Increasing Subsequence for detailed article.
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