Longest Bitonic Subsequence
Created: March 3, 2020 by [lek-tin]
Last updated: March 3, 2020
Given an array arr[0 … n-1] containing n positive integers, a subsequence of arr[] is called Bitonic if it is first increasing, then decreasing. Write a function that takes an array as argument and returns the length of the longest bitonic subsequence.
A sequence, sorted in increasing order is considered Bitonic with the decreasing part as empty. Similarly, decreasing order sequence is considered Bitonic with the increasing part as empty.
Example 1
Input arr[] = {1, 11, 2, 10, 4, 5, 2, 1};
Output: 6 (A Longest Bitonic Subsequence of length 6 is 1, 2, 10, 4, 2, 1)
Example 2
Input arr[] = {12, 11, 40, 5, 3, 1}
Output: 5 (A Longest Bitonic Subsequence of length 5 is 12, 11, 5, 3, 1)
Example 3
Input arr[] = {80, 60, 30, 40, 20, 10}
Output: 5 (A Longest Bitonic Subsequence of length 5 is 80, 60, 30, 20, 10)
Solution
def lbs(nums):
n = len(nums)
# allocate memory for LIS[] and initialize LIS values as 1
# for all indexes
lis = [1 for i in range(n+1)]
# Compute LIS values from left to right
for i in range(1 , n):
for j in range(0 , i):
if ((nums[i] > nums[j]) and (lis[i] < lis[j] +1)):
lis[i] = lis[j] + 1
# allocate memory for LDS and initialize LDS values for
# all indexes
lds = [1 for i in range(n+1)]
# Compute LDS values from right to left
for i in reversed(range(n-1)): #loop from n-2 downto 0
for j in reversed(range(i-1 ,n)): #loop from n-1 downto i-1
if(nums[i] > nums[j] and lds[i] < lds[j] + 1):
lds[i] = lds[j] + 1
# Return the maximum value of (lis[i] + lds[i] - 1)
maximum = lis[0] + lds[0] - 1
for i in range(1 , n):
maximum = max((lis[i] + lds[i]-1), maximum)
return maximum