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.
Examples:
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) Input arr[] = {12, 11, 40, 5, 3, 1} Output: 5 (A Longest Bitonic Subsequence of length 5 is 12, 11, 5, 3, 1) Input arr[] = {80, 60, 30, 40, 20, 10} Output: 5 (A Longest Bitonic Subsequence of length 5 is 80, 60, 30, 20, 10)
Source: Microsoft Interview Question
Solution
This problem is a variation of standard Longest Increasing Subsequence (LIS) problem. Let the input array be arr[] of length n. We need to construct two arrays lis[] and lds[] using Dynamic Programming solution of LIS problem. lis[i] stores the length of the Longest Increasing subsequence ending with arr[i]. lds[i] stores the length of the longest Decreasing subsequence starting from arr[i]. Finally, we need to return the max value of lis[i] + lds[i] – 1 where i is from 0 to n-1.
Following is the implementation of the above Dynamic Programming solution.
# Dynamic Programming implementation of longest bitonic subsequence problem """ lbs() returns the length of the Longest Bitonic Subsequence in arr[] of size n. The function mainly creates two temporary arrays lis[] and lds[] and returns the maximum lis[i] + lds[i] - 1. lis[i] ==> Longest Increasing subsequence ending with arr[i] lds[i] ==> Longest decreasing subsequence starting with arr[i] """ def lbs(arr):
n = len (arr)
# 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 ((arr[i] > arr[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 (arr[i] > arr[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
# Driver program to test the above function arr = [ 0 , 8 , 4 , 12 , 2 , 10 , 6 , 14 , 1 , 9 , 5 , 13 ,
3 , 11 , 7 , 15 ]
print "Length of LBS is" ,lbs(arr)
# This code is contributed by Nikhil Kumar Singh(nickzuck_007) |
Output:
Length of LBS is 7
Time Complexity: O(n^2)
Auxiliary Space: O(n)
Please refer complete article on Longest Bitonic Subsequence | DP-15 for more details!