Problem Statement

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)

Problem Link

Longest Bitonic Subsequence | DP-15 - GeeksforGeeks

Code

class Solution{
	public:
	int LongestBitonicSequence(vector<int>nums)
	{
	    int n = nums.size();
	    
	    vector<int> lis(n),lisr(n);
	    
	    for(int i=0;i<n;i++)
	    {
	        lis[i] = 1;
	        lisr[i] = 1;
	    }
	    
	    for(int i=1;i<n;i++)
	    {
	        for(int j=0;j<i;j++)
	        {
	            if(nums[i]>nums[j] and lis[i]<1+lis[j])
	            lis[i] = 1+lis[j];
	        }
	    }
	    
	    for(int i=n-2;i>=0;i--)
	    {
	        for(int j=n-1;j>i;j--)
	        {
	            if(nums[i]>nums[j] and lisr[i]<1+lisr[j])
	            lisr[i] = 1+lisr[j];
	        }
	    }
	    int res = 0;
	    
	    for(int i=0;i<n;i++)
	    {
	        res = max(res,lis[i]+lisr[i]-1);
	    }
	    return res;
	}
};
// Time Complexity : O(n^2)