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 C++ implementation of the above Dynamic Programming solution.

## C++

/* Dynamic Programming implementation of longest bitonic subsequence problem */ #include<stdio.h> #include<stdlib.h> /* 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] */ int lbs( int arr[], int n ) { int i, j; /* Allocate memory for LIS[] and initialize LIS values as 1 for all indexes */ int *lis = new int[n]; for (i = 0; i < n; i++) lis[i] = 1; /* Compute LIS values from left to right */ for (i = 1; i < n; i++) for (j = 0; j < i; j++) if (arr[i] > arr[j] && lis[i] < lis[j] + 1) lis[i] = lis[j] + 1; /* Allocate memory for lds and initialize LDS values for all indexes */ int *lds = new int [n]; for (i = 0; i < n; i++) lds[i] = 1; /* Compute LDS values from right to left */ for (i = n-2; i >= 0; i--) for (j = n-1; j > i; j--) if (arr[i] > arr[j] && lds[i] < lds[j] + 1) lds[i] = lds[j] + 1; /* Return the maximum value of lis[i] + lds[i] - 1*/ int max = lis[0] + lds[0] - 1; for (i = 1; i < n; i++) if (lis[i] + lds[i] - 1 > max) max = lis[i] + lds[i] - 1; return max; } /* Driver program to test above function */ int main() { int arr[] = {0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15}; int n = sizeof(arr)/sizeof(arr[0]); printf("Length of LBS is %d\n", lbs( arr, n ) ); return 0; }

## Java

/* Dynamic Programming implementation in Java for longest bitonic subsequence problem */ import java.util.*; import java.lang.*; import java.io.*; class LBS { /* 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] */ static int lbs( int arr[], int n ) { int i, j; /* Allocate memory for LIS[] and initialize LIS values as 1 for all indexes */ int[] lis = new int[n]; for (i = 0; i < n; i++) lis[i] = 1; /* Compute LIS values from left to right */ for (i = 1; i < n; i++) for (j = 0; j < i; j++) if (arr[i] > arr[j] && lis[i] < lis[j] + 1) lis[i] = lis[j] + 1; /* Allocate memory for lds and initialize LDS values for all indexes */ int[] lds = new int [n]; for (i = 0; i < n; i++) lds[i] = 1; /* Compute LDS values from right to left */ for (i = n-2; i >= 0; i--) for (j = n-1; j > i; j--) if (arr[i] > arr[j] && lds[i] < lds[j] + 1) lds[i] = lds[j] + 1; /* Return the maximum value of lis[i] + lds[i] - 1*/ int max = lis[0] + lds[0] - 1; for (i = 1; i < n; i++) if (lis[i] + lds[i] - 1 > max) max = lis[i] + lds[i] - 1; return max; } public static void main (String[] args) { int arr[] = {0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15}; int n = arr.length; System.out.println("Length of LBS is "+ lbs( arr, n )); } }

## Python

# 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)

## C#

/* Dynamic Programming implementation in C# for longest bitonic subsequence problem */ using System; class LBS { /* 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] */ static int lbs(int[] arr, int n) { int i, j; /* Allocate memory for LIS[] and initialize LIS values as 1 for all indexes */ int[] lis = new int[n]; for (i = 0; i < n; i++) lis[i] = 1; /* Compute LIS values from left to right */ for (i = 1; i < n; i++) for (j = 0; j < i; j++) if (arr[i] > arr[j] && lis[i] < lis[j] + 1) lis[i] = lis[j] + 1; /* Allocate memory for lds and initialize LDS values for all indexes */ int[] lds = new int[n]; for (i = 0; i < n; i++) lds[i] = 1; /* Compute LDS values from right to left */ for (i = n - 2; i >= 0; i--) for (j = n - 1; j > i; j--) if (arr[i] > arr[j] && lds[i] < lds[j] + 1) lds[i] = lds[j] + 1; /* Return the maximum value of lis[i] + lds[i] - 1*/ int max = lis[0] + lds[0] - 1; for (i = 1; i < n; i++) if (lis[i] + lds[i] - 1 > max) max = lis[i] + lds[i] - 1; return max; } // Driver code public static void Main() { int[] arr = { 0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15 }; int n = arr.Length; Console.WriteLine("Length of LBS is " + lbs(arr, n)); } } // This code is contributed by vt_m.

Output:

Length of LBS is 7

Time Complexity: O(n^2)

Auxiliary Space: O(n)

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