# Longest Bitonic Subsequence | DP-15

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.

## C++

 `/* Dynamic Programming implementation of longest bitonic subsequence problem */` `#include ` `#include ` ` `  `/* 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 + lds - 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); ` `  ``printf``(````"Length of LBS is %d "````, 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 + lds - 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. `

## PHP

 ` Longest Increasing subsequence ` `              ``ending with arr[i] ` `   ``lds[i] ==> Longest decreasing subsequence  ` `              ``starting with arr[i] ` `*/` `function` `lbs(&``\$arr``, ``\$n``) ` `{ ` ` `  `    ``/* Allocate memory for LIS[] and initialize  ` `       ``LIS values as 1 for all indexes */` `    ``\$lis` `= ``array_fill``(0, ``\$n``, NULL); ` `    ``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 */` `    ``\$lds` `= ``array_fill``(0, ``\$n``, NULL); ` `    ``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*/` `    ``\$max` `= ``\$lis`` + ``\$lds`` - 1; ` `    ``for` `(``\$i` `= 1; ``\$i` `< ``\$n``; ``\$i``++) ` `        ``if` `(``\$lis``[``\$i``] + ``\$lds``[``\$i``] - 1 > ``\$max``) ` `            ``\$max` `= ``\$lis``[``\$i``] + ``\$lds``[``\$i``] - 1; ` `    ``return` `\$max``; ` `} ` ` `  `// Driver Code ` `\$arr` `= ``array``(0, 8, 4, 12, 2, 10, 6, 14,  ` `             ``1, 9, 5, 13, 3, 11, 7, 15); ` `\$n` `= sizeof(``\$arr``); ` `echo` `"Length of LBS is "` `. lbs( ``\$arr``, ``\$n` `); ` ` `  `// This code is contributed by ita_c ` `?> `

Output:

``` Length of LBS is 7
```

Time Complexity: O(n^2)
Auxiliary Space: O(n)