The Longest Bitonic Subsequence problem is to find the longest subsequence of a given sequence such that it is first increasing and then decreasing. 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:[1, 11, 2, 10, 4, 5, 2, 1]Output:[1, 2, 10, 4, 2, 1] OR [1, 11, 10, 5, 2, 1] OR [1, 2, 4, 5, 2, 1]Input:[12, 11, 40, 5, 3, 1]Output:[12, 11, 5, 3, 1] OR [12, 40, 5, 3, 1]Input:[80, 60, 30, 40, 20, 10]Output:[80, 60, 30, 20, 10] OR [80, 60, 40, 20, 10]

In previous post, we have discussed about Longest Bitonic Subsequence problem. However, the post only covered code related to finding maximum sum of increasing subsequence, but not to the construction of subsequence. In this post, we will discuss how to construct Longest Bitonic Subsequence itself.

Let arr[0..n-1] be the input array. We define vector LIS such that LIS[i] is itself is a vector that stores Longest Increasing Subsequence of arr[0..i] that ends with arr[i]. Therefore for an index i, LIS[i] can be recursively written as –

LIS[0] = {arr[O]} LIS[i] = {Max(LIS[j])} + arr[i] wherej < iand arr[j] < arr[i] = arr[i], if there is no such j

We also define a vector LDS such that LDS[i] is itself is a vector that stores Longest Decreasing Subsequence of arr[i..n] that starts with arr[i]. Therefore for an index i, LDS[i] can be recursively written as –

LDS[n] = {arr[n]} LDS[i] = arr[i] + {Max(LDS[j])} wherej > iand arr[j] < arr[i] = arr[i], if there is no such j

For example, for array [1 11 2 10 4 5 2 1],

LIS[0]: 1 LIS[1]: 1 11 LIS[2]: 1 2 LIS[3]: 1 2 10 LIS[4]: 1 2 4 LIS[5]: 1 2 4 5 LIS[6]: 1 2 LIS[7]: 1

LDS[0]: 1 LDS[1]: 11 10 5 2 1 LDS[2]: 2 1 LDS[3]: 10 5 2 1 LDS[4]: 4 2 1 LDS[5]: 5 2 1 LDS[6]: 2 1 LDS[7]: 1

Therefore, Longest Bitonic Subsequence can be

LIS[1] + LDS[1] = [1 11 10 5 2 1] OR LIS[3] + LDS[3] = [1 2 10 5 2 1] OR LIS[5] + LDS[5] = [1 2 4 5 2 1]

Below is C++ implementation of above idea –

`/* Dynamic Programming solution to print Longest ` ` ` `Bitonic Subsequence */` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Utility function to print Longest Bitonic ` `// Subsequence ` `void` `print(vector<` `int` `>& arr, ` `int` `size) ` `{ ` ` ` `for` `(` `int` `i = 0; i < size; i++) ` ` ` `cout << arr[i] << ` `" "` `; ` `} ` ` ` `// Function to construct and print Longest ` `// Bitonic Subsequence ` `void` `printLBS(` `int` `arr[], ` `int` `n) ` `{ ` ` ` `// LIS[i] stores the length of the longest ` ` ` `// increasing subsequence ending with arr[i] ` ` ` `vector<vector<` `int` `>> LIS(n); ` ` ` ` ` `// initialize LIS[0] to arr[0] ` ` ` `LIS[0].push_back(arr[0]); ` ` ` ` ` `// Compute LIS values from left to right ` ` ` `for` `(` `int` `i = 1; i < n; i++) ` ` ` `{ ` ` ` `// for every j less than i ` ` ` `for` `(` `int` `j = 0; j < i; j++) ` ` ` `{ ` ` ` `if` `((arr[j] < arr[i]) && ` ` ` `(LIS[j].size() > LIS[i].size())) ` ` ` `LIS[i] = LIS[j]; ` ` ` `} ` ` ` `LIS[i].push_back(arr[i]); ` ` ` `} ` ` ` ` ` `/* LIS[i] now stores Maximum Increasing ` ` ` `Subsequence of arr[0..i] that ends with ` ` ` `arr[i] */` ` ` ` ` `// LDS[i] stores the length of the longest ` ` ` `// decreasing subsequence starting with arr[i] ` ` ` `vector<vector<` `int` `>> LDS(n); ` ` ` ` ` `// initialize LDS[n-1] to arr[n-1] ` ` ` `LDS[n - 1].push_back(arr[n - 1]); ` ` ` ` ` `// Compute LDS values from right to left ` ` ` `for` `(` `int` `i = n - 2; i >= 0; i--) ` ` ` `{ ` ` ` `// for every j greater than i ` ` ` `for` `(` `int` `j = n - 1; j > i; j--) ` ` ` `{ ` ` ` `if` `((arr[j] < arr[i]) && ` ` ` `(LDS[j].size() > LDS[i].size())) ` ` ` `LDS[i] = LDS[j]; ` ` ` `} ` ` ` `LDS[i].push_back(arr[i]); ` ` ` `} ` ` ` ` ` `// reverse as vector as we're inserting at end ` ` ` `for` `(` `int` `i = 0; i < n; i++) ` ` ` `reverse(LDS[i].begin(), LDS[i].end()); ` ` ` ` ` `/* LDS[i] now stores Maximum Decreasing Subsequence ` ` ` `of arr[i..n] that starts with arr[i] */` ` ` ` ` `int` `max = 0; ` ` ` `int` `maxIndex = -1; ` ` ` ` ` `for` `(` `int` `i = 0; i < n; i++) ` ` ` `{ ` ` ` `// Find maximum value of size of LIS[i] + size ` ` ` `// of LDS[i] - 1 ` ` ` `if` `(LIS[i].size() + LDS[i].size() - 1 > max) ` ` ` `{ ` ` ` `max = LIS[i].size() + LDS[i].size() - 1; ` ` ` `maxIndex = i; ` ` ` `} ` ` ` `} ` ` ` ` ` `// print all but last element of LIS[maxIndex] vector ` ` ` `print(LIS[maxIndex], LIS[maxIndex].size() - 1); ` ` ` ` ` `// print all elements of LDS[maxIndex] vector ` ` ` `print(LDS[maxIndex], LDS[maxIndex].size()); ` `} ` ` ` `// Driver program ` `int` `main() ` `{ ` ` ` `int` `arr[] = { 1, 11, 2, 10, 4, 5, 2, 1 }; ` ` ` `int` `n = ` `sizeof` `(arr) / ` `sizeof` `(arr[0]); ` ` ` ` ` `printLBS(arr, n); ` ` ` `return` `0; ` `} ` |

*chevron_right*

*filter_none*

Output:

1 11 10 5 2 1

**Time complexity** of above Dynamic Programming solution is O(n^{2}).

**Auxiliary space** used by the program is O(n^{2}).

This article is contributed by **Aditya Goel**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

## Recommended Posts:

- Printing Longest Common Subsequence | Set 2 (Printing All)
- Longest Bitonic Subsequence | DP-15
- Longest Bitonic Subsequence in O(n log n)
- Printing Longest Common Subsequence
- Printing longest Increasing consecutive subsequence
- Printing Maximum Sum Increasing Subsequence
- Find longest bitonic sequence such that increasing and decreasing parts are from two different arrays
- Longest subsequence with no 0 after 1
- Longest Zig-Zag Subsequence
- Longest Consecutive Subsequence
- Longest Repeating Subsequence
- Longest Repeated Subsequence
- Longest dividing subsequence
- Longest Uncommon Subsequence
- Longest Decreasing Subsequence