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; }

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}).

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