We are given two arrays, we need to find the longest possible bitonic sequence such that increasing part must be from first array and should be a subsequence of first array. Similarly, decreasing part of must be from second array and should be a subsequence of it.

Examples:

Input : arr1[] = {1, 5, 2, 4, 3, 5}, arr2[] = {8, 6, 4, 7, 3, 2} Output : 1, 2, 4, 5, 8, 6, 4, 3, 2 Input : arr1[] = {2, 0, 1, 3, 4}, arr2[] = {5, 3, 2, 1} Output : 0, 1, 2, 3, 4, 5, 3, 2, 1

The idea is to use largest increasing sequence from array1 and largest decreasing sequence from array2 and then combine both to get our result.

`// CPP to find largest bitonic sequence such that ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `vector<` `int` `> res; ` ` ` `// utility Binary search ` `int` `GetCeilIndex(` `int` `arr[], vector<` `int` `>& T, ` `int` `l, ` ` ` `int` `r, ` `int` `key) ` `{ ` ` ` `while` `(r - l > 1) { ` ` ` `int` `m = l + (r - l) / 2; ` ` ` `if` `(arr[T[m]] >= key) ` ` ` `r = m; ` ` ` `else` ` ` `l = m; ` ` ` `} ` ` ` `return` `r; ` `} ` ` ` `// function to find LIS in reverse form ` `void` `LIS(` `int` `arr[], ` `int` `n) ` `{ ` ` ` `// Add boundary case, when array n is zero ` ` ` `// Depend on smart pointers ` ` ` `vector<` `int` `> tailIndices(n, 0); ` `// Initialized with 0 ` ` ` `vector<` `int` `> prevIndices(n, -1); ` `// initialized with -1 ` ` ` ` ` `int` `len = 1; ` `// it will always point to empty location ` ` ` `for` `(` `int` `i = 1; i < n; i++) { ` ` ` ` ` `// new smallest value ` ` ` `if` `(arr[i] < arr[tailIndices[0]]) ` ` ` `tailIndices[0] = i; ` ` ` ` ` `// arr[i] wants to extend largest subsequence ` ` ` `else` `if` `(arr[i] > arr[tailIndices[len - 1]]) { ` ` ` `prevIndices[i] = tailIndices[len - 1]; ` ` ` `tailIndices[len++] = i; ` ` ` `} ` ` ` ` ` `// arr[i] wants to be a potential candidate of ` ` ` `// future subsequence ` ` ` `// It will replace ceil value in tailIndices ` ` ` `else` `{ ` ` ` `int` `pos = GetCeilIndex(arr, tailIndices, -1, ` ` ` `len - 1, arr[i]); ` ` ` `prevIndices[i] = tailIndices[pos - 1]; ` ` ` `tailIndices[pos] = i; ` ` ` `} ` ` ` `} ` ` ` ` ` `// put LIS into vector ` ` ` `for` `(` `int` `i = tailIndices[len - 1]; i >= 0; i = prevIndices[i]) ` ` ` `res.push_back(arr[i]); ` `} ` ` ` `// function for finding longest bitonic seq ` `void` `longestBitonic(` `int` `arr1[], ` `int` `n1, ` `int` `arr2[], ` `int` `n2) ` `{ ` ` ` `// find LIS of array 1 in reverse form ` ` ` `LIS(arr1, n1); ` ` ` ` ` `// reverse res to get LIS of first array ` ` ` `reverse(res.begin(), res.end()); ` ` ` ` ` `// reverse array2 and find its LIS ` ` ` `reverse(arr2, arr2 + n2); ` ` ` `LIS(arr2, n2); ` ` ` ` ` `// print result ` ` ` `for` `(` `int` `i = 0; i < res.size(); i++) ` ` ` `cout << res[i] << ` `" "` `; ` `} ` ` ` `// driver preogram ` `int` `main() ` `{ ` ` ` `int` `arr1[] = { 1, 2, 4, 3, 2 }; ` ` ` `int` `arr2[] = { 8, 6, 4, 7, 8, 9 }; ` ` ` `int` `n1 = ` `sizeof` `(arr1) / ` `sizeof` `(arr1[0]); ` ` ` `int` `n2 = ` `sizeof` `(arr2) / ` `sizeof` `(arr2[0]); ` ` ` `longestBitonic(arr1, n1, arr2, n2); ` ` ` `return` `0; ` `} ` |

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Output:

1 2 3 8 6 4

Time Complexity : O(n Log n)

Please note that O(n Log n) implementation of LIS is used

This article is contributed by **Shivam Pradhan (anuj_charm)**. 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.

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