# Queries on insertion of an element in a Bitonic Sequence

Given a Bitonic sequence **‘S’** and **‘Q’** no. of queries. Each query contain an integer **x _{i}**, 1 <= i <= Q. The task is to print the length of bitonic sequence after inserting the integer for each query. Also, print the bitonic sequence after all the queries.

**Examples:**

Input:S = { 1, 2, 5, 2 }, Q = 4, x = { 5, 1, 3, 2 }

Output:

4

5

6

6

Bitonic Sequence: 1 2 3 5 2 1

Explanation:

- For the 1st query, we need to insert x
_{1}= 5 but since 5 is the maximum element and we can have only one occurrence of the maximum element in S, so we won’t insert 5 in S. Hence, size = 4.- For the 2nd query, we need to insert x
_{2}= 1, so we insert it in decreasing part as increasing part already has 1 in it. Hence, size = 5.- For 3rd query, we insert x
_{3}= 3 in increasing side so size becomes 6.- For 4th query, we cannot insert x
_{2}= 2 since 2 is in both increasing and decreasing side.

Final sequence:1, 2, 3, 5, 2, 1

Input:S = { 1, 2, 5, 2 }, Q = 4, x = { 5, 6, 4, 4 }

Output:

4

5

6

7

Bitonic Sequence: 1 2 4 5 6 4 2

**Approach:** The idea is to make two sets, one for increasing sequence and other for decreasing sequence.

- Now, for each query, first check if the x
_{i}is greater than the maximum element in the bitonic sequence or not. - If yes, update the maximum sequence and include that element in the set for increasing sequence.
- If no, then check if that element is present in the increasing sequence set, if not include it in increasing sequence set, else include it into decreasing sequence set.

The benefit of using the set is it won’t allow multiple same values. So, if the element is new in the set, it will be included and increase the size of sequence else it will remain the size of sequence same.

**Below is the implementation of this approach:**

`// C++ implementation of above approach ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Function to find the bitonic sequence ` `void` `solveQuery(` `int` `a[], ` `int` `n, ` `int` `x[], ` `int` `q) ` `{ ` ` ` `int` `maxx = INT_MIN; ` ` ` ` ` `// Finding the maximum element ` ` ` `for` `(` `int` `i = 0; i < n; i++) ` ` ` `maxx = max(maxx, a[i]); ` ` ` ` ` `set<` `int` `> s1, s2; ` ` ` ` ` `s1.insert(a[0]); ` ` ` `s2.insert(a[n - 1]); ` ` ` ` ` `// set to include increasing sequence element ` ` ` `for` `(` `int` `i = 1; i < n; i++) ` ` ` `if` `(a[i] > a[i - 1]) ` ` ` `s1.insert(a[i]); ` ` ` ` ` `// set to include decreasing sequence element ` ` ` `for` `(` `int` `i = n - 2; i >= 0; i--) ` ` ` `if` `(a[i] > a[i + 1]) ` ` ` `s2.insert(a[i]); ` ` ` ` ` `// removing maximum element from ` ` ` `// decreasing sequence set ` ` ` `s2.erase(s2.find(maxx)); ` ` ` ` ` `// for each query ` ` ` `for` `(` `int` `i = 0; i < q; i++) { ` ` ` ` ` `// checking if x is greater than ` ` ` `// maximum element or not. ` ` ` `if` `(maxx <= x[i]) { ` ` ` `maxx = x[i]; ` ` ` `s1.insert(x[i]); ` ` ` `} ` ` ` ` ` `else` `{ ` ` ` ` ` `// checking if x lie in increasing sequence ` ` ` `if` `(s1.find(x[i]) == s1.end()) ` ` ` `s1.insert(x[i]); ` ` ` ` ` `// else insert into decreasing sequence set ` ` ` `else` ` ` `s2.insert(x[i]); ` ` ` `} ` ` ` ` ` `// finding the length ` ` ` `int` `ans = s1.size() + s2.size(); ` ` ` `cout << ans << ` `"\n"` `; ` ` ` `} ` ` ` ` ` `// printing the sequence ` ` ` `set<` `int` `>::iterator it; ` ` ` `for` `(it = s1.begin(); it != s1.end(); it++) ` ` ` `cout << (*it) << ` `" "` `; ` ` ` ` ` `set<` `int` `>::reverse_iterator rit; ` ` ` ` ` `for` `(rit = s2.rbegin(); rit != s2.rend(); rit++) ` ` ` `cout << (*rit) << ` `" "` `; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `int` `a[] = { 1, 2, 5, 2 }; ` ` ` `int` `n = ` `sizeof` `(a) / ` `sizeof` `(a[0]); ` ` ` `int` `x[] = { 5, 1, 3, 2 }; ` ` ` `int` `q = ` `sizeof` `(x) / ` `sizeof` `(x[0]); ` ` ` ` ` `solveQuery(a, n, x, q); ` ` ` `return` `0; ` `} ` |

*chevron_right*

*filter_none*

**Output:**

4 5 6 6 1 2 3 5 2 1

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