# Queries on insertion of an element in a Bitonic Sequence

• Last Updated : 01 Sep, 2022

Given a Bitonic sequence ‘S’ and ‘Q’ no. of queries. Each query contain an integer xi, 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:

Bitonic Sequence: 1 2 3 5 2 1

Explanation:

• For the 1st query, we need to insert x1 = 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 x2 = 1, so we insert it in decreasing part as increasing part already has 1 in it. Hence, size = 5.
• For 3rd query, we insert x3 = 3 in increasing side so size becomes 6.
• For 4th query, we cannot insert x2 = 2 since 2 is in both increasing and decreasing side.

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

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.

1. Now, for each query, first check if the xi is greater than the maximum element in the bitonic sequence or not.
2. If yes, update the maximum sequence and include that element in the set for increasing sequence.
3. 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.

Below is the implementation of this approach:

## C++

 // C++ implementation of above approach#include using namespace std; // Function to find the bitonic sequencevoid 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 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::iterator it;    for (it = s1.begin(); it != s1.end(); it++)        cout << (*it) << " ";     set::reverse_iterator rit;     for (rit = s2.rbegin(); rit != s2.rend(); rit++)        cout << (*rit) << " ";} // Driver codeint 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;}

## Java

 // Java implementation of above approachimport java.util.*; class GFG{ // Function to find the bitonic sequencestatic void solveQuery(int a[], int n, int x[], int q){    int maxx = Integer.MIN_VALUE;     // Finding the maximum element    for (int i = 0; i < n; i++)        maxx = Math.max(maxx, a[i]);     Set s1 = new HashSet<>(), s2 = new HashSet<>();     s1.add(a[0]);    s2.add(a[n - 1]);     // set to include increasing sequence element    for (int i = 1; i < n; i++)        if (a[i] > a[i - 1])            s1.add(a[i]);     // set to include decreasing sequence element    for (int i = n - 2; i >= 0; i--)        if (a[i] > a[i + 1])            s2.add(a[i]);     // removing maximum element from    // decreasing sequence set    s2.remove(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.add(x[i]);        }         else        {             // checking if x lie in increasing sequence            if (!s1.contains(x[i]))                s1.add(x[i]);             // else insert into decreasing sequence set            else                s2.add(x[i]);        }         // finding the length        int ans = s1.size() + s2.size();        System.out.print(ans + "\n");    }     // printing the sequence    for (Integer it : s1)        System.out.print((it) + " ");     for (Integer it : s2)        System.out.print((it) + " ");} // Driver codepublic static void main(String[] args){    int a[] = { 1, 2, 5, 2 };    int n = a.length;    int x[] = { 5, 1, 3, 2 };    int q = x.length;     solveQuery(a, n, x, q);}} // This code is contributed by Rajput-Ji

## Python3

 # Python3 implementation of above approachimport sys # Function to find the bitonic sequencedef solveQuery(a, n, x, q):    maxx = -sys.maxsize     # Finding the maximum element    for i in range(n):        maxx = max(maxx, a[i])     s1, s2 = set(), set()     s1.add(a[0])    s2.add(a[n - 1])     # set to include increasing sequence element    for i in range(1, n):        if a[i] > a[i - 1]:            s1.add(a[i])     # set to include decreasing sequence element    for i in range(n - 2, -1, -1):        if a[i] > a[i + 1]:            s2.add(a[i])     # removing maximum element from    # decreasing sequence set    s2.remove(maxx)     # for each query    for i in range(q):         # checking if x is greater than        # maximum element or not.        if maxx <= x[i]:            maxx = x[i]            s1.add(x[i])        else:             # checking if x lie in increasing sequence            if x[i] not in s1:                s1.add(x[i])             # else insert into decreasing sequence set            else:                s2.add(x[i])         # finding the length        ans = len(s1) + len(s2)        print(ans)     # printing the sequence    for i in s1:        print(i, end = " ")     for i in reversed(list(s2)):        print(i, end = " ") # Driver Codeif __name__ == "__main__":     a = [1, 2, 5, 2]    n = len(a)    x = [5, 1, 3, 2]    q = len(x)     solveQuery(a, n, x, q) # This code is contributed by# sanjeev2552

## C#

 // C# implementation of above approachusing System;using System.Collections.Generic; class GFG{ // Function to find the bitonic sequencestatic void solveQuery(int []a, int n, int []x, int q){    int maxx = int.MinValue;     // Finding the maximum element    for (int i = 0; i < n; i++)        maxx = Math.Max(maxx, a[i]);     HashSet s1 = new HashSet(), s2 = new HashSet();     s1.Add(a[0]);    s2.Add(a[n - 1]);     // set to include increasing sequence element    for (int i = 1; i < n; i++)        if (a[i] > a[i - 1])            s1.Add(a[i]);     // set to include decreasing sequence element    for (int i = n - 2; i >= 0; i--)        if (a[i] > a[i + 1])            s2.Add(a[i]);     // removing maximum element from    // decreasing sequence set    s2.Remove(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.Add(x[i]);        }         else        {             // checking if x lie in increasing sequence            if (!s1.Contains(x[i]))                s1.Add(x[i]);             // else insert into decreasing sequence set            else                s2.Add(x[i]);        }         // finding the length        int ans = s1.Count + s2.Count;        Console.Write(ans + "\n");    }     // printing the sequence    foreach (int it in s1)        Console.Write((it) + " ");     foreach (int it in s2)        Console.Write((it) + " ");} // Driver codepublic static void Main(String[] args){    int []a = { 1, 2, 5, 2 };    int n = a.Length;    int []x = { 5, 1, 3, 2 };    int q = x.Length;     solveQuery(a, n, x, q);}} // This code is contributed by Rajput-Ji

## Javascript



Output

4
5
6
6
1 2 3 5 2 1

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