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Queries on insertion of an element in a Bitonic Sequence

  • Last Updated : 15 Jan, 2020

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:
4
5
6
6
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.

    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.

  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.

    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++




// 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;
}

Java




// Java implementation of above approach
import java.util.*;
  
class GFG
{
  
// Function to find the bitonic sequence
static 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<Integer> 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 code
public 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 approach
import sys
  
# Function to find the bitonic sequence
def 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 Code
if __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 approach
using System;
using System.Collections.Generic;
  
class GFG
{
  
// Function to find the bitonic sequence
static 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<int> s1 = new HashSet<int>(), s2 = new HashSet<int>();
  
    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 code
public 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
Output:
4
5
6
6
1 2 3 5 2 1

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