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Lexicographically largest N-length Bitonic sequence made up of elements from given range
  • Last Updated : 15 Dec, 2020

Given three integers N, low and high, the task is to find the lexicographically largest bitonic sequence consisting of N elements lying in the range [low, high]. If it is not possible to generate such a sequence, then print “Not Possible”.

Examples:

Input: N = 5, low = 2, high = 6
Output: 5 6 5 4 3
Explanation:
The sequence {arr[0], arr[1]} is strictly increasing followed by strictly decreasing sequence of the remaining elements. This sequence is the lexicographically largest possible having all elements in the range [2, 6] and length of this sequence is 5.

Input: N = 10, low = 4, high = 10
Output: 7 8 9 10 9 8 7 6 5 4

Approach: The idea is to find the suitable index of high in the resultant sequence and then maintain a difference of 1 between adjacent elements in the sequence such that the bitonic sequence formed is the lexicographically largest possible. Follow the steps below to solve the problem:



  • Initialize an array A[] of size N to store the resultant sequence.
  • Initialize a variable high_index = -1 to store the index of high in A[] and set high_index = N – (high – low + 1).
  • If high_index > (N – 1) / 2, then the remaining N/2 elements cannot be placed in strictly increasing order. So, print “Not Possible”.
  • Otherwise perform the following steps:
    • If high_index ≤ 0, then set high_index = 1 as there has to be a strictly increasing sequence at the beginning.
    • Maintain a strictly decreasing sequence with a difference of 1 from the range [high_index, 0], starting with a value high.
    • Maintain a strictly decreasing sequence with a difference of 1 from the range[high_index + 1, N – 1] starting with a value (high – 1).
  • After completing the above steps, print all the elements in the array A[].

Below is the implementation of the above approach:

C++

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// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the lexicographically
// largest bitonic sequence of size N
// elements lies in the range[low, high]
void LargestArray(int N, int low, int high)
{
    // Store index of highest element
    int high_index = N - (high - low + 1);
 
    // If high_index > (N-1)/2, then
    // remaining N/2 elements cannot
    // be placed in bitonic order
    if (high_index > (N - 1) / 2) {
        cout << "Not Possible";
        return;
    }
 
    // If high_index <= 0, then
    // set high_index as 1
    if (high_index <= 0)
        high_index = 1;
 
    // Stores the resultant sequence
    int A[N];
 
    // Store the high value
    int temp = high;
 
    // Maintain strictly decreasing
    // sequence from index high_index
    // to 0 starting with temp
    for (int i = high_index; i >= 0; i--) {
 
        // Store the value and decrement
        // the temp variable by 1
        A[i] = temp--;
    }
 
    // Maintain the strictly decreasing
    // sequence from index high_index + 1
    // to N - 1 starting with high - 1
    high -= 1;
 
    for (int i = high_index + 1; i < N; i++)
 
        // Store the value and decrement
        // high by 1
        A[i] = high--;
 
    // Print the resultant sequence
    for (int i = 0; i < N; i++) {
        cout << A[i] << ' ';
    }
}
 
// Driver Code
int main()
{
    int N = 5, low = 2, high = 6;
 
    // Function Call
    LargestArray(N, low, high);
 
    return 0;
}

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Java

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// Java program for the above approach
import java.util.*;
   
class GFG{
     
// Function to find the lexicographically
// largest bitonic sequence of size N
// elements lies in the range[low, high]
static void LargestArray(int N, int low,
                         int high)
{
     
    // Store index of highest element
    int high_index = N - (high - low + 1);
     
    // If high_index > (N-1)/2, then
    // remaining N/2 elements cannot
    // be placed in bitonic order
    if (high_index > (N - 1) / 2)
    {
        System.out.print("Not Possible");
        return;
    }
     
    // If high_index <= 0, then
    // set high_index as 1
    if (high_index <= 0)
        high_index = 1;
         
    // Stores the resultant sequence
    int[] A = new int[N];
  
    // Store the high value
    int temp = high;
  
    // Maintain strictly decreasing
    // sequence from index high_index
    // to 0 starting with temp
    for(int i = high_index; i >= 0; i--)
    {
         
        // Store the value and decrement
        // the temp variable by 1
        A[i] = temp--;
    }
  
    // Maintain the strictly decreasing
    // sequence from index high_index + 1
    // to N - 1 starting with high - 1
    high -= 1;
  
    for(int i = high_index + 1; i < N; i++)
     
        // Store the value and decrement
        // high by 1
        A[i] = high--;
  
    // Print the resultant sequence
    for(int i = 0; i < N; i++)
    {
        System.out.print(A[i] + " ");
    }
}
   
// Driver Code
public static void main(String[] args)
{
    int N = 5, low = 2, high = 6;
     
    // Function Call
    LargestArray(N, low, high);
}
}
 
// This code is contributed by susmitakundugoaldanga

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Python3

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# Python3 program for the above approach
  
# Function to find the lexicographically
# largest bitonic sequence of size N
# elements lies in the range[low, high]
def LargestArray(N, low, high):
     
    # Store index of highest element
    high_index = N - (high - low + 1)
     
    # If high_index > (N-1)/2, then
    # remaining N/2 elements cannot
    # be placed in bitonic order
    if (high_index > (N - 1) // 2):
        print("Not Possible")
        return
     
    # If high_index <= 0, then
    # set high_index as 1
    if (high_index <= 0):
        high_index = 1
  
    # Stores the resultant sequence
    A = [0] * N
  
    # Store the high value
    temp = high
  
    # Maintain strictly decreasing
    # sequence from index high_index
    # to 0 starting with temp
    for i in range(high_index, -1, -1):
  
        # Store the value and decrement
        # the temp variable by 1
        A[i] = temp
        temp = temp - 1
     
    # Maintain the strictly decreasing
    # sequence from index high_index + 1
    # to N - 1 starting with high - 1
    high -= 1
  
    for i in range(high_index + 1, N):
         
        # Store the value and decrement
        # high by 1
        A[i] = high
        high = high - 1
  
    # Print the resultant sequence
    for i in range(N):
        print(A[i], end = " ")
 
# Driver Code
N = 5
low = 2
high = 6
  
# Function Call
LargestArray(N, low, high)
 
# This code is contributed by code_hunt

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C#

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// C# program for the above approach
using System;
 
class GFG{
     
// Function to find the lexicographically
// largest bitonic sequence of size N
// elements lies in the range[low, high]
static void LargestArray(int N, int low,
                        int high)
{
     
    // Store index of highest element
    int high_index = N - (high - low + 1);
     
    // If high_index > (N-1)/2, then
    // remaining N/2 elements cannot
    // be placed in bitonic order
    if (high_index > (N - 1) / 2)
    {
        Console.Write("Not Possible");
        return;
    }
     
    // If high_index <= 0, then
    // set high_index as 1
    if (high_index <= 0)
        high_index = 1;
         
    // Stores the resultant sequence
    int[] A = new int[N];
 
    // Store the high value
    int temp = high;
 
    // Maintain strictly decreasing
    // sequence from index high_index
    // to 0 starting with temp
    for(int i = high_index; i >= 0; i--)
    {
         
        // Store the value and decrement
        // the temp variable by 1
        A[i] = temp--;
    }
 
    // Maintain the strictly decreasing
    // sequence from index high_index + 1
    // to N - 1 starting with high - 1
    high -= 1;
 
    for(int i = high_index + 1; i < N; i++)
     
        // Store the value and decrement
        // high by 1
        A[i] = high--;
 
    // Print the resultant sequence
    for(int i = 0; i < N; i++)
    {
        Console.Write(A[i] + " ");
    }
}
 
// Driver Code
public static void Main(String[] args)
{
    int N = 5, low = 2, high = 6;
     
    // Function Call
    LargestArray(N, low, high);
}
}
 
// This code is contributed by shivanisinghss2110

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Output: 

5 6 5 4 3

 

Time Complexity: O(N)
Auxiliary Space: O(1)

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