Generate Bitonic Sequence of length N from integers in a given range
Last Updated : 22 Aug, 2022
Given integers N, L and R, the task is to generate a Bitonic Sequence of length N from the integers in the range [L, R] such that the first element is the maximum. If it is not possible to create such a sequence, then print "-1".
A Bitonic Sequence is a sequence that must be strictly increasing at first and then strictly decreasing.
Examples:
Input: N = 5, L = 3, R = 10
Output: 9, 10, 9, 8, 7
Explanation: The sequence {9, 10, 9, 8, 7} is first strictly increasing and then strictly decreasing.
Input: N = 5, L = 2, R = 5
Output: 4, 5, 4, 3, 2
Explanation:
[ The sequence {4, 5, 4, 3, 2} is first strictly increasing and then strictly decreasing.
Approach: The idea is to use a Deque so that elements can be added from the end and the beginning. Follow the steps below to solve the problem:
- Initialize a deque to store the element of the resultant bitonic sequence.
- Initialize a variable i as 0 and start adding elements in the resultant list starting from (R - i) until i less than the minimum of (R - L + 1) and (N - 1).
- After the above steps if the size of the resultant list is less than N then add elements from (R - 1) to L from the starting of the list until the size of the resultant list does not become N.
- After the above steps, if N is greater than (R - L)*2 + 1, then it is not possible to construct such a sequence then print "-1" else print the sequence stored in deque.
Below is the implementation of the above approach:
C++ // C++ program for the above approach #include <bits/stdc++.h> using namespace std; // Function to construct bitonic // sequence of length N from // integers in the range [L, R] void bitonicSequence(int num, int lower, int upper) { // If sequence is not possible if (num > (upper - lower) * 2 + 1) { cout << -1; return; } // Store the resultant list deque<int> ans; deque<int>::iterator j = ans.begin(); for(int i = 0; i < min(upper - lower + 1, num - 1); i++) ans.push_back(upper - i); // If size of deque < n for(int i = 0; i < num - ans.size(); i++) // Add elements from start ans.push_front(upper - i - 1); // Print the stored in the list cout << '['; for(j = ans.begin(); j != ans.end(); ++j) cout << ' ' << *j; cout << ' ' << ']'; } // Driver Code int main() { int N = 5, L = 3, R = 10; // Function Call bitonicSequence(N, L, R); return 0; } // This code is contributed by jana_sayantan // Java program for the above approach import java.util.*; class GFG { // Function to construct bitonic // sequence of length N from // integers in the range [L, R] public static void bitonicSequence( int num, int lower, int upper) { // If sequence is not possible if (num > (upper - lower) * 2 + 1) { System.out.println(-1); return; } // Store the resultant list Deque<Integer> ans = new ArrayDeque<>(); for (int i = 0; i < Math.min(upper - lower + 1, num - 1); i++) ans.add(upper - i); // If size of deque < n for (int i = 0; i < num - ans.size(); i++) // Add elements from start ans.addFirst(upper - i - 1); // Print the stored in the list System.out.println(ans); } // Driver Code public static void main(String[] args) { int N = 5, L = 3, R = 10; // Function Call bitonicSequence(N, L, R); } } # Python3 program for the above approach from collections import deque # Function to construct bitonic # sequence of length N from # integers in the range [L, R] def bitonicSequence(num, lower, upper): # If sequence is not possible if (num > (upper - lower) * 2 + 1): print(-1) return # Store the resultant list ans = deque() for i in range(min(upper - lower + 1, num - 1)): ans.append(upper - i) # If size of deque < n for i in range(num - len(ans)): # Add elements from start ans.appendleft(upper - i - 1) # Print the stored in the list print(list(ans)) # Driver Code if __name__ == '__main__': N = 5 L = 3 R = 10 # Function Call bitonicSequence(N, L, R) # This code is contributed by mohit kumar 29
// C# program for the above approach using System; using System.Collections.Generic; class GFG{ // Function to construct bitonic // sequence of length N from // integers in the range [L, R] public static void bitonicSequence(int num, int lower, int upper) { // If sequence is not possible if (num > (upper - lower) * 2 + 1) { Console.WriteLine(-1); return; } // Store the resultant list List<int> ans = new List<int>(); for(int i = 0; i < Math.Min(upper - lower + 1, num - 1); i++) ans.Add(upper - i); // If size of deque < n for(int i = 0; i < num - ans.Count; i++) // Add elements from start ans.Insert(0,upper - i - 1); // Print the stored in the list Console.Write("["); foreach(int x in ans) Console.Write(x + ", "); Console.Write("]"); } // Driver Code public static void Main(String[] args) { int N = 5, L = 3, R = 10; // Function Call bitonicSequence(N, L, R); } } // This code is contributed by Rajput-Ji <script> // Javascript program for the above approach // Function to construct bitonic // sequence of length N from // integers in the range [L, R] function bitonicSequence(num, lower, upper) { // If sequence is not possible if (num > (upper - lower) * 2 + 1) { document.write( -1); return; } // Store the resultant list var ans = []; for(var i = 0; i < Math.min(upper - lower + 1, num - 1); i++) ans.push(upper - i); // If size of deque < n for(var i = 0; i < num - ans.length; i++) { // Add elements from start ans.splice(0, 0, upper -i - 1) } // Print the stored in the list document.write( '['); ans.forEach(element => { document.write(" "+element); }); document.write( ' ' + ']'); } // Driver Code var N = 5, L = 3, R = 10; // Function Call bitonicSequence(N, L, R); </script> Time Complexity: O(N)
Auxiliary Space: O(1)
Similar Reads
Lexicographically largest N-length Bitonic sequence made up of elements from given range 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 = 6Output: 5 6 5 4 3Ex
8 min read
Find a sequence of distinct integers whose Harmonic Mean is N Given an integer N, the task is to find a sequence of distinct integers whose Harmonic Mean is N itself. In case such a sequence doesn't exist, print -1. Examples: Input: N = 5Output: {3, 4, 5, 6, 20} Explanation: Given sequence is the correct answer because 5/(1/3​+1/4+1/5​+1/6​+1/20)​=5. Note that
6 min read
Generate a N length Permutation having equal sized LIS from both ends Given an integer N, the task is to generate a permutation of elements in the range [1, N] in such order that the length of LIS from starting is equal to LIS from the end of the array. Print -1 if no such array exists. Examples: Input: N = 5Output: [1, 3, 5, 4, 2]Explanation:LIS from left side: [1, 3
12 min read
Maximum Length of Sequence of Sums of prime factors generated by the given operations Given two integers N and M, the task is to perform the following operations: For every value in the range [N, M], calculate the sum of its prime factors followed by the sum of prime factors of that sum and so on.Generate the above sequence for each array element and calculate the length of the seque
12 min read
Minimum number of coins that can generate all the values in the given range Given an integer N, the task is to find the minimum number of coins required to create all the values in the range [1, N]. Examples: Input: N = 5Output: 3The coins {1, 2, 4} can be used to generate all the values in the range [1, 5].1 = 12 = 23 = 1 + 24 = 45 = 1 + 4 Input: N = 10Output: 4 Approach:
6 min read