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Construct an array of first N natural numbers having no triplet (i, j, k) such that a[i] + a[j] = 2* a[k] where i < j< k
  • Last Updated : 24 Dec, 2020

Given a positive integer N, the task is to construct an array a[] using first N natural numbers which contains no such triplet (i, j, k) satisfying a[k] * 2 = a[i] + a[j] and i < j < k.

Examples:

Input: N = 3 
Output: {2, 3, 1 } 
Explanation: 
Since no such triplet exists in the array satisfying the condition, the required output is { 2, 3, 1 }.

Input: N = 10 
Output: { 8, 4, 6, 10, 2, 7, 3, 5, 9, 1 }

Approach: The problem can be solved using Greedy technique. Follow the steps below to solve the problem:

  • Recursively find the first (N / 2) elements of the resultant array and the last (N / 2) elements of the resultant array.
  • Merge both halves of the array such that the first half of the array contains even numbers and the last half of the array contains the odd numbers.
  • Finally, print the resultant array.

C++




// C++ program to implement
// the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to construct the array of size N
// that contains no such triplet satisfying
// the given conditions
vector<int> constructArray(int N)
{
 
    // Base case
    if (N == 1) {
 
        return { 1 };
    }
 
    // Stores the first half
    // of the array
    vector<int> first
        = constructArray(N / 2);
 
    // Stores the last half
    // of the array
    vector<int> last
        = constructArray(N - (N / 2));
 
    // Stores the merged array
    vector<int> ans;
 
    // Insert even numbers
    for (auto e : first) {
 
        // Insert 2 * e
        ans.push_back(2 * e);
    }
 
    // Insert odd numbers
    for (auto o : last) {
 
        // Insert (2 * o - 1)
        ans.push_back((2 * o) - 1);
    }
 
    return ans;
}
 
// Function to print the resultant array
void printArray(vector<int> ans, int N)
{
 
    // Print resultant array
    cout << "{ ";
    for (int i = 0; i < N; i++) {
 
        // Print current element
        cout << ans[i];
 
        // If i is not the last index
        // of the resultant array
        if (i != N - 1) {
            cout << ", ";
        }
    }
 
    cout << " }";
}
 
// Driver Code
int main()
{
    int N = 10;
 
    // Store the resultant array
    vector<int> ans
        = constructArray(N);
 
    printArray(ans, N);
 
    return 0;
}

Java




// Java program to implement
// the above approach
import java.io.*;
import java.util.*;
 
class GFG{
 
// Function to construct the array of size N
// that contains no such triplet satisfying
// the given conditions
static ArrayList<Integer> constructArray(int N)
{
     
    // Base case
    if (N == 1)
    {
        ArrayList<Integer> a = new ArrayList<Integer>(1);
        a.add(1);
        return a;
    }
     
    // Stores the first half
    // of the array
    ArrayList<Integer> first = new ArrayList<Integer>(N);
    first = constructArray(N / 2);
     
    // Stores the last half
    // of the array
    ArrayList<Integer> last = new ArrayList<Integer>(N);
    last = constructArray(N - N / 2);
     
    ArrayList<Integer> ans = new ArrayList<Integer>(N);
     
    // Insert even numbers
    for(int i = 0; i < first.size(); i++)
    {
         
        // Insert 2 * first[i]
        ans.add(2 * first.get(i));
    }
     
    // Insert odd numbers
    for(int i = 0; i < last.size(); i++)
    {
         
        // Insert (2 * last[i] - 1)
        ans.add(2 * last.get(i) - 1);
    }
    return ans;
}
 
// Driver code
public static void main(String[] args)
{
    int N = 10;
     
    ArrayList<Integer> answer = new ArrayList<Integer>(N);
    answer = constructArray(N);
     
    System.out.print("{");
    for(int i = 0; i < answer.size(); i++)
    {
        System.out.print(answer.get(i));
        System.out.print(", ");
    }
    System.out.print("}");
}   
}
 
// This code is contributed by koulick_sadhu

Python3




# Python3 program to implement
# the above approach
 
# Function to construct the array of size N
# that contains no such triplet satisfying
# the given conditions
def constructArray(N) :
 
    # Base case
    if (N == 1) :
        a = []
        a.append(1)
        return a;
 
    # Stores the first half
    # of the array
    first = constructArray(N // 2);
 
    # Stores the last half
    # of the array
    last = constructArray(N - (N // 2));
 
    # Stores the merged array
    ans = [];
 
    # Insert even numbers
    for e in first :
 
        # Insert 2 * e
        ans.append(2 * e);
 
    # Insert odd numbers
    for o in last:
 
        # Insert (2 * o - 1)
        ans.append((2 * o) - 1);
 
    return ans;
 
# Function to print the resultant array
def printArray(ans, N) :
 
    # Print resultant array
    print("{ ", end = "");
    for i in range(N) :
 
        # Print current element
        print(ans[i], end = "");
 
        # If i is not the last index
        # of the resultant array
        if (i != N - 1) :
            print(", ",end = "");
 
    print(" }", end = "");
 
# Driver Code
if __name__ == "__main__" :
    N = 10;
 
    # Store the resultant array
    ans = constructArray(N);
 
    printArray(ans, N);
 
       # This code is contributed by AnkThon

C#




// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
  
class GFG{
  
// Function to construct the array of size N
// that contains no such triplet satisfying
// the given conditions
static List<int> constructArray(int N)
{
     
    // Base case
    if (N == 1)
    {
        List<int> a = new List<int>(1);
        a.Add(1);
        return a;
    }
      
    // Stores the first half
    // of the array
    List<int> first = new List<int>();
    first = constructArray(N / 2);
      
    // Stores the last half
    // of the array
    List<int> last = new List<int>();
 
    last = constructArray(N - N / 2);
      
    List<int> ans = new List<int>();
      
    // Insert even numbers
    for(int i = 0; i < first.Count; i++)
    {
         
        // Insert 2 * first[i]
        ans.Add(2 * first[i]);
    }
      
    // Insert odd numbers
    for(int i = 0; i < last.Count; i++)
    {
         
        // Insert (2 * last[i] - 1)
        ans.Add(2 * last[i] - 1);
    }
    return ans;
}
  
// Driver code
public static void Main()
{
    int N = 10;
     
    List<int> answer = new List<int>(N);
    answer = constructArray(N);
      
    Console.Write("{");
    for(int i = 0; i < answer.Count; i++)
    {
        Console.Write(answer[i]);
        Console.Write(", ");
    }
    Console.Write("}");
}   
}
 
// This code is contributed by sanjoy_62
Output: 
{ 8, 4, 6, 10, 2, 7, 3, 5, 9, 1 }

 

Time Complexity : O(N * log(N)) 
Auxiliary Space: O(N)

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