Generar secuencia de suma mínima de enteros con elementos pares mayores

Dado un número entero N , la tarea es generar una secuencia de N números enteros positivos tal que:  

• Cada elemento en la posición par debe ser mayor que el elemento que le sigue y el elemento que le precede, es decir, arr[i – 1] < arr[i] > arr[i + 1]
• La suma de los elementos debe ser par y la mínima posible (entre todas las sucesiones posibles).

Ejemplos:  

Entrada: N = 4 
Salida: 1 2 1 2

Entrada: N = 5 
Salida: 1 3 1 2 1 

Planteamiento: Para obtener la sucesión con la mínima suma posible, la sucesión debe ser de la forma 1, 2, 1, 2, 1, 2, 1… y para los casos en que la suma de la sucesión no sea par, cualquier 2 de la secuencia se puede cambiar a 3 para que la suma de la secuencia sea pareja.

A continuación se muestra la implementación del enfoque anterior: 

// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to print the required sequence
void make_sequence(int N)
{
 
    // arr[] will hold the sequence
    // sum variable will store the sum
    // of the sequence
    int arr[N + 1], sum = 0;
 
    for (int i = 1; i <= N; i++) {
 
        if (i % 2 == 1)
            arr[i] = 1;
        else
            arr[i] = 2;
 
        sum += arr[i];
    }
 
    // If sum of the sequence is odd
    if (sum % 2 == 1)
        arr[2] = 3;
 
    // Print the sequence
    for (int i = 1; i <= N; i++)
        cout << arr[i] << " ";
}
 
// Driver Code
int main()
{
    int N = 9;
 
    make_sequence(N);
 
    return 0;
}

// Java implementation of above approach
 
class GFG
{
 
// Function to print the required sequence
static void make_sequence(int N)
{
 
    // arr[] will hold the sequence
    // sum variable will store the sum
    // of the sequence
    int[] arr = new int[N + 1];
    int sum = 0;
 
    for (int i = 1; i <= N; i++)
    {
        if (i % 2 == 1)
            arr[i] = 1;
        else
            arr[i] = 2;
 
        sum += arr[i];
    }
 
    // If sum of the sequence is odd
    if (sum % 2 == 1)
        arr[2] = 3;
 
    // Print the sequence
    for (int i = 1; i <= N; i++)
        System.out.print(arr[i] + " ");
}
 
// Driver Code
public static void main(String[] args)
{
    int N = 9;
    make_sequence(N);
}
}
 
// This code is contributed by iAyushRaj.

# Python 3 implementation of above approach
 
# Function to print the required sequence
def make_sequence( N):
 
    # arr[] will hold the sequence
    # sum variable will store the sum
    # of the sequence
    arr = [0] * (N + 1)
    sum = 0
 
    for i in range(1, N + 1):
 
        if (i % 2 == 1):
            arr[i] = 1
        else:
            arr[i] = 2
 
        sum += arr[i]
 
    # If sum of the sequence is odd
    if (sum % 2 == 1):
        arr[2] = 3
 
    # Print the sequence
    for i in range(1, N + 1):
        print(arr[i], end = " ")
 
# Driver Code
if __name__ == "__main__":
     
    N = 9
    make_sequence(N)
 
# This code is contributed by ita_c

// C# implementation of above approach
using System;
 
class GFG
{
 
// Function to print the required sequence
public static void make_sequence(int N)
{
 
    // arr will hold the sequence
    // sum variable will store the sum
    // of the sequence
    int[] arr = new int[N + 1];
    int sum = 0;
 
    for (int i = 1; i <= N; i++)
    {
        if (i % 2 == 1)
            arr[i] = 1;
        else
            arr[i] = 2;
 
        sum += arr[i];
    }
 
    // If sum of the sequence is odd
    if (sum % 2 == 1)
        arr[2] = 3;
 
    // Print the sequence
    for (int i = 1; i <= N; i++)
        Console.Write(arr[i] + " ");
}
 
// Driver Code
public static void Main()
{
    int N = 9;
 
    make_sequence(N);
}
}
 
// This code is contributed by iAyushRaj.

<?php
// PHP implementation of above approach
 
// Function to print the required sequence
function make_sequence($N)
{
 
    // arr[] will hold the sequence
    // sum variable will store the sum
    // of the sequence
    $arr = array();
    $sum = 0;
 
    for ($i = 1; $i <= $N; $i++)
    {
        if ($i % 2 == 1)
            $arr[$i] = 1;
        else
            $arr[$i] = 2;
 
        $sum += $arr[$i];
    }
 
    // If sum of the sequence is odd
    if ($sum % 2 == 1)
        $arr[2] = 3;
 
    // Print the sequence
    for ($i = 1; $i <= $N; $i++)
        echo $arr[$i], " ";
}
 
// Driver Code
$N = 9;
make_sequence($N);
 
// This code is contributed by Ryuga
?>

<script>
 
// Javascript implementation of above approach
 
// Function to print the required sequence
function make_sequence(N)
{
 
    // arr[] will hold the sequence
    // sum variable will store the sum
    // of the sequence
    var arr = Array(N+1), sum = 0;
 
    for (var i = 1; i <= N; i++) {
 
        if (i % 2 == 1)
            arr[i] = 1;
        else
            arr[i] = 2;
 
        sum += arr[i];
    }
 
    // If sum of the sequence is odd
    if (sum % 2 == 1)
        arr[2] = 3;
 
    // Print the sequence
    for (var i = 1; i <= N; i++)
        document.write( arr[i] + " ");
}
 
// Driver Code
var N = 9;
make_sequence(N);
 
</script>

1 3 1 2 1 2 1 2 1

Complejidad de tiempo: O(N)

Espacio Auxiliar: O(N)