Cuente todas las arrays de longitud N formadas por distintos elementos consecutivos cuyo primer y último elemento son iguales

Dados dos enteros M y N , la tarea es encontrar el número de arrays de N longitudes posibles que tengan elementos adyacentes no iguales que se encuentren en el rango [1, M] que tengan elementos en el primer y último índice iguales.

Ejemplos: 
 

Entrada: N = 3, M = 3
Salida: 6
Explicación:
Las arrays posibles son {1, 2, 1}, {1, 3, 1}, {2, 1, 2}, {2, 3, 2}, {3, 1, 3}, {3, 2, 3}.

Entrada: N = 5, M = 4
Salida: 84

Enfoque: siga los pasos a continuación para resolver el problema:

• Primero fije arr[0] y arr[N-1] igual a 1.
• Ahora encuentre el número de arreglos posibles de tamaño i que termina en 1 (es decir, arr[i] = 1 ). Almacene este resultado en end_with_one[i] .
• Ahora, encuentra el número de arreglos posibles de tamaño i que no terminen en 1 ( arr[i] ≠ 1 ). Almacene este resultado en end_not_with_one[i] .
• Dado que, el número de formas de formar la array hasta el i ^-ésimo índice con arr[i] = 1 , es el mismo que el número de formas de formar la array hasta el (i – 1) ^ésimo índice con arr[i – 1] ≠ 1 , establecer end_with_one[i] = end_not_with_one[i – 1] .
• Ahora, el número de formas de formar el arreglo hasta el i ^-ésimo índice con arr[i] ≠ 1 es el siguiente:
  • Si arr[i – 1]= 1, hay (M – 1) números para ser colocados en el i ^-ésimo índice.
  • Si arr[i – 1] ≠ 1, entonces (M – 2) números pueden colocarse en el índice i , ya que arr[i] no puede ser 1 y arr[i] no puede ser igual a arr[i – 1] .
  • Por lo tanto, establezca end_not_with_one[i] = end_with_one[i-1] * (M – 1) + end_not_with_one[i-1]* (M – 2) .
• Por lo tanto, el número de formas de formar arreglos de tamaño N con arr[0] y arr[N – 1] igual a 1 es end_with_one[N – 1] .
• De manera similar, arr[0] y arr[N – 1] se pueden establecer en cualquier elemento de 1 a M.
• Por lo tanto, el número total de arreglos posibles es M * end_with_one[N-1] .

A continuación se muestra la implementación del enfoque anterior:

// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to print the count of
// arrays satisfying given condition
int totalArrays(int N, int M)
{
 
    int end_with_one[N + 1];
    int end_not_with_one[N + 1];
 
    // First element of
    // array is set as 1
    end_with_one[0] = 1;
    end_not_with_one[0] = 0;
 
    // Since the first element
    // of arr[] is 1, the
    // second element can't be 1
    end_with_one[1] = 0;
    end_not_with_one[1] = M - 1;
 
    // Traverse the remaining indices
    for (int i = 2; i < N; i++) {
 
        // If arr[i] = 1
        end_with_one[i]
            = end_not_with_one[i - 1];
 
        // If arr[i] ≠ 1
        end_not_with_one[i]
            = end_with_one[i - 1] * (M - 1)
              + end_not_with_one[i - 1] * (M - 2);
    }
 
    // Since last element needs to be 1
    return end_with_one[N - 1];
}
 
// Driver Code
int main()
{
 
    int N = 3, M = 3;
 
    // Stores the count of arrays
    // where arr[0] = arr[N - 1] = 1
    int temp = totalArrays(N, M);
 
    // Since arr[0] and arr[N - 1]
    // can be any number from 1 to M
    int ans = M * temp;
 
    // Print answer
    cout << ans << "\n";
 
    return 0;
}

// Java program for the above approach
import java.util.*;
class GFG
{
 
// Function to print the count of
// arrays satisfying given condition
static int totalArrays(int N, int M)
{
    int []end_with_one = new int[N + 1];
    int []end_not_with_one = new int[N + 1];
 
    // First element of
    // array is set as 1
    end_with_one[0] = 1;
    end_not_with_one[0] = 0;
 
    // Since the first element
    // of arr[] is 1, the
    // second element can't be 1
    end_with_one[1] = 0;
    end_not_with_one[1] = M - 1;
 
    // Traverse the remaining indices
    for (int i = 2; i < N; i++)
    {
 
        // If arr[i] = 1
        end_with_one[i]
            = end_not_with_one[i - 1];
 
        // If arr[i] ≠ 1
        end_not_with_one[i]
            = end_with_one[i - 1] * (M - 1)
              + end_not_with_one[i - 1] * (M - 2);
    }
 
    // Since last element needs to be 1
    return end_with_one[N - 1];
}
 
// Driver Code
public static void main(String[] args)
{
    int N = 3, M = 3;
 
    // Stores the count of arrays
    // where arr[0] = arr[N - 1] = 1
    int temp = totalArrays(N, M);
 
    // Since arr[0] and arr[N - 1]
    // can be any number from 1 to M
    int ans = M * temp;
 
    // Print answer
    System.out.print(ans+ "\n");
}
}
 
// This code is contributed by 29AjayKumar

# Python program for the above approach
 
# Function to print the count of
# arrays satisfying given condition
def totalArrays(N, M):
    end_with_one = [0] * (N + 1);
    end_not_with_one = [0] * (N + 1);
 
    # First element of
    # array is set as 1
    end_with_one[0] = 1;
    end_not_with_one[0] = 0;
 
    # Since the first element
    # of arr is 1, the
    # second element can't be 1
    end_with_one[1] = 0;
    end_not_with_one[1] = M - 1;
 
    # Traverse the remaining indices
    for i in range(2, N):
       
        # If arr[i] = 1
        end_with_one[i] = end_not_with_one[i - 1];
 
        # If arr[i] ≠ 1
        end_not_with_one[i] = end_with_one[i - 1] * (M - 1) + end_not_with_one[i - 1] * (M - 2);
 
    # Since last element needs to be 1
    return end_with_one[N - 1];
 
# Driver Code
if __name__ == '__main__':
    N = 3;
    M = 3;
 
    # Stores the count of arrays
    # where arr[0] = arr[N - 1] = 1
    temp = totalArrays(N, M);
 
    # Since arr[0] and arr[N - 1]
    # can be any number from 1 to M
    ans = M * temp;
 
    # Print answer
    print(ans);
     
# This code is contributed by 29AjayKumar

// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG
{
     
    // Function to print the count of
    // arrays satisfying given condition
    static int totalArrays(int N, int M)
    {
       
        int[] end_with_one = new int[N + 1];
        int[] end_not_with_one = new int[N + 1];
       
        // First element of
        // array is set as 1
        end_with_one[0] = 1;
        end_not_with_one[0] = 0;
       
        // Since the first element
        // of arr[] is 1, the
        // second element can't be 1
        end_with_one[1] = 0;
        end_not_with_one[1] = M - 1;
       
        // Traverse the remaining indices
        for (int i = 2; i < N; i++) {
       
            // If arr[i] = 1
            end_with_one[i]
                = end_not_with_one[i - 1];
       
            // If arr[i] ≠ 1
            end_not_with_one[i]
                = end_with_one[i - 1] * (M - 1)
                  + end_not_with_one[i - 1] * (M - 2);
        }
       
        // Since last element needs to be 1
        return end_with_one[N - 1];
    } 
 
  // Driver code
  static void Main()
  {
       
    int N = 3, M = 3;
   
    // Stores the count of arrays
    // where arr[0] = arr[N - 1] = 1
    int temp = totalArrays(N, M);
   
    // Since arr[0] and arr[N - 1]
    // can be any number from 1 to M
    int ans = M * temp;
   
    // Print answer
    Console.WriteLine(ans);
  }
}
 
// This code is contributed by divyeshrabadiya07

<script>
 
// Javascript program for the above approach
 
// Function to print the count of
// arrays satisfying given condition
function totalArrays(N, M)
{
 
    var end_with_one = Array(N+1);
    var end_not_with_one = Array(N+1);
 
    // First element of
    // array is set as 1
    end_with_one[0] = 1;
    end_not_with_one[0] = 0;
 
    // Since the first element
    // of arr[] is 1, the
    // second element can't be 1
    end_with_one[1] = 0;
    end_not_with_one[1] = M - 1;
 
    // Traverse the remaining indices
    for (var i = 2; i < N; i++) {
 
        // If arr[i] = 1
        end_with_one[i]
            = end_not_with_one[i - 1];
 
        // If arr[i] ≠ 1
        end_not_with_one[i]
            = end_with_one[i - 1] * (M - 1)
            + end_not_with_one[i - 1] * (M - 2);
    }
 
    // Since last element needs to be 1
    return end_with_one[N - 1];
}
 
// Driver Code
 
var N = 3, M = 3;
 
// Stores the count of arrays
// where arr[0] = arr[N - 1] = 1
var temp = totalArrays(N, M);
 
// Since arr[0] and arr[N - 1]
// can be any number from 1 to M
var ans = M * temp;
 
// Print answer
document.write( ans + "<br>");
 
 
</script>

6

Complejidad temporal: O(N)
Espacio auxiliar: O(N)