Cuente el número de strings binarias sin 1 consecutivos: Conjunto 2

Posted on by Rudeus Greyrat

Dado un entero positivo N , la tarea es contar todas las posibles strings binarias distintas de longitud N de modo que no haya unos consecutivos.

Ejemplos: 

Entrada: N = 5 
Salida: 5 
Explicación: 
Los enteros no negativos <= 5 con sus correspondientes representaciones binarias son: 
0 : 0 
1 : 1 
2 : 10 
3 : 11 
4 : 100 
5 : 101 
Entre ellos, solo el 3 tiene dos 1 consecutivos. Por lo tanto cuenta requerida = 5

Entrada: N = 12 
Salida: 8 
 

Enfoque de programación dinámica: Ya se ha discutidoun enfoque de programación dinámica en este artículo . 

Enfoque: En este artículo, se analiza  un enfoque que utiliza el concepto de digit-dp .

  • Similar al problema digit-dp, aquí se crea una tabla tridimensional para almacenar los valores calculados. Se supone que el N < 2 31 – 1, y el rango de cada número es solo 2 (O 0 o 1). Por lo tanto, las dimensiones de la mesa se toman como 32 x 2 x 2 .
  • Después de construir la tabla, el número dado se convierte en una string binaria .
  • Luego, se itera el número. Para cada iteración: 
    1. Compruebe si el dígito anterior es un 0 o un 1.
    2. Si es un 0, entonces el número actual puede ser un 0 o un 1.
    3. Pero si el número anterior es 1, entonces el número actual tiene que ser 0 porque no podemos tener dos 1 consecutivos en la representación binaria.
  • Ahora, la tabla se llena exactamente como el problema digit-dp .

A continuación se muestra la implementación del enfoque anterior.  

C++

// C++ program to count number of
// binary strings without consecutive 1’s
 
#include <bits/stdc++.h>
using namespace std;
 
// Table to store the solution of
// every sub problem
int memo[32][2][2];
 
// Function to fill the table
/* Here,
   pos: keeps track of current position.
   f1: is the flag to check if current
         number is less than N or not.
   pr: represents the previous digit
*/
int dp(int pos, int fl, int pr, string& bin)
{
    // Base case
    if (pos == bin.length())
        return 1;
 
    // Check if this subproblem
    // has already been solved
    if (memo[pos][fl][pr] != -1)
        return memo[pos][fl][pr];
 
    int val = 0;
 
    // Placing 0 at the current position
    // as it does not violate the condition
    if (bin[pos] == '0')
        val = val + dp(pos + 1, fl, 0, bin);
 
    // Here flag will be 1 for the
    // next recursive call
    else if (bin[pos] == '1')
        val = val + dp(pos + 1, 1, 0, bin);
 
    // Placing 1 at this position only if
    // the previously inserted number is 0
    if (pr == 0) {
 
        // If the number is smaller than N
        if (fl == 1) {
            val += dp(pos + 1, fl, 1, bin);
        }
 
        // If the digit at current position is 1
        else if (bin[pos] == '1') {
            val += dp(pos + 1, fl, 1, bin);
        }
    }
 
    // Storing the solution to this subproblem
    return memo[pos][fl][pr] = val;
}
 
// Function to find the number of integers
// less than or equal to N with no
// consecutive 1’s in binary representation
int findIntegers(int num)
{
    // Convert N to binary form
    string bin;
 
    // Loop to convert N
    // from Decimal to binary
    while (num > 0) {
        if (num % 2)
            bin += "1";
        else
            bin += "0";
        num /= 2;
    }
    reverse(bin.begin(), bin.end());
 
    // Initialising the table with -1.
    memset(memo, -1, sizeof(memo));
 
    // Calling the function
    return dp(0, 0, 0, bin);
}
 
// Driver code
int main()
{
    int N = 12;
    cout << findIntegers(N);
 
    return 0;
}

Java

// Java program to count number of
// binary Strings without consecutive 1’s
class GFG{
  
// Table to store the solution of
// every sub problem
static int [][][]memo = new int[32][2][2];
  
// Function to fill the table
/* Here,
   pos: keeps track of current position.
   f1: is the flag to check if current
         number is less than N or not.
   pr: represents the previous digit
*/
static int dp(int pos, int fl, int pr, String bin)
{
    // Base case
    if (pos == bin.length())
        return 1;
  
    // Check if this subproblem
    // has already been solved
    if (memo[pos][fl][pr] != -1)
        return memo[pos][fl][pr];
  
    int val = 0;
  
    // Placing 0 at the current position
    // as it does not violate the condition
    if (bin.charAt(pos) == '0')
        val = val + dp(pos + 1, fl, 0, bin);
  
    // Here flag will be 1 for the
    // next recursive call
    else if (bin.charAt(pos) == '1')
        val = val + dp(pos + 1, 1, 0, bin);
  
    // Placing 1 at this position only if
    // the previously inserted number is 0
    if (pr == 0) {
  
        // If the number is smaller than N
        if (fl == 1) {
            val += dp(pos + 1, fl, 1, bin);
        }
  
        // If the digit at current position is 1
        else if (bin.charAt(pos) == '1') {
            val += dp(pos + 1, fl, 1, bin);
        }
    }
  
    // Storing the solution to this subproblem
    return memo[pos][fl][pr] = val;
}
  
// Function to find the number of integers
// less than or equal to N with no
// consecutive 1’s in binary representation
static int findIntegers(int num)
{
    // Convert N to binary form
    String bin = "";
  
    // Loop to convert N
    // from Decimal to binary
    while (num > 0) {
        if (num % 2 == 1)
            bin += "1";
        else
            bin += "0";
        num /= 2;
    }
    bin = reverse(bin);
  
    // Initialising the table with -1.
    for(int i = 0; i < 32; i++){
        for(int j = 0; j < 2; j++){
            for(int l = 0; l < 2; l++)
                memo[i][j][l] = -1;
        }
    }
  
    // Calling the function
    return dp(0, 0, 0, bin);
}
static String reverse(String input) {
    char[] a = input.toCharArray();
    int l, r = a.length - 1;
    for (l = 0; l < r; l++, r--) {
        char temp = a[l];
        a[l] = a[r];
        a[r] = temp;
    }
    return String.valueOf(a);
}
 
// Driver code
public static void main(String[] args)
{
    int N = 12;
    System.out.print(findIntegers(N));
  
}
}
 
// This code is contributed by sapnasingh4991

Python3

# Python program to count number of
# binary strings without consecutive 1’s
 
 
 
# Table to store the solution of
# every sub problem
memo=[[[-1 for i in range(2)] for j in range(2)] for k in range(32)]
 
# Function to fill the table
''' Here,
pos: keeps track of current position.
f1: is the flag to check if current
        number is less than N or not.
pr: represents the previous digit
'''
def dp(pos,fl,pr,bin):
    # Base case
    if (pos == len(bin)):
        return 1;
 
    # Check if this subproblem
    # has already been solved
    if (memo[pos][fl][pr] != -1):
        return memo[pos][fl][pr];
 
    val = 0
 
    # Placing 0 at the current position
    # as it does not violate the condition
    if (bin[pos] == '0'):
        val = val + dp(pos + 1, fl, 0, bin)
 
    # Here flag will be 1 for the
    # next recursive call
    elif (bin[pos] == '1'):
        val = val + dp(pos + 1, 1, 0, bin)
 
    # Placing 1 at this position only if
    # the previously inserted number is 0
    if (pr == 0):
 
        # If the number is smaller than N
        if (fl == 1):
            val += dp(pos + 1, fl, 1, bin)
 
        # If the digit at current position is 1
        elif (bin[pos] == '1'):
            val += dp(pos + 1, fl, 1, bin)
         
    # Storing the solution to this subproblem
    memo[pos][fl][pr] = val
    return val
 
# Function to find the number of integers
# less than or equal to N with no
# consecutive 1’s in binary representation
def findIntegers(num):
    # Convert N to binary form
    bin=""
 
    # Loop to convert N
    # from Decimal to binary
    while (num > 0):
        if (num % 2):
            bin += "1"
        else:
            bin += "0"
        num //= 2
     
    bin=bin[::-1];
 
     
 
    # Calling the function
    return dp(0, 0, 0, bin)
 
# Driver code
if __name__ == "__main__":
 
    N = 12
    print(findIntegers(N))

C#

// C# program to count number of
// binary Strings without consecutive 1’s
using System;
 
public class GFG{
   
// Table to store the solution of
// every sub problem
static int [,,]memo = new int[32,2,2];
   
// Function to fill the table
/* Here,
   pos: keeps track of current position.
   f1: is the flag to check if current
         number is less than N or not.
   pr: represents the previous digit
*/
static int dp(int pos, int fl, int pr, String bin)
{
    // Base case
    if (pos == bin.Length)
        return 1;
   
    // Check if this subproblem
    // has already been solved
    if (memo[pos,fl,pr] != -1)
        return memo[pos,fl,pr];
   
    int val = 0;
   
    // Placing 0 at the current position
    // as it does not violate the condition
    if (bin[pos] == '0')
        val = val + dp(pos + 1, fl, 0, bin);
   
    // Here flag will be 1 for the
    // next recursive call
    else if (bin[pos] == '1')
        val = val + dp(pos + 1, 1, 0, bin);
   
    // Placing 1 at this position only if
    // the previously inserted number is 0
    if (pr == 0) {
   
        // If the number is smaller than N
        if (fl == 1) {
            val += dp(pos + 1, fl, 1, bin);
        }
   
        // If the digit at current position is 1
        else if (bin[pos] == '1') {
            val += dp(pos + 1, fl, 1, bin);
        }
    }
   
    // Storing the solution to this subproblem
    return memo[pos,fl,pr] = val;
}
   
// Function to find the number of integers
// less than or equal to N with no
// consecutive 1’s in binary representation
static int findints(int num)
{
    // Convert N to binary form
    String bin = "";
   
    // Loop to convert N
    // from Decimal to binary
    while (num > 0) {
        if (num % 2 == 1)
            bin += "1";
        else
            bin += "0";
        num /= 2;
    }
    bin = reverse(bin);
   
    // Initialising the table with -1.
    for(int i = 0; i < 32; i++){
        for(int j = 0; j < 2; j++){
            for(int l = 0; l < 2; l++)
                memo[i,j,l] = -1;
        }
    }
   
    // Calling the function
    return dp(0, 0, 0, bin);
}
static String reverse(String input) {
    char[] a = input.ToCharArray();
    int l, r = a.Length - 1;
    for (l = 0; l < r; l++, r--) {
        char temp = a[l];
        a[l] = a[r];
        a[r] = temp;
    }
    return String.Join("",a);
}
  
// Driver code
public static void Main(String[] args)
{
    int N = 12;
    Console.Write(findints(N));
   
}
}
  
// This code contributed by Rajput-Ji

Javascript

<script>
 
// JavaScript program to count number of
// binary strings without consecutive 1’s
 
// Table to store the solution of
// every sub problem
var memo = Array.from(Array(32), ()=>Array(2));
 
for(var i =0;i<32;i++)
{
    for(var j =0; j<2; j++)
    {
        memo[i][j] = new Array(2).fill(-1);
    }
}
 
// Function to fill the table
/* Here,
   pos: keeps track of current position.
   f1: is the flag to check if current
         number is less than N or not.
   pr: represents the previous digit
*/
function dp(pos, fl, pr, bin)
{
    // Base case
    if (pos == bin.length)
        return 1;
 
    // Check if this subproblem
    // has already been solved
    if (memo[pos][fl][pr] != -1)
        return memo[pos][fl][pr];
 
    var val = 0;
 
    // Placing 0 at the current position
    // as it does not violate the condition
    if (bin[pos] == '0')
        val = val + dp(pos + 1, fl, 0, bin);
 
    // Here flag will be 1 for the
    // next recursive call
    else if (bin[pos] == '1')
        val = val + dp(pos + 1, 1, 0, bin);
 
    // Placing 1 at this position only if
    // the previously inserted number is 0
    if (pr == 0) {
 
        // If the number is smaller than N
        if (fl == 1) {
            val += dp(pos + 1, fl, 1, bin);
        }
 
        // If the digit at current position is 1
        else if (bin[pos] == '1') {
            val += dp(pos + 1, fl, 1, bin);
        }
    }
 
    // Storing the solution to this subproblem
    return memo[pos][fl][pr] = val;
}
 
// Function to find the number of integers
// less than or equal to N with no
// consecutive 1’s in binary representation
function findIntegers(num)
{
    // Convert N to binary form
    var bin = "";
 
    // Loop to convert N
    // from Decimal to binary
    while (num > 0) {
        if (num % 2)
            bin += "1";
        else
            bin += "0";
        num =parseInt(num/2);
    }
 
    bin = bin.split('').reverse().join('')
 
 
    // Calling the function
    return dp(0, 0, 0, bin);
}
 
// Driver code
var N = 12;
document.write( findIntegers(N));
 
</script>
Producción: 

8

 

Complejidad de tiempo: O(L * log(N)) 

  • O(log(N)) para convertir el número de decimal a binario.
  • O(L) para llenar la tabla, donde L es la longitud de la forma binaria.

Espacio Auxiliar: O(32 * 2 * 2)
 

Publicación traducida automáticamente

Artículo escrito por king_tsar y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA

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