Encuentre el número de strings binarias de longitud N con al menos 3 1 consecutivos

Dado un número entero N . La tarea es encontrar el número de todas las posibles strings binarias distintas de longitud N que tengan al menos 3 1 consecutivos.
Ejemplos: 
 

Entrada: N = 3 
Salida: 1 
La única string de longitud 3 posible es «111».
Entrada: N = 4 
Salida: 3 
Las 3 strings son “1110”, “0111” y “1111”. 
 

Enfoque ingenuo: considere todas las strings posibles.
A continuación se muestra la implementación del enfoque anterior: 
 

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function that returns true if s contains
// three consecutive 1's
bool check(string& s)
{
    int n = s.length();
    for (int i = 2; i < n; i++) {
        if (s[i] == '1' && s[i - 1] == '1' && s[i - 2] == '1')
            return 1;
    }
    return 0;
}
 
// Function to return the count
// of required strings
int countStr(int i, string& s)
{
    if (i < 0) {
        if (check(s))
            return 1;
        return 0;
    }
    s[i] = '0';
    int ans = countStr(i - 1, s);
    s[i] = '1';
    ans += countStr(i - 1, s);
    s[i] = '0';
    return ans;
}
 
// Driver code
int main()
{
    int N = 4;
    string s(N, '0');
    cout << countStr(N - 1, s);
 
    return 0;
}

// Java implementation of the approach
import java.util.*;
 
class GFG
{
 
// Function that returns true if s contains
// three consecutive 1's
static boolean check(String s)
{
    int n = s.length();
    for (int i = 2; i < n; i++)
    {
        if (s.charAt(i) == '1' &&
          s.charAt(i-1) == '1' &&
          s.charAt(i-2) == '1')
            return true;
    }
    return false;
}
 
// Function to return the count
// of required strings
static int countStr(int i, String s)
{
    if (i < 0)
    {
        if (check(s))
            return 1;
        return 0;
    }
    char[] myNameChars = s.toCharArray();
    myNameChars[i] = '0';
    s = String.valueOf(myNameChars);
 
    int ans = countStr(i - 1, s);
    char[] myChar = s.toCharArray();
    myChar[i] = '1';
    s = String.valueOf(myChar);
 
    ans += countStr(i - 1, s);
    char[]myChar1 = s.toCharArray();
    myChar1[i] = '0';
    s = String.valueOf(myChar1);
 
    return ans;
}
 
// Driver code
public static void main(String args[])
{
    int N = 4;
    String s = "0000";
    System.out.println(countStr(N - 1, s));
}
}
 
// This code is contributed by
// Surendra_Gangwar

# Python3 implementation of the approach
 
# Function that returns true if s contains
# three consecutive 1's
def check(s) :
    n = len(s);
    for i in range(2, n) :
        if (s[i] == '1' and s[i - 1] == '1' and
                            s[i - 2] == '1') :
            return 1;
 
# Function to return the count
# of required strings
def countStr(i, s) :
     
    if (i < 0) :
        if (check(s)) :
            return 1;
        return 0;
     
    s[i] = '0';
    ans = countStr(i - 1, s);
     
    s[i] = '1';
    ans += countStr(i - 1, s);
     
    s[i] = '0';
    return ans;
 
# Driver code
if __name__ == "__main__" :
    N = 4;
    s = list('0' * N);
     
    print(countStr(N - 1, s));
 
# This code is contributed by Ryuga

// C# implementation of the approach
// value x
using System;
 
class GFG
{
 
// Function that returns true if s contains
// three consecutive 1's
static bool check(String s)
{
    int n = s.Length;
    for (int i = 2; i < n; i++)
    {
        if (s[i] == '1' && s[i - 1] == '1' && s[i - 2] == '1')
            return true;
    }
    return false;
}
 
// Function to return the count
// of required strings
static int countStr(int i, String s)
{
    if (i < 0)
    {
        if (check(s))
            return 1;
        return 0;
    }
    char[] myNameChars = s.ToCharArray();
    myNameChars[i] = '0';
    s = String.Join("", myNameChars);
 
    int ans = countStr(i - 1, s);
    char[] myChar = s.ToCharArray();
    myChar[i] = '1';
    s = String.Join("", myChar);
 
    ans += countStr(i - 1, s);
    char[]myChar1 = s.ToCharArray();
    myChar1[i] = '0';
    s = String.Join("", myChar1);
 
    return ans;
}
 
// Driver code
public static void Main(String []args)
{
    int N = 4;
    String s = "0000";
    Console.WriteLine(countStr(N - 1, s));
}
}
 
/* This code contributed by PrinciRaj1992 */

<script>
 
// JavaScript implementation of the approach
// value x
 
// Function that returns true if s contains
// three consecutive 1's
function check(s)
{
    let n = s.length;
    for (let i = 2; i < n; i++)
    {
        if (s[i] == '1' &&
          s[i-1] == '1' &&
          s[i-2] == '1')
            return true;
    }
    return false;
}
 
// Function to return the count
// of required strings
function countStr(i,s)
{
    if (i < 0)
    {
        if (check(s))
            return 1;
        return 0;
    }
    let myNameChars = s.split("");
    myNameChars[i] = '0';
    s = (myNameChars).join("");
  
    let ans = countStr(i - 1, s);
    let myChar = s.split("");
    myChar[i] = '1';
    s = (myChar).join("");
  
    ans += countStr(i - 1, s);
    let myChar1 = s.split("");
    myChar1[i] = '0';
    s = (myChar1).join("");
  
    return ans;
}
 
// Driver code
let N = 4;
let s = "0000";
document.write(countStr(N - 1, s));
 
 
// This code is contributed by avanitrachhadiya2155
 
</script>

3

Complejidad de tiempo: O(2 ^N ) 
Complejidad de espacio: O(N) debido al espacio de pila de recurrencia.
Enfoque eficiente: utilizamos programación dinámica para calcular el número de strings. 
Estado de dp: dp(i, x) : indica el número de strings de longitud i con x 1s consecutivos en la posición i + 1 a i + x. 
Recurrencia: dp(i, x) = dp(i – 1, 0) + dp(i – 1, x + 1) 
La recurrencia se basa en el hecho de que la string puede tener un ‘0’ en la posición i o una ‘1’. 
 

1. Si tiene un ‘0’ en la posición i, entonces para (i-1) el valor de la posición x = 0.
2. Si tiene un ‘1’ en la posición i, el valor de la (i-1)-ésima posición de x = valor de x en la posición i + 1.

Condición base: dp(i, 3) = 2 ⁱ . Porque una vez que tiene 3 ‘1’ consecutivos, no le importa qué caracteres hay en la posición 1, 2… i de la string, ya que todas las strings son válidas.
A continuación se muestra la implementación del enfoque anterior: 
 

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
int n;
 
// Function to return the count
// of required strings
int solve(int i, int x, int dp[][4])
{
    if (i < 0)
        return x == 3;
    if (dp[i][x] != -1)
        return dp[i][x];
 
    // '0' at ith position
    dp[i][x] = solve(i - 1, 0, dp);
 
    // '1' at ith position
    dp[i][x] += solve(i - 1, x + 1, dp);
    return dp[i][x];
}
 
// Driver code
int main()
{
    n = 4;
    int dp[n][4];
 
    for (int i = 0; i < n; i++)
        for (int j = 0; j < 4; j++)
            dp[i][j] = -1;
 
    for (int i = 0; i < n; i++) {
 
        // Base condition:
        // 2^(i+1) because of 0 indexing
        dp[i][3] = (1 << (i + 1));
    }
    cout << solve(n - 1, 0, dp);
 
    return 0;
}

// Java implementation of the above approach
import java.util.*;
 
class GFG
{
 
    static int n;
 
    // Function to return the count
    // of required strings
    static int solve(int i, int x, int dp[][])
    {
        if (i < 0)
        {
            return x == 3 ? 1 : 0;
        }
        if (dp[i][x] != -1)
        {
            return dp[i][x];
        }
 
        // '0' at ith position
        dp[i][x] = solve(i - 1, 0, dp);
 
        // '1' at ith position
        dp[i][x] += solve(i - 1, x + 1, dp);
        return dp[i][x];
    }
 
    // Driver code
    public static void main(String[] args)
    {
        n = 4;
        int dp[][] = new int[n][4];
 
        for (int i = 0; i < n; i++)
        {
            for (int j = 0; j < 4; j++)
            {
                dp[i][j] = -1;
            }
        }
 
        for (int i = 0; i < n; i++)
        {
 
            // Base condition:
            // 2^(i+1) because of 0 indexing
            dp[i][3] = (1 << (i + 1));
        }
        System.out.print(solve(n - 1, 0, dp));
    }
}
 
// This code has been contributed by 29AjayKumar

# Python 3 implementation of the approach
 
# Function to return the count
# of required strings
def solve(i, x, dp):
    if (i < 0):
        return x == 3
    if (dp[i][x] != -1):
        return dp[i][x]
 
    # '0' at ith position
    dp[i][x] = solve(i - 1, 0, dp)
 
    # '1' at ith position
    dp[i][x] += solve(i - 1, x + 1, dp)
    return dp[i][x]
 
 
# Driver code
if __name__ == "__main__":
 
    n = 4;
    dp = [[0 for i in range(n)] for j in range(4)]
 
    for i in range(n):
        for j in range(4):
            dp[i][j] = -1
 
    for i in range(n) :
 
        # Base condition:
        # 2^(i+1) because of 0 indexing
        dp[i][3] = (1 << (i + 1))
     
    print(solve(n - 1, 0, dp))
 
# This code is contributed by ChitraNayal

// C# implementation of the above approach
using System;
 
class GFG
{
 
    static int n;
 
    // Function to return the count
    // of required strings
    static int solve(int i, int x, int [,]dp)
    {
        if (i < 0)
        {
            return x == 3 ? 1 : 0;
        }
        if (dp[i,x] != -1)
        {
            return dp[i,x];
        }
 
        // '0' at ith position
        dp[i,x] = solve(i - 1, 0, dp);
 
        // '1' at ith position
        dp[i,x] += solve(i - 1, x + 1, dp);
        return dp[i,x];
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        n = 4;
        int [,]dp = new int[n, 4];
 
        for (int i = 0; i < n; i++)
        {
            for (int j = 0; j < 4; j++)
            {
                dp[i, j] = -1;
            }
        }
 
        for (int i = 0; i < n; i++)
        {
 
            // Base condition:
            // 2^(i+1) because of 0 indexing
            dp[i,3] = (1 << (i + 1));
        }
        Console.Write(solve(n - 1, 0, dp));
    }
}
 
// This code contributed by Rajput-Ji

<script>
 
// Javascript implementation of the above approach
 
    var n;
 
    // Function to return the count
    // of required strings
    function solve(i , x , dp)
    {
        if (i < 0) {
            return x == 3 ? 1 : 0;
        }
        if (dp[i][x] != -1) {
            return dp[i][x];
        }
 
        // '0' at ith position
        dp[i][x] = solve(i - 1, 0, dp);
 
        // '1' at ith position
        dp[i][x] += solve(i - 1, x + 1, dp);
        return dp[i][x];
    }
 
    // Driver code
     
        n = 4;
        var dp = Array(n).fill().map(()=>Array(4).fill(0));
 
        for (i = 0; i < n; i++) {
            for (j = 0; j < 4; j++) {
                dp[i][j] = -1;
            }
        }
 
        for (i = 0; i < n; i++) {
 
            // Base condition:
            // 2^(i+1) because of 0 indexing
            dp[i][3] = (1 << (i + 1));
        }
        document.write(solve(n - 1, 0, dp));
 
// This code contributed by gauravrajput1
 
</script>

3

Complejidad temporal: O(N) 
Complejidad espacial: O(N)