计算由“11”组成的 N 长度二进制字符串作为子字符串

给定一个正整数N ，任务是找到长度为N的二进制字符串的数量，其中包含“11”作为子字符串。

例子：

Input: N = 2
Output: 1
Explanation: The only string of length 2 that has “11” as a substring is “11”.

Input: N = 12
Output: 3719

方法：这个想法是基于以下观察得出将“11”作为以0或1开头的二进制表示的子串的可能性数量：

• 如果第一位为0 ，则起始位对具有“11”作为子串的字符串没有贡献。因此，剩余(N – 1)的比特具有以形成具有“11”作为一个子字符串的字符串。
• 如果第一位是1，位以下也为1，则存在^{2(N – 2)，}其具有“11”作为一个子字符串。
• 如果第一位是1，但以下的位为0，则具有“11”作为一个子字符串可以与剩余的(N – 2)形成的位字符串。
• 因此，生成所有长度为N的二进制字符串的递推关系为：

dp[i] = dp[i – 1] + dp[i – 2] + 2^{(i – 2)}
where, 
dp[i] is the string of length i having “11” as a substring.
and dp[0] = dp[1] = 0.

请按照以下步骤解决问题：

• 初始化一个数组，比如dp[] ，大小为(N + 1)并将dp[0]指定为0 ，将dp[1] 指定为0 。
• 预先计算2的前N 个幂并将其存储在一个数组中，比如power[] 。
• 迭代范围[2, N]并将dp[i]更新为(dp[i – 1] + dp[i – 2] + power[i – 2]) 。
• 完成以上步骤后，打印dp[N]的值作为结果。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to count binary strings
// of length N having substring "11"
void binaryStrings(int N)
{
    // Initialize dp[] of size N + 1
    int dp[N + 1];
 
    // Base Cases
    dp[0] = 0;
    dp[1] = 0;
 
 
 
    // Iterate over the range [2, N]
    for (int i = 2; i <= N; i++) {
        dp[i] = dp[i - 1]
                + dp[i - 2]       
                + (1<<(i-2));   // 1<<(i-2) means power of 2^(i-2)
    }
 
    // Print total count of substrings
    cout << dp[N];
}
 
// Driver Code
int main()
{
    int N = 12;
    binaryStrings(N);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
 
class GFG{
 
// Function to count binary strings
// of length N having substring "11"
static void binaryStrings(int N)
{
     
    // Initialize dp[] of size N + 1
    int[] dp = new int[N + 1];
 
    // Base Cases
    dp[0] = 0;
    dp[1] = 0;
 
    // Stores the first N powers of 2
    int[] power = new int[N + 1];
    power[0] = 1;
 
    // Generate
    for(int i = 1; i <= N; i++)
    {
        power[i] = 2 * power[i - 1];
    }
 
    // Iterate over the range [2, N]
    for(int i = 2; i <= N; i++)
    {
        dp[i] = dp[i - 1] + dp[i - 2] + power[i - 2];
    }
     
    // Print total count of substrings
    System.out.println(dp[N]);
}
 
// Driver Code
public static void main(String[] args)
{
    int N = 12;
     
    binaryStrings(N);
}
}
 
// This code is contributed by ukasp

Python3

# Python3 program for the above approach
 
# Function to count binary strings
# of length N having substring "11"
def binaryStrings(N):
     
    # Initialize dp[] of size N + 1
    dp = [0]*(N + 1)
 
    # Base Cases
    dp[0] = 0
    dp[1] = 0
 
    # Stores the first N powers of 2
    power = [0]*(N + 1)
    power[0] = 1
 
    # Generate
    for i in range(1, N + 1):
        power[i] = 2 * power[i - 1]
 
    # Iterate over the range [2, N]
    for i in range(2, N + 1):
        dp[i] = dp[i - 1] + dp[i - 2] + power[i - 2]
 
    # Prtotal count of substrings
    print (dp[N])
 
# Driver Code
if __name__ == '__main__':
    N = 12
    binaryStrings(N)
 
    # This code is contributed by mohit kumar 29.

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG
{
 
  // Function to count binary strings
  // of length N having substring "11"
  static void binaryStrings(int N)
  {
 
    // Initialize dp[] of size N + 1
    int []dp = new int[N + 1];
 
    // Base Cases
    dp[0] = 0;
    dp[1] = 0;
 
    // Stores the first N powers of 2
    int []power = new int[N + 1];
    power[0] = 1;
 
    // Generate
    for (int i = 1; i <= N; i++) {
      power[i] = 2 * power[i - 1];
    }
 
    // Iterate over the range [2, N]
    for (int i = 2; i <= N; i++) {
      dp[i] = dp[i - 1]
        + dp[i - 2]
        + power[i - 2];
    }
 
    // Print total count of substrings
    Console.WriteLine(dp[N]);
  }
 
  // Driver Code
  public static void Main()
  {
    int N = 12;
    binaryStrings(N);
  }
}
 
// This code is contributed by bgangwar59.

Javascript

 
// JavaScript program for the above approach
 
    // Function to count binary strings
    // of length N having substring "11"
    function binaryStrings(N) {
 
        // Initialize dp of size N + 1
        var dp = Array(N + 1).fill(0);
 
        // Base Cases
        dp[0] = 0;
        dp[1] = 0;
 
        // Stores the first N powers of 2
        var power = Array(N+1).fill(0);
        power[0] = 1;
 
        // Generate
        for (i = 1; i <= N; i++) {
            power[i] = 2 * power[i - 1];
        }
 
        // Iterate over the range [2, N]
        for (i = 2; i <= N; i++) {
            dp[i] = dp[i - 1] + dp[i - 2] + power[i - 2];
        }
 
        // Print total count of substrings
        document.write(dp[N]);
    }
 
    // Driver Code
     
        var N = 12;
 
        binaryStrings(N);
 
// This code contributed by aashish1995
 

输出

3719

时间复杂度： O(N)
辅助空间： O(N)