计算可能的长度为N的二进制字符串，而没有P个连续的0和Q个连续的1

给定三个整数N ， P和Q ，任务是对长度为N的所有可能的不同二进制字符串进行计数，以使每个二进制字符串不包含P乘以连续的0和Q乘以连续的1。

例子：

Input: N = 5, P = 2, Q = 3
Output: 7
Explanation: Binary strings that satisfy the given conditions are { “01010”, “01011”, “01101”, “10101”, “10110”, “11010”, “11011”}. Therefore, the required output is 7.

Input: N = 5, P = 3, Q = 4
Output: 21

天真的方法：可以使用递归解决问题。以下是递归关系及其基本情况：

At each possible index of a Binary String, either place the value ‘0‘ or place the value ‘1‘

Therefore, cntBinStr(str, N, P, Q) = cntBinStr(str + ‘0’, N, P, Q) + cntBinStr(str + ‘1’, N, P, Q)
where cntBinStr(str, N, P, Q) stores the count of distinct binary strings which does not contain P consecutive 1s and Q consecutive 0s.

Base Case: If length(str) == N, check if str satisfy the given condition or not. If found to be true, return 1. Otherwise, return 0.

下面是上述方法的实现：

C++

// C++ program to implement
// the above approach
 
#include 
using namespace std;
 
// Function to check if a
// string satisfy the given
// condition or not
bool checkStr(string str,
              int P, int Q)
{
    // Stores the length
    // of string
    int N = str.size();
 
    // Stores the previous
    // character of the string
    char prev = str[0];
 
    // Stores the count of
    // consecutive equal characters
    int cnt = 0;
 
    // Traverse the string
    for (int i = 0; i < N;
         i++) {
 
        // If current character
        // is equal to the
        // previous character
        if (str[i] == prev) {
 
            cnt++;
        }
        else {
 
            // If count of consecutive
            // 1s is more than Q
            if (prev == '1' && cnt >= Q) {
 
                return false;
            }
 
            // If count of consecutive
            // 0s is more than P
            if (prev == '0' && cnt >= P) {
 
                return false;
            }
 
            // Reset value of cnt
            cnt = 1;
        }
 
        prev = str[i];
    }
 
    // If count of consecutive
    // 1s is more than Q
    if (prev == '1' && cnt >= Q) {
        return false;
    }
 
    // If count of consecutive
    // 0s is more than P
    if (prev == '0' && cnt >= P) {
        return false;
    }
 
    return true;
}
 
// Function to count all distinct
// binary strings that satisfy
// the given condition
int cntBinStr(string str, int N,
              int P, int Q)
{
    // Stores the length of str
    int len = str.size();
 
    // If length of str is N
    if (len == N) {
 
        // If str satisfy
        // the given condition
        if (checkStr(str, P, Q))
            return 1;
 
        // If str does not satisfy
        // the given condition
        return 0;
    }
 
    // Append a character '0' at
    // end of str
    int X = cntBinStr(str + '0',
                      N, P, Q);
 
    // Append a character '1' at
    // end of str
    int Y = cntBinStr(str + '1',
                      N, P, Q);
 
    // Return total count
    // of binary strings
    return X + Y;
}
 
// Driver Code
int main()
{
    int N = 5, P = 2, Q = 3;
    cout << cntBinStr("", N, P, Q);
 
    return 0;
}

Java

// Java program to implement
// the above approach 
class GFG{
   
// Function to check if a
// string satisfy the given
// condition or not
static boolean checkStr(String str,
                        int P, int Q)
{
     
    // Stores the length
    // of string
    int N = str.length();
   
    // Stores the previous
    // character of the string
    char prev = str.charAt(0);
   
    // Stores the count of
    // consecutive equal characters
    int cnt = 0;
   
    // Traverse the string
    for(int i = 0; i < N; i++)
    {
          
        // If current character
        // is equal to the
        // previous character
        if (str.charAt(i) == prev)
        {
            cnt++;
        }
        else
        {
              
            // If count of consecutive
            // 1s is more than Q
            if (prev == '1' && cnt >= Q)
            {
                return false;
            }
   
            // If count of consecutive
            // 0s is more than P
            if (prev == '0' && cnt >= P)
            {
                return false;
            }
   
            // Reset value of cnt
            cnt = 1;
        }
        prev = str.charAt(i);
    }
   
    // If count of consecutive
    // 1s is more than Q
    if (prev == '1' && cnt >= Q)
    {
        return false;
    }
   
    // If count of consecutive
    // 0s is more than P
    if (prev == '0' && cnt >= P)
    {
        return false;
    }
    return true;
}
   
// Function to count all distinct
// binary strings that satisfy
// the given condition
static int cntBinStr(String str, int N,
                          int P, int Q)
{
     
    // Stores the length of str
    int len = str.length();
   
    // If length of str is N
    if (len == N)
    {
          
        // If str satisfy
        // the given condition
        if (checkStr(str, P, Q))
            return 1;
   
        // If str does not satisfy
        // the given condition
        return 0;
    }
   
    // Append a character '0' at
    // end of str
    int X = cntBinStr(str + '0',
                      N, P, Q);
   
    // Append a character '1' at
    // end of str
    int Y = cntBinStr(str + '1',
                      N, P, Q);
   
    // Return total count
    // of binary strings
    return X + Y;
}
   
// Driver Code
public static void main (String[] args)
{
    int N = 5, P = 2, Q = 3;
      
    System.out.println(cntBinStr("", N, P, Q));
}
}
 
// This code is contributed by code_hunt

Python3

# Python3 program to implement
# the above approach
 
# Function to check if a
# satisfy the given
# condition or not
def checkStr(str, P, Q):
     
    # Stores the lenngth
    # of string
    N = len(str)
 
    # Stores the previous
    # character of the string
    prev = str[0]
 
    # Stores the count of
    # consecutive equal
    # characters
    cnt = 0
 
    # Traverse the string
    for i in range(N):
         
        # If current character
        # is equal to the
        # previous character
        if (str[i] == prev):
            cnt += 1
 
        else:
             
            # If count of consecutive
            # 1s is more than Q
            if (prev == '1' and
                cnt >= Q):
                return False
 
            # If count of consecutive
            # 0s is more than P
            if (prev == '0' and
                cnt >= P):
                return False
 
            # Reset value of cnt
            cnt = 1
 
        prev = str[i]
 
    # If count of consecutive
    # 1s is more than Q
    if (prev == '1'and
        cnt >= Q):
        return False
 
    # If count of consecutive
    # 0s is more than P
    if (prev == '0' and
        cnt >= P):
        return False
 
    return True
 
# Function to count all
# distinct binary strings
# that satisfy the given
# condition
def cntBinStr(str, N,
              P, Q):
     
    # Stores the length
    # of str
    lenn = len(str)
 
    # If lenngth of str
    # is N
    if (lenn == N):
 
        # If str satisfy
        # the given condition
        if (checkStr(str, P, Q)):
            return 1
 
        # If str does not satisfy
        # the given condition
        return 0
 
    # Append a character '0'
    # at end of str
    X = cntBinStr(str + '0',
                  N, P, Q)
 
    # Append a character
    # '1' at end of str
    Y = cntBinStr(str + '1',
                  N, P, Q)
 
    # Return total count
    # of binary strings
    return X + Y
 
# Driver Code
if __name__ == '__main__':
     
    N = 5
    P = 2
    Q = 3   
    print(cntBinStr("", N,
                    P, Q))
 
# This code is contributed by mohit kumar 29

C#

// C# program to implement
// the above approach 
using System;
 
class GFG{
  
// Function to check if a
// string satisfy the given
// condition or not
static bool checkStr(string str,
                     int P, int Q)
{
     
    // Stores the length
    // of string
    int N = str.Length;
  
    // Stores the previous
    // character of the string
    char prev = str[0];
  
    // Stores the count of
    // consecutive equal characters
    int cnt = 0;
  
    // Traverse the string
    for(int i = 0; i < N; i++)
    {
         
        // If current character
        // is equal to the
        // previous character
        if (str[i] == prev)
        {
            cnt++;
        }
        else
        {
             
            // If count of consecutive
            // 1s is more than Q
            if (prev == '1' && cnt >= Q)
            {
                return false;
            }
  
            // If count of consecutive
            // 0s is more than P
            if (prev == '0' && cnt >= P)
            {
                return false;
            }
  
            // Reset value of cnt
            cnt = 1;
        }
  
        prev = str[i];
    }
  
    // If count of consecutive
    // 1s is more than Q
    if (prev == '1' && cnt >= Q)
    {
        return false;
    }
  
    // If count of consecutive
    // 0s is more than P
    if (prev == '0' && cnt >= P)
    {
        return false;
    }
     
    return true;
}
  
// Function to count all distinct
// binary strings that satisfy
// the given condition
static int cntBinStr(string str, int N,
                          int P, int Q)
{
     
    // Stores the length of str
    int len = str.Length;
  
    // If length of str is N
    if (len == N)
    {
         
        // If str satisfy
        // the given condition
        if (checkStr(str, P, Q))
            return 1;
  
        // If str does not satisfy
        // the given condition
        return 0;
    }
  
    // Append a character '0' at
    // end of str
    int X = cntBinStr(str + '0',
                      N, P, Q);
  
    // Append a character '1' at
    // end of str
    int Y = cntBinStr(str + '1',
                      N, P, Q);
  
    // Return total count
    // of binary strings
    return X + Y;
}
  
// Driver Code
public static void Main ()
{
    int N = 5, P = 2, Q = 3;
     
    Console.WriteLine(cntBinStr("", N, P, Q));
}
}
 
// This code is contributed by sanjoy_62

输出

7

时间复杂度： O(2 ^N )
辅助空间： O(1)

高效方法：为了优化上述方法，其思想是使用动态编程。请按照以下步骤解决问题：

• 初始化两个2D数组，比如0 [N] [P]和one [N] [Q] 。
• zero [i] [j]存储长度为i的j个连续0的二进制字符串的计数。以自下而上的方式填充所有零[i] [j]的值。

Insert 0 at the i^th index. 
Case 1: If (i – 1)^th index of string contains 1. 
 

$zero[i][1] = \sum\limits_{j=1}^{Q - 1}one[i - 1][j]$

Case 2: If (i – 1)^th index of string contains 0.

$zero[i][j] = zero[i-1][j-1]$

for all r in the range [2, P – 1]. 

• one [i] [j]存储长度为i且具有j个连续的1的二进制字符串的计数。以自下而上的方式填充所有零[i] [j]的值。

Insert 1 at the i^th index. 
Case 1: If (i-1)^th index of string contains 0. 
 

$one[i][1] = \sum\limits_{j=1}^{P - 1}zero[i - 1][j]$

Case 2: If (i-1)^th index of string contains 1.

$one[i][j] = one[i-1][j-1]$

for all j in the range [2, Q – 1]. 

• 最后，打印满足给定条件的子数组的计数。

C++

// C++ program to implement
// the above approach
#include 
using namespace std;
 
// Function to count binary strings
// that satisfy the given condition
int cntBinStr(int N, int P, int Q)
{
    // zero[i][j] stores count
    // of binary strings of length i
    // having j consecutive 0s
    int zero[N + 1][P];
 
    // one[i][j] stores count
    // of binary strings of length i
    // having j consecutive 1s
    int one[N + 1][Q];
 
    // Set all values of
    // zero[][] array to 0
    memset(zero, 0, sizeof(zero));
 
    // Set all values of
    // one[i][j] array to 0
    memset(one, 0, sizeof(one));
 
    // Base case
    zero[1][1] = one[1][1] = 1;
 
    // Fill all the values of zero[i][j]
    // and one[i][j] in bottom up manner
    for (int i = 2; i <= N; i++) {
 
        for (int j = 2; j < P;
             j++) {
            zero[i][j] = zero[i - 1][j - 1];
        }
 
        for (int j = 1; j < Q;
             j++) {
            zero[i][1] = zero[i][1] + one[i - 1][j];
        }
 
        for (int j = 2; j < Q;
             j++) {
            one[i][j] = one[i - 1][j - 1];
        }
 
        for (int j = 1; j < P;
             j++) {
            one[i][1] = one[i][1] + zero[i - 1][j];
        }
    }
 
    // Stores count of binary strings
    // that satisfy the given condition
    int res = 0;
 
    // Count binary strings of
    // length N having less than
    // P consecutive 0s
    for (int i = 1; i < P; i++) {
        res = res + zero[N][i];
    }
 
    // Count binary strings of
    // length N having less than
    // Q consecutive 1s
    for (int i = 1; i < Q; i++) {
        res = res + one[N][i];
    }
 
    return res;
}
 
// Driver Code
int main()
{
    int N = 5, P = 2, Q = 3;
    cout << cntBinStr(N, P, Q);
    return 0;
}

Java

// Java program to implement
// the above approach
import java.util.*;
 
class GFG{
 
// Function to count binary Strings
// that satisfy the given condition
static int cntBinStr(int N, int P, int Q)
{
     
    // zero[i][j] stores count
    // of binary Strings of length i
    // having j consecutive 0s
    int [][]zero = new int[N + 1][P];
 
    // one[i][j] stores count
    // of binary Strings of length i
    // having j consecutive 1s
    int [][]one = new int[N + 1][Q];
 
    // Base case
    zero[1][1] = one[1][1] = 1;
 
    // Fill all the values of zero[i][j]
    // and one[i][j] in bottom up manner
    for(int i = 2; i <= N; i++)
    {
        for(int j = 2; j < P; j++)
        {
            zero[i][j] = zero[i - 1][j - 1];
        }
 
        for(int j = 1; j < Q; j++)
        {
            zero[i][1] = zero[i][1] +
                          one[i - 1][j];
        }
 
        for(int j = 2; j < Q; j++)
        {
            one[i][j] = one[i - 1][j - 1];
        }
 
        for(int j = 1; j < P; j++)
        {
            one[i][1] = one[i][1] +
                       zero[i - 1][j];
        }
    }
 
    // Stores count of binary Strings
    // that satisfy the given condition
    int res = 0;
 
    // Count binary Strings of
    // length N having less than
    // P consecutive 0s
    for(int i = 1; i < P; i++)
    {
        res = res + zero[N][i];
    }
 
    // Count binary Strings of
    // length N having less than
    // Q consecutive 1s
    for(int i = 1; i < Q; i++)
    {
        res = res + one[N][i];
    }
    return res;
}
 
// Driver Code
public static void main(String[] args)
{
    int N = 5, P = 2, Q = 3;
     
    System.out.print(cntBinStr(N, P, Q));
}
}
 
// This code is contributed by Amit Katiyar

Python3

# Python3 program to implement
# the above approach
 
# Function to count binary
# Strings that satisfy the
# given condition
def cntBinStr(N, P, Q):
   
    # zero[i][j] stores count
    # of binary Strings of length i
    # having j consecutive 0s
    zero = [[0 for i in range(P)]
               for j in range(N + 1)];
 
    # one[i][j] stores count
    # of binary Strings of length i
    # having j consecutive 1s
    one = [[0 for i in range(Q)]
              for j in range(N + 1)];
 
    # Base case
    zero[1][1] = one[1][1] = 1;
 
    # Fill all the values of
    # zero[i][j] and one[i][j]
    # in bottom up manner
    for i in range(2, N + 1):
        for j in range(2, P):
            zero[i][j] = zero[i - 1][j - 1];
 
        for j in range(1, Q):
            zero[i][1] = (zero[i][1] +
                          one[i - 1][j]);
 
        for j in range(2, Q):
            one[i][j] = one[i - 1][j - 1];
 
        for j in range(1, P):
            one[i][1] = one[i][1] + zero[i - 1][j];
 
    # Stores count of binary
    # Strings that satisfy
    # the given condition
    res = 0;
 
    # Count binary Strings of
    # length N having less than
    # P consecutive 0s
    for i in range(1, P):
        res = res + zero[N][i];
 
    # Count binary Strings of
    # length N having less than
    # Q consecutive 1s
    for i in range(1, Q):
        res = res + one[N][i];
 
    return res;
 
# Driver Code
if __name__ == '__main__':
   
    N = 5;
    P = 2;
    Q = 3;
    print(cntBinStr(N, P, Q));
 
# This code is contributed by gauravrajput1

C＃

// C# program to implement
// the above approach
using System;
 
class GFG{
 
// Function to count binary Strings
// that satisfy the given condition
static int cntBinStr(int N, int P, int Q)
{
     
    // zero[i,j] stores count
    // of binary Strings of length i
    // having j consecutive 0s
    int [,]zero = new int[N + 1, P];
 
    // one[i,j] stores count
    // of binary Strings of length i
    // having j consecutive 1s
    int [,]one = new int[N + 1, Q];
 
    // Base case
    zero[1, 1] = one[1, 1] = 1;
 
    // Fill all the values of zero[i,j]
    // and one[i,j] in bottom up manner
    for(int i = 2; i <= N; i++)
    {
        for(int j = 2; j < P; j++)
        {
            zero[i, j] = zero[i - 1, j - 1];
        }
 
        for(int j = 1; j < Q; j++)
        {
            zero[i, 1] = zero[i, 1] +
                          one[i - 1, j];
        }
 
        for(int j = 2; j < Q; j++)
        {
            one[i, j] = one[i - 1, j - 1];
        }
 
        for(int j = 1; j < P; j++)
        {
            one[i, 1] = one[i, 1] +
                       zero[i - 1, j];
        }
    }
 
    // Stores count of binary Strings
    // that satisfy the given condition
    int res = 0;
 
    // Count binary Strings of
    // length N having less than
    // P consecutive 0s
    for(int i = 1; i < P; i++)
    {
        res = res + zero[N, i];
    }
 
    // Count binary Strings of
    // length N having less than
    // Q consecutive 1s
    for(int i = 1; i < Q; i++)
    {
        res = res + one[N, i];
    }
    return res;
}
 
// Driver Code
public static void Main(String[] args)
{
    int N = 5, P = 2, Q = 3;
     
    Console.Write(cntBinStr(N, P, Q));
}
}
 
// This code is contributed by gauravrajput1

输出

7

时间复杂度： O(N *(P + Q))
辅助空间： O(N *(P + Q))