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Count possible binary strings of length N without P consecutive 0s and Q consecutive 1s

Given three integers N, P, and Q, the task is to count all possible distinct binary strings of length N such that each binary string does not contain P times consecutive 0’s and Q times consecutive 1’s.

Examples:

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

Naive Approach: The problem can be solved using Recursion. Following are the recurrence relations and their base cases :

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.

Below is the implementation of the above approach:

C++

 // C++ program to implement// the above approach #include using namespace std; // Function to check if a// string satisfy the given// condition or notbool 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 conditionint 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 Codeint 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.io.*;class GFG{   // Function to check if a// string satisfy the given// condition or notstatic 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 conditionstatic 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 Codepublic 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 notdef checkStr(str, P, Q):         # Stores the length    # 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# conditiondef cntBinStr(str, N,              P, Q):         # Stores the length    # of str    lenn = len(str)     # If length 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 Codeif __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 notstatic 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 conditionstatic 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 Codepublic static void Main (){    int N = 5, P = 2, Q = 3;         Console.WriteLine(cntBinStr("", N, P, Q));}} // This code is contributed by sanjoy_62

Javascript

 

Output

7

Time Complexity: O(2N)
Auxiliary Space: O(1)

Efficient Approach: To optimize the above approach the idea is to use Dynamic Programming. Follow the steps below to solve the problem:

• Initialize two 2D arrays, say zero[N][P] and one[N][Q].
• zero[i][j] stores the count of binary strings of length i having j consecutive 0’s. Fill all the value of zero[i][j] in bottom-up manner.

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

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

for all r in the range [2, P – 1].

• one[i][j] stores the count of binary strings of length i having j consecutive 1’s. Fill all the value of zero[i][j] in bottom-up manner.

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

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

for all j in the range [2, Q – 1].

• Finally, print count of subarrays that satisfy the given condition.

C++

 // C++ program to implement// the above approach#include using namespace std; // Function to count binary strings// that satisfy the given conditionint 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 Codeint main(){    int N = 5, P = 2, Q = 3;    cout << cntBinStr(N, P, Q);    return 0;}

Java

 // Java program to implement// the above approachimport java.io.*; class GFG{ // Function to count binary Strings// that satisfy the given conditionstatic 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 Codepublic 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 conditiondef 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 Codeif __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 approachusing System; class GFG{ // Function to count binary Strings// that satisfy the given conditionstatic 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 Codepublic static void Main(String[] args){    int N = 5, P = 2, Q = 3;         Console.Write(cntBinStr(N, P, Q));}} // This code is contributed by gauravrajput1

Javascript

 

Output

7

Time complexity: O(N * (P + Q))
Auxiliary Space: O(N * (P + Q))

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