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 <bits/stdc++.h>
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
import java.io.*;
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 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
# condition
def 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 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

Javascript

<script>

// Javascript program to implement
// the above approach

// Function to check if a
// string satisfy the given
// condition or not
function checkStr(str, P, Q)
{

    // Stores the length
    // of string
    let N = str.length;

    // Stores the previous
    // character of the string
    let prev = str[0];

    // Stores the count of
    // consecutive equal characters
    let cnt = 0;

    // Traverse the string
    for(let 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
function cntBinStr(str, N, P, Q)
{

    // Stores the length of str
    let 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
    let X = cntBinStr(str + '0',
                      N, P, Q);

    // Append a character '1' at
    // end of str
    let Y = cntBinStr(str + '1',
                      N, P, Q);

    // Return total count
    // of binary strings
    return X + Y;
}

// Driver Code

    let N = 5, P = 2, Q = 3;

    document.write(cntBinStr("", N, P, Q));

</script>

Output

7

Time Complexity: O(2^N)
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 i^th 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 i^th 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 <bits/stdc++.h>
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.io.*;

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

Javascript

<script>

// JavaScript program to implement
// the above approach

// Function to count binary strings
// that satisfy the given condition
function cntBinStr(N, P, Q)
{
    // zero[i][j] stores count
    // of binary strings of length i
    // having j consecutive 0s
    //and
    // Set all values of
    // zero[][] array to 0
    var zero = new Array(N+1).fill(0).
    map(item=>(new Array(P).fill(0)));

    // one[i][j] stores count
    // of binary strings of length i
    // having j consecutive 1s
    //and
    // Set all values of
    // one[i][j] array to 0
    var one  = new Array(N+1).fill(0).
    map(item=>(new Array(Q).fill(0)));;

    // 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 (var i = 2; i <= N; i++) {

        for (var j = 2; j < P;
             j++) {
            zero[i][j] = zero[i - 1][j - 1];
        }

        for (var j = 1; j < Q;
             j++) {
            zero[i][1] = zero[i][1] + one[i - 1][j];
        }

        for (var j = 2; j < Q;
             j++) {
            one[i][j] = one[i - 1][j - 1];
        }

        for (var 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
    var res = 0;

    // Count binary strings of
    // length N having less than
    // P consecutive 0s
    for (var i = 1; i < P; i++) {
        res = res + zero[N][i];
    }

    // Count binary strings of
    // length N having less than
    // Q consecutive 1s
    for (var i = 1; i < Q; i++) {
        res = res + one[N][i];
    }

    return res;
}

var N = 5, P = 2, Q = 3;
 document.write( cntBinStr(N, P, Q));

// This code in contributed by SoumikMondal

</script>

Output

7

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