Count possible binary strings of length N without P consecutive 0s and Q consecutive 1s
Last Updated :
13 Dec, 2022
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 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 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# 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
<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>
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.
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] 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.
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].
- 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 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 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# 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
<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>
Time complexity: O(N * (P + Q))
Auxiliary Space: O(N * (P + Q))
Similar Reads
Count of Binary strings having at most X consecutive 1s and Y consecutive 0s
Given two integers N and M (1 ≤ N, M ≤ 100) denoting the total number of 1s and 0s respectively. The task is to count the number of possible arrangements of these 0s and 1s such that any arrangement has at most X consecutive 1s and Y consecutive 0s (1 ≤ X, Y ≤ 10). As the number of arrangements can
7 min read
Generate a Binary String without any consecutive 0's and at most K consecutive 1's
Given two integers N and M, the task is to construct a binary string with the following conditions : The Binary String consists of N 0's and M 1'sThe Binary String has at most K consecutive 1's.The Binary String does not contain any adjacent 0's. If it is not possible to construct such a binary stri
7 min read
Count of binary strings of length N with even set bit count and at most K consecutive 1s
Given two integers N and K, the task is to find the number of binary strings of length N having an even number of 1's out of which less than K are consecutive.Examples: Input: N = 4, K = 2 Output: 4 Explanation: The possible binary strings are 0000, 0101, 1001, 1010. They all have even number of 1's
12 min read
Count number of binary strings without consecutive 1’s : Set 2
Given a positive integer N, the task is to count all possible distinct binary strings of length N such that there are no consecutive 1’s. Examples: Input: N = 5 Output: 5 Explanation: The non-negative integers <= 5 with their corresponding binary representations are: 0 : 0 1 : 1 2 : 10 3 : 11 4 :
15+ min read
Count number of binary strings without consecutive 1's
Given a positive integer n, the task is to count all possible distinct binary strings of length n such that there are no consecutive 1's.Examples: Input: n = 3Output: 5Explanation: 5 strings are ("000", "001", "010", "100", "101").Input: n = 2Output: 3Explanation: 3 strings are ("00", "01", "10").Ta
15+ min read
Generate all Binary Strings of length N with equal count of 0s and 1s
Given an integer N, the task is to generate all the binary strings with equal 0s and 1s. If no strings are possible, print -1 Examples: Input: N = 2 Output: “01â€, “10â€Explanation: All possible binary strings of length 2 are: 01, 10, 11, 00. Out of these, only 2 have equal number of 0s and 1s Input:
6 min read
Count of binary string of length N with X 0s and Y 1s
Given positive integers N, X and Y. The task is to find the count of unique binary strings of length N having X 0s and Y 1s. Examples: Input: N=5, X=3, Y=2Output: 10Explanation: There are 10 binary strings of length 5 with 3 0s and 2 1s, such as: 00011, 00101, 01001, 10001, 00110, 01010, 10010, 0110
4 min read
Count of binary strings of length N having equal count of 0's and 1's
Given an integer N, the task is to find the number of binary strings possible of length N having same frequency of 0s and 1s. If such string is possible of length N, print -1. Note: Since the count can be very large, return the answer modulo 109+7.Examples: Input: N = 2 Output: 2 Explanation: All po
6 min read
Count of Binary Strings possible as per given conditions
Given two integers N and M, where N denotes the count of '0' and M denotes the count of '1', and an integer K, the task is to find the maximum number of binary strings that can be generated of the following two types: A string can consist of K '0's and a single '1'.A string can consist of K '1's and
4 min read
Count N-length Binary Strings consisting of "11" as substring
Given a positive integer N, the task is to find the number of binary strings of length N which contains "11" as a substring. Examples: Input: N = 2Output: 1Explanation: The only string of length 2 that has "11" as a substring is "11". Input: N = 12Output: 3719 Approach: The idea is to derive the num
8 min read