Count possible binary strings of length N without P consecutive 0s and Q consecutive 1s
Last Updated :
26 Jul, 2025
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))
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Greedy AlgorithmsGreedy algorithms are a class of algorithms that make locally optimal choices at each step with the hope of finding a global optimum solution. At every step of the algorithm, we make a choice that looks the best at the moment. To make the choice, we sometimes sort the array so that we can always get
3 min read
Graph AlgorithmsGraph is a non-linear data structure like tree data structure. The limitation of tree is, it can only represent hierarchical data. For situations where nodes or vertices are randomly connected with each other other, we use Graph. Example situations where we use graph data structure are, a social net
3 min read
Dynamic Programming or DPDynamic Programming is an algorithmic technique with the following properties.It is mainly an optimization over plain recursion. Wherever we see a recursive solution that has repeated calls for the same inputs, we can optimize it using Dynamic Programming. The idea is to simply store the results of
3 min read
Bitwise AlgorithmsBitwise algorithms in Data Structures and Algorithms (DSA) involve manipulating individual bits of binary representations of numbers to perform operations efficiently. These algorithms utilize bitwise operators like AND, OR, XOR, NOT, Left Shift, and Right Shift.BasicsIntroduction to Bitwise Algorit
4 min read
Advanced
Segment TreeSegment Tree is a data structure that allows efficient querying and updating of intervals or segments of an array. It is particularly useful for problems involving range queries, such as finding the sum, minimum, maximum, or any other operation over a specific range of elements in an array. The tree
3 min read
Pattern SearchingPattern searching algorithms are essential tools in computer science and data processing. These algorithms are designed to efficiently find a particular pattern within a larger set of data. Patten SearchingImportant Pattern Searching Algorithms:Naive String Matching : A Simple Algorithm that works i
2 min read
GeometryGeometry is a branch of mathematics that studies the properties, measurements, and relationships of points, lines, angles, surfaces, and solids. From basic lines and angles to complex structures, it helps us understand the world around us.Geometry for Students and BeginnersThis section covers key br
2 min read
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