Count number of binary strings without consecutive 1's
Last Updated :
23 Jul, 2025
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 = 3
Output: 5
Explanation: 5 strings are ("000", "001", "010", "100", "101").
Input: n = 2
Output: 3
Explanation: 3 strings are ("00", "01", "10").
Using Recursion - O(2^n) Time and O(n) Space
The idea is to explore two possible choices at each step of building the binary string. When constructing a string of length n, at each index, we have two choices: either place a 0 or place a 1. If we choose to place a 0, we can proceed to the next index. However, if we choose to place a 1, we must ensure that the next index does not contain a 1, so we skip the next index.
Mathematically the recurrence relation will look like the following:
countStrings(i) = countStrings(i+1) + countStrings(i+2)
Base Case:
- countStrings(i) = 1, if i >= n.
C++
// C++ program to count number of binary
// strings without consecutive 1's using recursion
#include <bits/stdc++.h>
using namespace std;
int countRecur(int i, int n) {
// Base case
if (i >= n) return 1;
// If we take 1 at ith index,
// we cannot have 1 at (i-1)
int take = countRecur(i+2, n);
// If we skip 1, we can consider
// 1 at i-1.
int noTake = countRecur(i+1, n);
return take + noTake;
}
int countStrings(int n) {
return countRecur(0, n);
}
int main() {
int n = 3;
cout << countStrings(n);
return 0;
}
// Java program to count number of binary
// strings without consecutive 1's using recursion
class GfG {
static int countRecur(int i, int n) {
// Base case
if (i >= n) return 1;
// If we take 1 at ith index,
// we cannot have 1 at (i-1)
int take = countRecur(i + 2, n);
// If we skip 1, we can consider
// 1 at i-1.
int noTake = countRecur(i + 1, n);
return take + noTake;
}
static int countStrings(int n) {
return countRecur(0, n);
}
public static void main(String[] args) {
int n = 3;
System.out.println(countStrings(n));
}
}
# Python program to count number of binary
# strings without consecutive 1's using recursion
def countRecur(i, n):
# Base case
if i >= n:
return 1
# If we take 1 at ith index,
# we cannot have 1 at (i-1)
take = countRecur(i + 2, n)
# If we skip 1, we can consider
# 1 at i-1.
noTake = countRecur(i + 1, n)
return take + noTake
def countStrings(n):
return countRecur(0, n)
if __name__ == "__main__":
n = 3
print(countStrings(n))
// C# program to count number of binary
// strings without consecutive 1's using recursion
using System;
class GfG {
static int countRecur(int i, int n) {
// Base case
if (i >= n) return 1;
// If we take 1 at ith index,
// we cannot have 1 at (i-1)
int take = countRecur(i + 2, n);
// If we skip 1, we can consider
// 1 at i-1.
int noTake = countRecur(i + 1, n);
return take + noTake;
}
static int countStrings(int n) {
return countRecur(0, n);
}
static void Main(string[] args) {
int n = 3;
Console.WriteLine(countStrings(n));
}
}
// JavaScript program to count number of binary
// strings without consecutive 1's using recursion
function countRecur(i, n) {
// Base case
if (i >= n) return 1;
// If we take 1 at ith index,
// we cannot have 1 at (i-1)
let take = countRecur(i + 2, n);
// If we skip 1, we can consider
// 1 at i-1.
let noTake = countRecur(i + 1, n);
return take + noTake;
}
function countStrings(n) {
return countRecur(0, n);
}
const n = 3;
console.log(countStrings(n));
Using Top-Down DP (Memoization) - O(n) Time and O(n) Space
If we notice carefully, we can observe that the above recursive solution holds the following two properties of Dynamic Programming:
1. Optimal Substructure: Number of ways to make binary string at i'th index depends on the optimal solutions of countStrings(i+1) and countStrings(i+2). By combining these substructures, we can efficiently calculate number of ways to make binary strings with consecutive 1's at index i.
2. Overlapping Subproblems: While applying a recursive approach in this problem, we notice that certain subproblems are computed multiple times. For example, while calculating countStrings(3), countStrings(4) and countStrings(5) are called. countStrings(4) will again call countStrings(5) which will lead to Overlapping Subproblems.
- There is only one parameter: i that changes in the recursive solution. So we create a 1D array of size n for memoization.
- We initialize this array as -1 to indicate nothing is computed initially.
- Now we modify our recursive solution to first check if the value is -1, then only make recursive calls. This way, we avoid re-computations of the same subproblems.
C++
// C++ program to count number of binary
// strings without consecutive 1's using memoization
#include <bits/stdc++.h>
using namespace std;
int countRecur(int i, int n, vector<int> &memo) {
// Base case
if (i >= n) return 1;
// If value is memoized
if (memo[i]!=-1) return memo[i];
// If we take 1 at ith index,
// we cannot have 1 at (i-1)
int take = countRecur(i+2, n, memo);
// If we skip 1, we can consider
// 1 at i-1.
int noTake = countRecur(i+1, n, memo);
return memo[i] = take + noTake;
}
int countStrings(int n) {
vector<int> memo(n, -1);
return countRecur(0, n, memo);
}
int main() {
int n = 3;
cout << countStrings(n);
return 0;
}
// Java program to count number of binary
// strings without consecutive 1's using memoization
class GfG {
static int countRecur(int i, int n, int[] memo) {
// Base case
if (i >= n) return 1;
// If value is memoized
if (memo[i] != -1) return memo[i];
// If we take 1 at ith index,
// we cannot have 1 at (i-1)
int take = countRecur(i + 2, n, memo);
// If we skip 1, we can consider
// 1 at i-1.
int noTake = countRecur(i + 1, n, memo);
return memo[i] = take + noTake;
}
static int countStrings(int n) {
int[] memo = new int[n];
for (int j = 0; j < n; j++) memo[j] = -1;
return countRecur(0, n, memo);
}
public static void main(String[] args) {
int n = 3;
System.out.println(countStrings(n));
}
}
# Python program to count number of binary
# strings without consecutive 1's using memoization
def countRecur(i, n, memo):
# Base case
if i >= n:
return 1
# If value is memoized
if memo[i] != -1:
return memo[i]
# If we take 1 at ith index,
# we cannot have 1 at (i-1)
take = countRecur(i + 2, n, memo)
# If we skip 1, we can consider
# 1 at i-1.
noTake = countRecur(i + 1, n, memo)
memo[i] = take + noTake
return memo[i]
def countStrings(n):
memo = [-1] * n
return countRecur(0, n, memo)
if __name__ == "__main__":
n = 3
print(countStrings(n))
// C# program to count number of binary
// strings without consecutive 1's using memoization
using System;
class GfG {
static int countRecur(int i, int n, int[] memo) {
// Base case
if (i >= n) return 1;
// If value is memoized
if (memo[i] != -1) return memo[i];
// If we take 1 at ith index,
// we cannot have 1 at (i-1)
int take = countRecur(i + 2, n, memo);
// If we skip 1, we can consider
// 1 at i-1.
int noTake = countRecur(i + 1, n, memo);
return memo[i] = take + noTake;
}
static int countStrings(int n) {
int[] memo = new int[n];
for (int j = 0; j < n; j++) memo[j] = -1;
return countRecur(0, n, memo);
}
static void Main(string[] args) {
int n = 3;
Console.WriteLine(countStrings(n));
}
}
// JavaScript program to count number of binary
// strings without consecutive 1's using memoization
function countRecur(i, n, memo) {
// Base case
if (i >= n) return 1;
// If value is memoized
if (memo[i] !== -1) return memo[i];
// If we take 1 at ith index,
// we cannot have 1 at (i-1)
let take = countRecur(i + 2, n, memo);
// If we skip 1, we can consider
// 1 at i-1.
let noTake = countRecur(i + 1, n, memo);
return memo[i] = take + noTake;
}
function countStrings(n) {
let memo = Array(n).fill(-1);
return countRecur(0, n, memo);
}
const n = 3;
console.log(countStrings(n));
Using Bottom-Up DP (Tabulation) - O(n) Time and O(n) Space
The idea is to fill the DP table based on next values. For each index, we can either place 1 or 0. The array is filled in an iterative manner from i = n-1 to i = 0.
The dynamic programming relation is as follows:
- dp[i] = dp[i+1] + dp[i+2]
C++
// C++ program to count number of binary
// strings without consecutive 1's using tabulation
#include <bits/stdc++.h>
using namespace std;
int countStrings(int n) {
if (n == 1)
return 2;
if (n == 2)
return 3;
vector<int> dp(n);
dp[n - 1] = 2;
dp[n - 2] = 3;
for (int i = n - 3; i >= 0; i--) {
dp[i] = dp[i + 1] + dp[i + 2];
}
return dp[0];
}
int main() {
int n = 3;
cout << countStrings(n);
return 0;
}
// Java program to count number of binary
// strings without consecutive 1's using tabulation
class GfG {
static int countStrings(int n) {
if (n == 1) return 2;
if (n == 2) return 3;
int[] dp = new int[n];
dp[n - 1] = 2;
dp[n - 2] = 3;
for (int i = n - 3; i >= 0; i--) {
dp[i] = dp[i + 1] + dp[i + 2];
}
return dp[0];
}
public static void main(String[] args) {
int n = 3;
System.out.println(countStrings(n));
}
}
# Python program to count number of binary
# strings without consecutive 1's using tabulation
def countStrings(n):
if n == 1:
return 2
if n == 2:
return 3
dp = [0] * n
dp[n - 1] = 2
dp[n - 2] = 3
for i in range(n - 3, -1, -1):
dp[i] = dp[i + 1] + dp[i + 2]
return dp[0]
if __name__ == "__main__":
n = 3
print(countStrings(n))
// C# program to count number of binary
// strings without consecutive 1's using tabulation
using System;
class GfG {
static int countStrings(int n) {
if (n == 1) return 2;
if (n == 2) return 3;
int[] dp = new int[n];
dp[n - 1] = 2;
dp[n - 2] = 3;
for (int i = n - 3; i >= 0; i--) {
dp[i] = dp[i + 1] + dp[i + 2];
}
return dp[0];
}
static void Main(string[] args) {
int n = 3;
Console.WriteLine(countStrings(n));
}
}
// JavaScript program to count number of binary
// strings without consecutive 1's using tabulation
function countStrings(n) {
if (n === 1) return 2;
if (n === 2) return 3;
const dp = new Array(n).fill(0);
dp[n - 1] = 2;
dp[n - 2] = 3;
for (let i = n - 3; i >= 0; i--) {
dp[i] = dp[i + 1] + dp[i + 2];
}
return dp[0];
}
const n = 3;
console.log(countStrings(n));
Using Space Optimized DP - O(n) Time and O(1) Space
In previous approach of dynamic programming we have derive the relation between states as given below:
- dp[i] = dp[i+1] + dp[i+2]
We observe that for calculating dp[i] state we only need dp[i+1] and dp[i+2]. There is no need to store all the next states.
C++
// C++ program to count number of binary
// strings without consecutive 1's
// using space optimised dp
#include <bits/stdc++.h>
using namespace std;
int countStrings(int n) {
if (n == 1)
return 2;
if (n == 2)
return 3;
int prev1 = 3, prev2 = 2;
for (int i = n - 3; i >= 0; i--) {
int curr = prev1 + prev2;
prev2 = prev1;
prev1 = curr;
}
return prev1;
}
int main() {
int n = 3;
cout << countStrings(n);
return 0;
}
// Java program to count number of binary
// strings without consecutive 1's
// using space optimised dp
class GfG {
static int countStrings(int n) {
if (n == 1)
return 2;
if (n == 2)
return 3;
int prev1 = 3, prev2 = 2;
for (int i = n - 3; i >= 0; i--) {
int curr = prev1 + prev2;
prev2 = prev1;
prev1 = curr;
}
return prev1;
}
public static void main(String[] args) {
int n = 3;
System.out.println(countStrings(n));
}
}
# Python program to count number of binary
# strings without consecutive 1's
# using space optimised dp
def countStrings(n):
if n == 1:
return 2
if n == 2:
return 3
prev1, prev2 = 3, 2
for i in range(n - 3, -1, -1):
curr = prev1 + prev2
prev2 = prev1
prev1 = curr
return prev1
if __name__ == "__main__":
n = 3
print(countStrings(n))
// C# program to count number of binary
// strings without consecutive 1's
// using space optimised dp
using System;
class GfG {
static int countStrings(int n) {
if (n == 1)
return 2;
if (n == 2)
return 3;
int prev1 = 3, prev2 = 2;
for (int i = n - 3; i >= 0; i--) {
int curr = prev1 + prev2;
prev2 = prev1;
prev1 = curr;
}
return prev1;
}
static void Main(string[] args) {
int n = 3;
Console.WriteLine(countStrings(n));
}
}
// JavaScript program to count number of binary
// strings without consecutive 1's
// using space optimised dp
function countStrings(n) {
if (n === 1)
return 2;
if (n === 2)
return 3;
let prev1 = 3, prev2 = 2;
for (let i = n - 3; i >= 0; i--) {
let curr = prev1 + prev2;
prev2 = prev1;
prev1 = curr;
}
return prev1;
}
const n = 3;
console.log(countStrings(n));
Using Matrix Exponentiation - O(logn) Time and O(logn) Space
The recurrence relation to count binary strings of length i is:
with bases cases F(1) = 2 and F(2) = 3.
Matrix exponentiation can be used to solve this problem in O(logn) time. Refer to Matrix Exponentiation for detailed approach.
C++
// C++ program to count number of binary
// strings without consecutive 1's
#include <bits/stdc++.h>
using namespace std;
// Function to mutiply two matrices
vector<vector<int>> multiply(vector<vector<int>>& v1, vector<vector<int>>& v2) {
vector<vector<int>> ans(2, vector<int>(2, 0));
for (int i = 0; i < 2; i++) {
for (int j = 0; j < 2; j++) {
for (int k = 0; k < 2; k++) {
ans[i][j] += v1[i][k] * v2[k][j];
}
}
}
return ans;
}
// Function to find matrix to the power n.
vector<vector<int>> power(vector<vector<int>> &v, int n) {
if (n == 0) {
return {{1, 0}, {0, 1}};
}
vector<vector<int>> res = power(v, n / 2);
res = multiply(res, res);
if (n % 2 == 1)
res = multiply(res, v);
return res;
}
int countStrings(int n) {
if (n == 2)
return 3;
if (n == 1)
return 2;
vector<vector<int>> v = {{1, 1}, {1, 0}};
vector<vector<int>> p = power(v, n - 2);
int ans = p[0][0] * 3 + p[0][1] * 2;
return ans;
}
int main() {
int n = 3;
cout << countStrings(n);
return 0;
}
// Java program to count number of binary
// strings without consecutive 1's
class GfG {
// Function to multiply two matrices
static int[][] multiply(int[][] v1, int[][] v2) {
int[][] ans = new int[2][2];
for (int i = 0; i < 2; i++) {
for (int j = 0; j < 2; j++) {
for (int k = 0; k < 2; k++) {
ans[i][j] += v1[i][k] * v2[k][j];
}
}
}
return ans;
}
// Function to find matrix to the power n
static int[][] power(int[][] v, int n) {
if (n == 0) {
return new int[][] { { 1, 0 }, { 0, 1 } };
}
int[][] res = power(v, n / 2);
res = multiply(res, res);
if (n % 2 == 1) {
res = multiply(res, v);
}
return res;
}
static int countStrings(int n) {
if (n == 2)
return 3;
if (n == 1)
return 2;
int[][] v = { { 1, 1 }, { 1, 0 } };
int[][] p = power(v, n - 2);
int ans = p[0][0] * 3 + p[0][1] * 2;
return ans;
}
public static void main(String[] args) {
int n = 3;
System.out.println(countStrings(n));
}
}
# Python program to count number of binary
# strings without consecutive 1's
# Function to multiply two matrices
def multiply(v1, v2):
ans = [[0, 0], [0, 0]]
for i in range(2):
for j in range(2):
for k in range(2):
ans[i][j] += v1[i][k] * v2[k][j]
return ans
# Function to find matrix to the power n
def power(v, n):
if n == 0:
return [[1, 0], [0, 1]]
res = power(v, n // 2)
res = multiply(res, res)
if n % 2 == 1:
res = multiply(res, v)
return res
def countStrings(n):
if n == 2:
return 3
if n == 1:
return 2
v = [[1, 1], [1, 0]]
p = power(v, n - 2)
ans = p[0][0] * 3 + p[0][1] * 2
return ans
if __name__ == "__main__":
n = 3
print(countStrings(n))
// C# program to count number of binary
// strings without consecutive 1's
using System;
class GfG {
// Function to multiply two matrices
static int[, ] multiply(int[, ] v1, int[, ] v2) {
int[, ] ans = new int[2, 2];
for (int i = 0; i < 2; i++) {
for (int j = 0; j < 2; j++) {
for (int k = 0; k < 2; k++) {
ans[i, j] += v1[i, k] * v2[k, j];
}
}
}
return ans;
}
// Function to find matrix to the power n
static int[, ] power(int[, ] v, int n) {
if (n == 0) {
return new int[, ] { { 1, 0 }, { 0, 1 } };
}
int[, ] res = power(v, n / 2);
res = multiply(res, res);
if (n % 2 == 1) {
res = multiply(res, v);
}
return res;
}
static int countStrings(int n) {
if (n == 2)
return 3;
if (n == 1)
return 2;
int[, ] v = { { 1, 1 }, { 1, 0 } };
int[, ] p = power(v, n - 2);
int ans = p[0, 0] * 3 + p[0, 1] * 2;
return ans;
}
static void Main(string[] args) {
int n = 3;
Console.WriteLine(countStrings(n));
}
}
// JavaScript program to count number of binary
// strings without consecutive 1's
// Function to multiply two matrices
function multiply(v1, v2) {
let ans = [ [ 0, 0 ], [ 0, 0 ] ];
for (let i = 0; i < 2; i++) {
for (let j = 0; j < 2; j++) {
for (let k = 0; k < 2; k++) {
ans[i][j] += v1[i][k] * v2[k][j];
}
}
}
return ans;
}
// Function to find matrix to the power n
function power(v, n) {
if (n === 0) {
return [ [ 1, 0 ], [ 0, 1 ] ];
}
let res = power(v, Math.floor(n / 2));
res = multiply(res, res);
if (n % 2 === 1) {
res = multiply(res, v);
}
return res;
}
function countStrings(n) {
if (n === 2)
return 3;
if (n === 1)
return 2;
let v = [ [ 1, 1 ], [ 1, 0 ] ];
let p = power(v, n - 2);
let ans = p[0][0] * 3 + p[0][1] * 2;
return ans;
}
const n = 3;
console.log(countStrings(n));
Related article:
Count Binary Strings With No Consecutive 1s | DSA Problem
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