Given the number N, count the number of ways to create a binary string (the string that contains characters as zero or one) of size N such that it does not contain “0100” as a substring.
A substring is a contiguous sequence of characters within a string.
Examples:
Input: N = 4
Output: 15
Explanation: The answer will contain all possible substrings of size 4 except 0100 itself.Input: N = 5
Output: 28
Naive approach: The article can be solved based on the following idea:
There are N positions to fill of binary string with either 1 or 0 while avoiding “0100” substring, for any position i recursive function will be called by setting that position with either 1 or 0 except when last three characters are 010 only recursive function is called for 1 (to form 0101) not for 0 since if 0 is called then 0100 will be present as a substring in given count of strings. Now to keep track of last three characters make use of bitmask.
Follow the steps below to solve the problem:
- Create a recursive function that takes two parameters one is the position that needs to fill and the other is the last three characters in form of a bitmask.
- Check the base cases. If the value of i is equal to N return 1.
- If the last three characters are not 010 Call the function recursively for 1 and 0, and sum up the values that are returned.
- else if the last three characters are 010 then call the function only for 1.
- return the value sum.
Below is the implementation of the above approach.
C++
// C++ code to implement the approach#include <bits/stdc++.h>using namespace std;// Returns number of strings that does// not contain 0100 in its substringint recur(int i, int lastThreeDigits, int N){ // Base Case if (i == N) return 1; int ans = 0LL; if (i >= 3 and lastThreeDigits == 2) { // If last three substring is 010 // then only one option we have to // choose for next character which // is 1 so lastThreeDigits will be // 5 which is 101 ans += recur(i + 1, 5, N); } else { // Choosing 1 as a next character by // left shifting lastThreeDigits by // 1 then taking its bitwise AND with // 7 so that only first three bits // remain active and others become zero ans += recur(i + 1, lastThreeDigits << 1 & 7 | 1, N); // Choosing 0 as a next character by // left shifting lastThreeDigits by // 1 then taking its bitwise AND with // 7 so that only first three bits // remain active and others become zero ans += recur(i + 1, lastThreeDigits << 1 & 7, N); } return ans;}// Function to count number of binary// strings that does not contain 0100// as substringvoid countBinStrings(int N){ cout << recur(0, 0, N) << endl;}// Driver Codeint main(){ int N = 4; // Function Call countBinStrings(N); int N1 = 5; // Function Call countBinStrings(N1); return 0;} |
Java
// Java code to implement the approachimport java.io.*;import java.util.*;class GFG { static int recur(int i, int lastThreeDigits, int N) { // Base Case if (i == N) return 1; int ans = 0; if (i >= 3 && lastThreeDigits == 2) { // If last three substring is 010 // then only one option we have to // choose for next character which // is 1 so lastThreeDigits will be // 5 which is 101 ans += recur(i + 1, 5, N); } else { // Choosing 1 as a next character by // left shifting lastThreeDigits by // 1 then taking its bitwise AND with // 7 so that only first three bits // remain active and others become zero ans += recur(i + 1, lastThreeDigits << 1 & 7 | 1, N); // Choosing 0 as a next character by // left shifting lastThreeDigits by // 1 then taking its bitwise AND with // 7 so that only first three bits // remain active and others become zero ans += recur(i + 1, lastThreeDigits << 1 & 7, N); } return ans; } // Function to count number of binary // strings that does not contain 0100 // as substring static void countBinStrings(int N) { System.out.println(recur(0, 0, N)); } // Driver Code public static void main(String[] args) { int N = 4; // Function Call countBinStrings(N); int N1 = 5; // Function Call countBinStrings(N1); }}// This code is contributed by lokesh. |
Python3
#Python code for the above approachdef recur(i, lastThreeDigits, N): # Base Case if i == N: return 1 ans = 0 if i >= 3 and lastThreeDigits == 2: # If last three substring is 010 # then only one option we have to # choose for next character which # is 1 so lastThreeDigits will be # 5 which is 101 ans += recur(i + 1, 5, N) else: # Choosing 1 as a next character by # left shifting lastThreeDigits by # 1 then taking its bitwise AND with # 7 so that only first three bits # remain active and others become zero ans += recur(i + 1, (lastThreeDigits << 1) & 7 | 1, N) # Choosing 0 as a next character by # left shifting lastThreeDigits by # 1 then taking its bitwise AND with # 7 so that only first three bits # remain active and others become zero ans += recur(i + 1, (lastThreeDigits << 1) & 7, N) return ans# Function to count number of binary# strings that does not contain 0100# as substringdef countBinStrings(N): print(recur(0, 0, N))# Driver codeN = 4# Function CallcountBinStrings(N)N1 = 5# Function CallcountBinStrings(N1)#This code is contributed by Potta Lokesh |
C#
// C# code to implement the approachusing System;using System.Collections.Generic;public class Gfg { // Returns number of strings that does // not contain 0100 in its substring static int recur(int i, int lastThreeDigits, int N) { // Base Case if (i == N) return 1; int ans = 0; if (i >= 3 && lastThreeDigits == 2) { // If last three substring is 010 // then only one option we have to // choose for next character which // is 1 so lastThreeDigits will be // 5 which is 101 ans += recur(i + 1, 5, N); } else { // Choosing 1 as a next character by // left shifting lastThreeDigits by // 1 then taking its bitwise AND with // 7 so that only first three bits // remain active and others become zero ans += recur(i + 1, lastThreeDigits << 1 & 7 | 1, N); // Choosing 0 as a next character by // left shifting lastThreeDigits by // 1 then taking its bitwise AND with // 7 so that only first three bits // remain active and others become zero ans += recur(i + 1, lastThreeDigits << 1 & 7, N); } return ans; } // Function to count number of binary // strings that does not contain 0100 // as substring static void countBinStrings(int N) { Console.WriteLine(recur(0, 0, N)); } // Driver Code public static void Main(string[] args) { int N = 4; // Function Call countBinStrings(N); int N1 = 5; // Function Call countBinStrings(N1); }} |
Javascript
// Javascript code for the above approach// Returns number of strings that does// not contain 0100 in its substringfunction recur(i, lastThreeDigits, N){ // Base Case if (i == N) return 1; let ans = 0; if (i >= 3 && lastThreeDigits == 2) { // If last three substring is 010 // then only one option we have to // choose for next character which // is 1 so lastThreeDigits will be // 5 which is 101 ans += recur(i + 1, 5, N); } else { // Choosing 1 as a next character by // left shifting lastThreeDigits by // 1 then taking its bitwise AND with // 7 so that only first three bits // remain active and others become zero ans += recur(i + 1, lastThreeDigits << 1 & 7 | 1, N); // Choosing 0 as a next character by // left shifting lastThreeDigits by // 1 then taking its bitwise AND with // 7 so that only first three bits // remain active and others become zero ans += recur(i + 1, lastThreeDigits << 1 & 7, N); } return ans;}// Function to count number of binary// strings that does not contain 0100// as substringfunction countBinStrings(N){ console.log(recur(0, 0, N));}// Driver Codelet N = 4; // Function Call countBinStrings(N); let N1 = 5; // Function Call countBinStrings(N1); // This code is contributed by poojaagarwal2. |
Output
15 28
Time Complexity: O(2N)
Auxiliary Space: O(1)
Efficient Approach: The above approach can be optimized based on the following idea:
The idea is similar, but it can be observed that there are N * 8 states but the recursive function is called several times. That means that some states are called repeatedly. So the idea is to store the value of states. This can be done using recursive structure intact and just store the value in a HashMap and whenever the function is called, return the value store without computing .
Follow the steps below to solve the problem:
- Create a 2d array of dp[N + 1][8] initially filled with -1.
- If the answer for a particular state is computed then save it in dp[i][lastThreeDigits].
- If the answer for a particular state is already computed then just return dp[i][lastThreeDigits].
Below is the implementation of the above approach.
C++
// C++ code to implement the approach#include <bits/stdc++.h>using namespace std;int dp[100001][8];// Returns number of strings that does// not contain 0100 in its substringint recur(int i, int lastThreeDigits, int N){ // Base Case if (i == N) return 1; // If answer for current state is // already calculated then just // return the answer. if (dp[i][lastThreeDigits] != -1) return dp[i][lastThreeDigits]; int ans = 0LL; if (i >= 3 and lastThreeDigits == 2) { // If last three substring is 010 // then only one option we have to // choose for next character which // is 1 so lastThreeDigits will be // 5 which is 101 ans += recur(i + 1, 5, N); } else { // Choosing 1 as a next character // by left shifting lastThreeDigits // by 1 then taking its bitwise // AND with 7 so that only first // three bits remain active and // others become zero ans += recur(i + 1, lastThreeDigits << 1 & 7 | 1, N); // Choosing 0 as a next character // by left shifting lastThreeDigits // by 1 then taking its bitwise AND // with 7 so that only first three // bits remain active and others // become zero ans += recur(i + 1, lastThreeDigits << 1 & 7, N); } // Saving and returning sum return dp[i][lastThreeDigits] = ans;}// Function to count number of binary// strings that does not contain 0100// as substringvoid countBinStrings(int N){ // Initializing value of dp with -1 memset(dp, -1, sizeof(dp)); cout << recur(0, 0, N) << endl;}// Driver Codeint main(){ int N = 4; // Function Call countBinStrings(N); int N1 = 5; // Function Call countBinStrings(N1); return 0;} |
Java
// Java code to implement the approachimport java.io.*;class GFG {static int[][] dp=new int[100001][8];// Returns number of strings that does// not contain 0100 in its substringstatic int recur(int i, int lastThreeDigits, int N){ // Base Case if (i == N) return 1; // If answer for current state is // already calculated then just // return the answer. if (dp[i][lastThreeDigits] != -1) return dp[i][lastThreeDigits]; int ans = 0; if (i >= 3 && lastThreeDigits == 2) { // If last three substring is 010 // then only one option we have to // choose for next character which // is 1 so lastThreeDigits will be // 5 which is 101 ans += recur(i + 1, 5, N); } else { // Choosing 1 as a next character // by left shifting lastThreeDigits // by 1 then taking its bitwise // AND with 7 so that only first // three bits remain active and // others become zero ans += recur(i + 1, lastThreeDigits << 1 & 7 | 1, N); // Choosing 0 as a next character // by left shifting lastThreeDigits // by 1 then taking its bitwise AND // with 7 so that only first three // bits remain active and others // become zero ans += recur(i + 1, lastThreeDigits << 1 & 7, N); } // Saving and returning sum return dp[i][lastThreeDigits] = ans;}// Function to count number of binary// strings that does not contain 0100// as substringstatic void countBinStrings(int N){ // Initializing value of dp with -1 for(int i=0;i<dp.length;i++) { for(int j=0;j<dp[0].length;j++) { dp[i][j]=-1; } } System.out.println(recur(0, 0, N));}// Driver Codepublic static void main (String[] args){ int N = 4; // Function Call countBinStrings(N); int N1 = 5; // Function Call countBinStrings(N1);}}// This code is contributed by Aman Kumar |
Python3
# Python code to implement the approach# Returns number of strings that does # not contain 0100 in its substringdef recur(i, lastThreeDigits, N, dp): # Base Case if i == N: return 1 # If answer for current state is already # calculated then just return the answer. if dp[i][lastThreeDigits] != -1: return dp[i][lastThreeDigits] ans = 0 if i >= 3 and lastThreeDigits == 2: # If last three substring is 010 then only one option # we have to choose for next character which is 1 so # lastThreeDigits will be 5 which is 101 ans += recur(i + 1, 5, N, dp) else: # Choosing 1 as a next character by left shifting # lastThreeDigits by 1 then taking its bitwise AND # with 7 so that only first three bits remain active # and others become zero ans += recur(i + 1, (lastThreeDigits << 1) & 7 | 1, N, dp) # Choosing 0 as a next character by left shifting # lastThreeDigits by 1 then taking its bitwise AND # with 7 so that only first three bits remain active # and others become zero ans += recur(i + 1, (lastThreeDigits << 1) & 7, N, dp) # Saving and returning sum dp[i][lastThreeDigits] = ans return ans# Function to count number of binary strings that does# not contain 0100 as substringdef countBinStrings(N): # Initializing value of dp with -1 dp = [[-1 for j in range(8)] for i in range(100001)] print(recur(0, 0, N, dp))# Driver Codeif __name__ == "__main__": N = 4 # Function Call countBinStrings(N) N1 = 5 # Function Call countBinStrings(N1)# This code is contributed by sankar. |
C#
// C# code implementation for the above approachusing System;public class GFG { static int[, ] dp = new int[100001, 8]; // Returns number of strings that does not contain 0100 // in its substring static int Recur(int i, int lastThreeDigits, int N) { // Base Case if (i == N) return 1; // If answer for current state is already calculated // then just return the answer. if (dp[i, lastThreeDigits] != -1) return dp[i, lastThreeDigits]; int ans = 0; if (i >= 3 && lastThreeDigits == 2) { // If last three substring is 010 then only one // option we have to choose for next character // which is 1 so lastThreeDigits will be 5 which // is 101 ans += Recur(i + 1, 5, N); } else { // Choosing 1 as a next character by left // shifting lastThreeDigits by 1 then taking its // bitwise AND with 7 so that only first three // bits remain active and others become zero ans += Recur(i + 1, (lastThreeDigits << 1) & 7 | 1, N); // Choosing 0 as a next character by left // shifting lastThreeDigits by 1 then taking its // bitwise AND with 7 so that only first three // bits remain active and others become zero ans += Recur(i + 1, (lastThreeDigits << 1) & 7, N); } // Saving and returning sum return dp[i, lastThreeDigits] = ans; } // Function to count number of binary strings that does // not contain 0100 as substring static void CountBinStrings(int N) { // Initializing value of dp with -1 for (int i = 0; i < dp.GetLength(0); i++) { for (int j = 0; j < dp.GetLength(1); j++) { dp[i, j] = -1; } } Console.WriteLine(Recur(0, 0, N)); } static public void Main() { // Code int N = 4; // Function Call CountBinStrings(N); int N1 = 5; // Function Call CountBinStrings(N1); }}// This code is contributed by karthik. |
Javascript
// JS code to implement the approach// dp table initialized with - 1let dp=new Array(100001);for(let i=0; i<100001; i++) dp[i]=new Array(8);// Returns number of strings that does// not contain 0100 in its substringfunction recur( i, lastThreeDigits, N){ // Base Case if (i == N) return 1; // If answer for current state is // already calculated then just // return the answer. if (dp[i][lastThreeDigits] != -1) return dp[i][lastThreeDigits]; let ans = 0; if (i >= 3 && lastThreeDigits == 2) { // If last three substring is 010 // then only one option we have to // choose for next character which // is 1 so lastThreeDigits will be // 5 which is 101 ans += recur(i + 1, 5, N); } else { // Choosing 1 as a next character // by left shifting lastThreeDigits // by 1 then taking its bitwise // AND with 7 so that only first // three bits remain active and // others become zero ans += recur(i + 1, lastThreeDigits << 1 & 7 | 1, N); // Choosing 0 as a next character // by left shifting lastThreeDigits // by 1 then taking its bitwise AND // with 7 so that only first three // bits remain active and others // become zero ans += recur(i + 1, lastThreeDigits << 1 & 7, N); } // Saving and returning sum return dp[i][lastThreeDigits] = ans;}// Function to count number of binary// strings that does not contain 0100// as substringfunction countBinStrings( N){ // Initializing value of dp with -1 for(let i=0; i<100001; i++) for(let j=0; j<8; j++) dp[i][j]=-1; console.log(recur(0, 0, N)+"<br>");}// Driver Codelet N = 4;// Function CallcountBinStrings(N);let N1 = 5;// Function CallcountBinStrings(N1); |
Output
15 28
Time Complexity: O(N)
Auxiliary Space: O(N)
Related Articles :
- Introduction to Strings – Data Structures and Algorithms Tutorials
- Introduction to Dynamic Programming – Data Structures and Algorithms Tutorials
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