Given an integer N. The task is to find the number of all possible distinct binary strings of length N which have at least 3 consecutive 1s.
Examples:
Input: N = 3
Output: 1
The only string of length 3 possible is “111”.
Input: N = 4
Output: 3
The 3 strings are “1110”, “0111” and “1111”.
Naive approach: Consider all possible strings.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function that returns true if s contains// three consecutive 1'sbool check(string& s){ int n = s.length(); for (int i = 2; i < n; i++) { if (s[i] == '1' && s[i - 1] == '1' && s[i - 2] == '1') return 1; } return 0;}// Function to return the count// of required stringsint countStr(int i, string& s){ if (i < 0) { if (check(s)) return 1; return 0; } s[i] = '0'; int ans = countStr(i - 1, s); s[i] = '1'; ans += countStr(i - 1, s); s[i] = '0'; return ans;}// Driver codeint main(){ int N = 4; string s(N, '0'); cout << countStr(N - 1, s); return 0;} |
Java
// Java implementation of the approachimport java.util.*;class GFG{// Function that returns true if s contains// three consecutive 1'sstatic boolean check(String s){ int n = s.length(); for (int i = 2; i < n; i++) { if (s.charAt(i) == '1' && s.charAt(i-1) == '1' && s.charAt(i-2) == '1') return true; } return false;}// Function to return the count// of required stringsstatic int countStr(int i, String s){ if (i < 0) { if (check(s)) return 1; return 0; } char[] myNameChars = s.toCharArray(); myNameChars[i] = '0'; s = String.valueOf(myNameChars); int ans = countStr(i - 1, s); char[] myChar = s.toCharArray(); myChar[i] = '1'; s = String.valueOf(myChar); ans += countStr(i - 1, s); char[]myChar1 = s.toCharArray(); myChar1[i] = '0'; s = String.valueOf(myChar1); return ans;}// Driver codepublic static void main(String args[]){ int N = 4; String s = "0000"; System.out.println(countStr(N - 1, s));}}// This code is contributed by// Surendra_Gangwar |
Python3
# Python3 implementation of the approach# Function that returns true if s contains# three consecutive 1'sdef check(s) : n = len(s); for i in range(2, n) : if (s[i] == '1' and s[i - 1] == '1' and s[i - 2] == '1') : return 1;# Function to return the count# of required stringsdef countStr(i, s) : if (i < 0) : if (check(s)) : return 1; return 0; s[i] = '0'; ans = countStr(i - 1, s); s[i] = '1'; ans += countStr(i - 1, s); s[i] = '0'; return ans;# Driver codeif __name__ == "__main__" : N = 4; s = list('0' * N); print(countStr(N - 1, s));# This code is contributed by Ryuga |
C#
// C# implementation of the approach// value xusing System; class GFG{// Function that returns true if s contains// three consecutive 1'sstatic bool check(String s){ int n = s.Length; for (int i = 2; i < n; i++) { if (s[i] == '1' && s[i - 1] == '1' && s[i - 2] == '1') return true; } return false;}// Function to return the count// of required stringsstatic int countStr(int i, String s){ if (i < 0) { if (check(s)) return 1; return 0; } char[] myNameChars = s.ToCharArray(); myNameChars[i] = '0'; s = String.Join("", myNameChars); int ans = countStr(i - 1, s); char[] myChar = s.ToCharArray(); myChar[i] = '1'; s = String.Join("", myChar); ans += countStr(i - 1, s); char[]myChar1 = s.ToCharArray(); myChar1[i] = '0'; s = String.Join("", myChar1); return ans;}// Driver codepublic static void Main(String []args){ int N = 4; String s = "0000"; Console.WriteLine(countStr(N - 1, s));}}/* This code contributed by PrinciRaj1992 */ |
Javascript
<script>// JavaScript implementation of the approach// value x// Function that returns true if s contains// three consecutive 1'sfunction check(s){ let n = s.length; for (let i = 2; i < n; i++) { if (s[i] == '1' && s[i-1] == '1' && s[i-2] == '1') return true; } return false;}// Function to return the count// of required stringsfunction countStr(i,s){ if (i < 0) { if (check(s)) return 1; return 0; } let myNameChars = s.split(""); myNameChars[i] = '0'; s = (myNameChars).join(""); let ans = countStr(i - 1, s); let myChar = s.split(""); myChar[i] = '1'; s = (myChar).join(""); ans += countStr(i - 1, s); let myChar1 = s.split(""); myChar1[i] = '0'; s = (myChar1).join(""); return ans;}// Driver codelet N = 4;let s = "0000";document.write(countStr(N - 1, s));// This code is contributed by avanitrachhadiya2155</script> |
Output:
3
Time Complexity: O(2N)
Space Complexity: O(N) because of recurrence stack space.
Efficient approach: We use dynamic programming for computing the number of strings.
State of dp: dp(i, x) : denotes number of strings of length i with x consecutive 1s in position i + 1 to i + x.
Recurrence: dp(i, x) = dp(i – 1, 0) + dp(i – 1, x + 1)
The recurrence is based on the fact that either the string can have a ‘0’ at position i or a ‘1’.
- If it has a ‘0’ at position i then for (i-1)th position value of x = 0.
- If it has a ‘1’ at position i the for (i-1)th position value of x = value of x at position i + 1.
Base Condition: dp(i, 3) = 2i. Because once you have 3 consecutive ‘1’s you don’t care what characters are there at position 1, 2…i of string as all the strings are valid.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;int n;// Function to return the count// of required stringsint solve(int i, int x, int dp[][4]){ if (i < 0) return x == 3; if (dp[i][x] != -1) return dp[i][x]; // '0' at ith position dp[i][x] = solve(i - 1, 0, dp); // '1' at ith position dp[i][x] += solve(i - 1, x + 1, dp); return dp[i][x];}// Driver codeint main(){ n = 4; int dp[n][4]; for (int i = 0; i < n; i++) for (int j = 0; j < 4; j++) dp[i][j] = -1; for (int i = 0; i < n; i++) { // Base condition: // 2^(i+1) because of 0 indexing dp[i][3] = (1 << (i + 1)); } cout << solve(n - 1, 0, dp); return 0;} |
Java
// Java implementation of the above approach import java.util.*;class GFG { static int n; // Function to return the count // of required strings static int solve(int i, int x, int dp[][]) { if (i < 0) { return x == 3 ? 1 : 0; } if (dp[i][x] != -1) { return dp[i][x]; } // '0' at ith position dp[i][x] = solve(i - 1, 0, dp); // '1' at ith position dp[i][x] += solve(i - 1, x + 1, dp); return dp[i][x]; } // Driver code public static void main(String[] args) { n = 4; int dp[][] = new int[n][4]; for (int i = 0; i < n; i++) { for (int j = 0; j < 4; j++) { dp[i][j] = -1; } } for (int i = 0; i < n; i++) { // Base condition: // 2^(i+1) because of 0 indexing dp[i][3] = (1 << (i + 1)); } System.out.print(solve(n - 1, 0, dp)); }}// This code has been contributed by 29AjayKumar |
Python 3
# Python 3 implementation of the approach# Function to return the count# of required stringsdef solve(i, x, dp): if (i < 0): return x == 3 if (dp[i][x] != -1): return dp[i][x] # '0' at ith position dp[i][x] = solve(i - 1, 0, dp) # '1' at ith position dp[i][x] += solve(i - 1, x + 1, dp) return dp[i][x]# Driver codeif __name__ == "__main__": n = 4; dp = [[0 for i in range(n)] for j in range(4)] for i in range(n): for j in range(4): dp[i][j] = -1 for i in range(n) : # Base condition: # 2^(i+1) because of 0 indexing dp[i][3] = (1 << (i + 1)) print(solve(n - 1, 0, dp))# This code is contributed by ChitraNayal |
C#
// C# implementation of the above approach using System; class GFG { static int n; // Function to return the count // of required strings static int solve(int i, int x, int [,]dp) { if (i < 0) { return x == 3 ? 1 : 0; } if (dp[i,x] != -1) { return dp[i,x]; } // '0' at ith position dp[i,x] = solve(i - 1, 0, dp); // '1' at ith position dp[i,x] += solve(i - 1, x + 1, dp); return dp[i,x]; } // Driver code public static void Main(String[] args) { n = 4; int [,]dp = new int[n, 4]; for (int i = 0; i < n; i++) { for (int j = 0; j < 4; j++) { dp[i, j] = -1; } } for (int i = 0; i < n; i++) { // Base condition: // 2^(i+1) because of 0 indexing dp[i,3] = (1 << (i + 1)); } Console.Write(solve(n - 1, 0, dp)); }}// This code contributed by Rajput-Ji |
Javascript
<script>// Javascript implementation of the above approach var n; // Function to return the count // of required strings function solve(i , x , dp) { if (i < 0) { return x == 3 ? 1 : 0; } if (dp[i][x] != -1) { return dp[i][x]; } // '0' at ith position dp[i][x] = solve(i - 1, 0, dp); // '1' at ith position dp[i][x] += solve(i - 1, x + 1, dp); return dp[i][x]; } // Driver code n = 4; var dp = Array(n).fill().map(()=>Array(4).fill(0)); for (i = 0; i < n; i++) { for (j = 0; j < 4; j++) { dp[i][j] = -1; } } for (i = 0; i < n; i++) { // Base condition: // 2^(i+1) because of 0 indexing dp[i][3] = (1 << (i + 1)); } document.write(solve(n - 1, 0, dp));// This code contributed by gauravrajput1</script> |
Output:
3
Time complexity: O(N)
Space Complexity: O(N)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
