Count strings with consecutive 1’s
Given a number n, count number of n length strings with consecutive 1’s in them.
Examples:
Input : n = 2 Output : 1 There are 4 strings of length 2, the strings are 00, 01, 10 and 11. Only the string 11 has consecutive 1's. Input : n = 3 Output : 3 There are 8 strings of length 3, the strings are 000, 001, 010, 011, 100, 101, 110 and 111. The strings with consecutive 1's are 011, 110 and 111. Input : n = 5 Output : 19
Naive Approach
We will follow the following steps-
- we made a variable “ans” which will tell how many strings have 2 consecutive ones
- We will find all strings with 0 and 1 of a given length using the pick and non-pick concept of recursion
- After that, if any string has maximum consecutive ones greater than or equal to 2 then increment “ans” by 1
- Then finally print the value stored in “ans” variable.
C++
// C++ program to count all distinct// binary strings with two consecutive 1's#include <iostream>using namespace std;// Gives count of n length binary// strings with consecutive 1'svoid countStrings(int n,int ind,string str,int &ans){ if(ind==n){ int count=0; int temp=0; for(int i=0;i<n;i++){ if(str[i]=='1'){ temp++; }else{ temp=0; } count=max(count,temp); } if(count>=2){ans++;} return; } countStrings(n,ind+1,str+"0",ans); countStrings(n,ind+1,str+"1",ans);}// Driver codeint main(){ int ans=0; countStrings(5,0,"",ans) ; cout <<ans<< endl; return 0;} |
Java
// Java program to count all distinct// binary strings with two consecutive 1'simport java.util.*;public class GFG { // Gives count of n length binary // strings with consecutive 1's public static void countStrings(int n, int ind, String str, int[] ans) { if (ind == n) { int count = 0; int temp = 0; for (int i = 0; i < n; i++) { if (str.charAt(i) == '1') { temp++; } else { temp = 0; } count = Math.max(count, temp); } if (count >= 2) { ans[0]++; } return; } countStrings(n, ind + 1, str + "0", ans); countStrings(n, ind + 1, str + "1", ans); } // Driver code public static void main(String[] args) { int[] ans = { 0 }; countStrings(5, 0, "", ans); System.out.println(ans[0]); }}// This code is contributed by Susobhan Akhuli |
Python3
# Python program to count all distinct# binary strings with two consecutive 1's# Gives count of n length binary# strings with consecutive 1'sdef countStrings(n, ind, s, ans): if ind == n: count = 0 temp = 0 for i in range(n): if s[i] == '1': temp += 1 else: temp = 0 count = max(count, temp) if count >= 2: ans[0] += 1 return countStrings(n, ind + 1, s + "0", ans) countStrings(n, ind + 1, s + "1", ans)# Driver codeif __name__ == '__main__': ans = [0] countStrings(3, 0, "", ans) print(ans[0])# This code is contributed by Susobhan Akhuli |
C#
// C# program to count all distinct// binary strings with two consecutive 1'susing System;public class GFG { // Gives count of n length binary // strings with consecutive 1's public static void CountStrings(int n, int ind, string str, ref int ans) { if (ind == n) { int count = 0; int temp = 0; for (int i = 0; i < n; i++) { if (str[i] == '1') { temp++; } else { temp = 0; } count = Math.Max(count, temp); } if (count >= 2) { ans++; } return; } CountStrings(n, ind + 1, str + "0", ref ans); CountStrings(n, ind + 1, str + "1", ref ans); } // Driver code public static void Main() { int ans = 0; CountStrings(5, 0, "", ref ans); Console.WriteLine(ans); }}// This code is contributed by Susobhan Akhuli |
Javascript
// Javascript code to count all distinct // binary strings with two consecutive 1's// Gives count of n length binary strings with consecutive 1'sfunction countStrings(n, ind, str, ans){ if (ind == n){ let count = 0; let temp = 0; for (let i = 0; i < n; i++){ if (str.charAt(i) == '1'){temp++; } else {temp = 0; } count = Math.max(count, temp); } if (count >= 2) { ans[0]++; } return; } countStrings(n, ind + 1, str + "0", ans); countStrings(n, ind + 1, str + "1", ans);}// Driver codelet ans = [0];countStrings(5, 0, "", ans);console.log(ans[0]);// This code is contributed by Susobhan Akhuli. |
Output
19
Time Complexity: O((2^n) * n), “2^n” for generating all strings, and “n” for traversing each string to count the maximum number of consecutive ones.
Auxiliary Space: O(n),Recursion Stack Space
Optimized Approach
The reverse problem of counting strings without consecutive 1’s can be solved using Dynamic Programming (See the solution here). We can use that solution and find the required count using below steps.
- Compute the number of binary strings without consecutive 1’s using the approach discussed here. Let this count be c.
- Count of all possible binary strings of length n is 2^n.
- Total binary strings with consecutive 1 is 2^n – c.
Below is the implementation of the above steps.
C++
// C++ program to count all distinct// binary strings with two consecutive 1's#include <iostream>using namespace std;// Returns count of n length binary// strings with consecutive 1'sint countStrings(int n){ // Count binary strings without consecutive 1's. // See the approach discussed on be // ( http://goo.gl/p8A3sW ) int a[n], b[n]; a[0] = b[0] = 1; for (int i = 1; i < n; i++) { a[i] = a[i - 1] + b[i - 1]; b[i] = a[i - 1]; } // Subtract a[n-1]+b[n-1] from 2^n return (1 << n) - a[n - 1] - b[n - 1];}// Driver codeint main(){ cout << countStrings(5) << endl; return 0;} |
Java
// Java program to count all distinct// binary strings with two consecutive 1'sclass GFG { // Returns count of n length binary // strings with consecutive 1's static int countStrings(int n) { // Count binary strings without consecutive 1's. // See the approach discussed on be // ( http://goo.gl/p8A3sW ) int a[] = new int[n], b[] = new int[n]; a[0] = b[0] = 1; for (int i = 1; i < n; i++) { a[i] = a[i - 1] + b[i - 1]; b[i] = a[i - 1]; } // Subtract a[n-1]+b[n-1] from 2 ^ n return (1 << n) - a[n - 1] - b[n - 1]; } // Driver code public static void main(String args[]) { System.out.println(countStrings(5)); }}// This code is contributed by Nikita tiwari. |
Python 3
# Python 3 program to count all# distinct binary strings with# two consecutive 1's# Returns count of n length# binary strings with# consecutive 1'sdef countStrings(n): # Count binary strings without # consecutive 1's. # See the approach discussed on be # ( http://goo.gl/p8A3sW ) a = [0] * n b = [0] * n a[0] = b[0] = 1 for i in range(1, n): a[i] = a[i - 1] + b[i - 1] b[i] = a[i - 1] # Subtract a[n-1]+b[n-1] from 2^n return (1 << n) - a[n - 1] - b[n - 1]# Driver codeprint(countStrings(5))# This code is contributed# by Nikita tiwari. |
C#
// program to count all distinct// binary strings with two// consecutive 1'susing System;class GFG { // Returns count of n length // binary strings with // consecutive 1's static int countStrings(int n) { // Count binary strings without // consecutive 1's. // See the approach discussed on // ( http://goo.gl/p8A3sW ) int[] a = new int[n]; int[] b = new int[n]; a[0] = b[0] = 1; for (int i = 1; i < n; i++) { a[i] = a[i - 1] + b[i - 1]; b[i] = a[i - 1]; } // Subtract a[n-1]+b[n-1] // from 2^n return (1 << n) - a[n - 1] - b[n - 1]; } // Driver code public static void Main() { Console.WriteLine(countStrings(5)); }}// This code is contributed by Anant Agarwal. |
PHP
<?php// PHP program to count all// distinct binary strings// with two consecutive 1's// Returns count of n length binary // strings with consecutive 1'sfunction countStrings($n){ // Count binary strings without consecutive 1's. // See the approach discussed on be // ( http://goo.gl/p8A3sW ) $a[$n] = 0; $b[$n] = 0; $a[0] = $b[0] = 1; for ($i = 1; $i < $n; $i++) { $a[$i] = $a[$i - 1] + $b[$i - 1]; $b[$i] = $a[$i - 1]; } // Subtract a[n-1]+b[n-1] from 2^n return (1 << $n) - $a[$n - 1] - $b[$n - 1];} // Driver Code echo countStrings(5), "\n";// This Code is contributed by Ajit?> |
Javascript
<script>// JavaScript program to count all distinct// binary strings with two// consecutive 1's // Returns count of n length binary // strings with consecutive 1's function countStrings(n) { // Count binary strings without consecutive 1's. // See the approach discussed on be // ( http://goo.gl/p8A3sW ) let a = [], b = []; a[0] = b[0] = 1; for (let i = 1; i < n; i++) { a[i] = a[i - 1] + b[i - 1]; b[i] = a[i - 1]; } // Subtract a[n-1]+b[n-1] // from 2 ^ n return (1 << n) - a[n - 1] - b[n - 1]; }// Driver Code document.write(countStrings(5));</script> |
Output
19
Time Complexity: O(n)
Auxiliary Space: O(n)
Optimization:
The time complexity of the above solution is O(n). We can optimize the above solution to work in O(Logn).
If we take a closer look at the pattern of counting strings without consecutive 1’s, we can observe that the count is actually (n+2)th Fibonacci number for n >= 1. The Fibonacci Numbers are 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 141, ….
n = 1, count = 0 = 21 - fib(3) n = 2, count = 1 = 22 - fib(4) n = 3, count = 3 = 23 - fib(5) n = 4, count = 8 = 24 - fib(6) n = 5, count = 19 = 25 - fib(7) ................
We can find n’th Fibonacci Number in O(Log n) time (See method 4 here).
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