Count number of binary strings without consecutive 1's - GeeksforGeeks

Count number of binary strings without consecutive 1's - GeeksforGeeks
Given a positive integer N, count all possible distinct binary strings of length N such that there are no consecutive 1's.
Examples:

Input:  N = 2  Output: 3  // The 3 strings are 00, 01, 10  

1) Compute number of binary strings without consecutive 1’s using the approach discussed here. Let this count be c.

2) Count of all possible binary strings with consecutive 1’s is 2^n where n is input length.

3) Total binary strings with consecutive 1 is 2^n – c.

This problem can be solved using Dynamic Programming. Let a[i] be the number of binary strings of length i which do not contain any two consecutive 1’s and which end in 0. Similarly, let b[i] be the number of such strings which end in 1. We can append either 0 or 1 to a string ending in 0, but we can only append 0 to a string ending in 1. This yields the recurrence relation:

a[i] = a[i - 1] + b[i - 1]
b[i] = a[i - 1] 

The base cases of above recurrence are a[1] = b[1] = 1. The total number of strings of length i is just a[i] + b[i].

int countStrings(int n)

{

    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];

    }

    return a[n-1] + b[n-1]

}

courses.csail.mit.edu/6.006/oldquizzes/solutions/q2-f2009-sol.pdf

If we take a closer look at the pattern, 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 = 2  = fib(3)
n = 2, count = 3  = fib(4)
n = 3, count = 5  = fib(5)
n = 4, count = 8  = fib(6)
n = 5, count = 13 = fib(7)
................

Therefore we can count the strings in O(Log n) time also using the method 5 here.

Read full article from Count number of binary strings without consecutive 1's - GeeksforGeeks

Count strings with consecutive 1's - GeeksforGeeks

Count strings with consecutive 1's - GeeksforGeeks
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.  

1) Compute number of binary strings without consecutive 1’s using the approach discussed here. Let this count be c.

2) Count of all possible binary strings with consecutive 1’s is 2^n where n is input length.

3) Total binary strings with consecutive 1 is 2^n – c.

int 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];

}

Time complexity of above solution is O(n). We can optimize 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 = 24 - fib(7)
................

We can find n’th Fibonacci Number in O(Log n) time (See method 4 here).

Read full article from Count strings with consecutive 1's - GeeksforGeeks

LeetCode 203 - Remove Linked List Elements

http://www.programcreek.com/2014/04/leetcode-remove-linked-list-elements-java/

Remove all elements from a linked list of integers that have value val.

Example

Given: 1 --> 2 --> 6 --> 3 --> 4 --> 5 --> 6, val = 6
Return: 1 --> 2 --> 3 --> 4 --> 5

The key to solve this problem is using a helper node to track the head of the list.
http://www.jiuzhang.com/solutions/remove-linked-list-elements/
http://www.programcreek.com/2014/04/leetcode-remove-linked-list-elements-java/

    public ListNode removeElements(ListNode head, int val) {
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        head = dummy;
        
        while (head.next != null) {
            if (head.next.val == val) {
                head.next = head.next.next;
            } else {
                head = head.next;
            }
        }
        
        return dummy.next;
    }

https://leetcode.com/discuss/33142/iterative-short-java-solution

public ListNode removeElements(ListNode head, int val) { while (head != null && head.val == val) head = head.next; ListNode curr = head; while (curr != null && curr.next != null) if (curr.next.val == val) curr.next = curr.next.next; else curr = curr.next; return head; }

X. Recursive Version

https://leetcode.com/discuss/33043/3-line-recursive-solution

public ListNode removeElements(ListNode head, int val) { if (head == null) return null; head.next = removeElements(head.next, val); return head.val == val ? head.next : head; }

public ListNode removeElements(ListNode head, int val) { if(head == null) return null; ListNode next = removeElements(head.next, val); if(head.val == val) return next; head.next = next; return head; }

http://www.sigmainfy.com/blog/leetcode-remove-linked-list-elements-iterative-vs-recursive.html

    public ListNode removeElements(ListNode head, int val) {
 if (head == null) {
  return null;
 }

 if (head.val == val){
  head =
                removeElements(head.next, val);
 } else {
  head.next =
                removeElements(head.next, val);
 }

 return head;
    }