Problem: Count And Say
The count-and-say sequence is the sequence of integers beginning as follows:
1, 11, 21, 1211, 111221, ...
1 is read off as one 1 or 11. 11 is read off as two 1s or 21. 21 is read off as one 2, then one 1 or 1211.
Given an integer n, generate the nth sequence.
Note: The sequence of integers will be represented as a string.
if n = 2, the sequence is 11.
The approach is to use a simple brute force.
For A = 1, ans1 = "1" // Base case For A = 2, start traversing ans, and keep count of the current character. => For the current case ans2 becomes "11" since the count of "1" in ans1 is ONE. For A = 3, ans2 has TWO "1", so ans3 = "21" For A = 4, in ans3 count of 2 is ONE, and count of 1 is ONE. So ans 4 becomes "1211" For A = 5, in ans4 count of 1 is ONE, then 2 comes with count ONE, then again 1 comes with count TWO. So ans5 = "111221".
Time & Space complexity:
Time Complexity: O(N)
- As you have to traverse from 1 to N, and the string would never become too big because the maximum continuous character that can occur is only 3. You can try it on paper to find out.
Space Complexity: O(1)
- We are storing only two strings, the current one and the previous one. So the space is almost constant.
Code in C++.