Count And Say

Problem Description

rough nb

The count-and-say sequence is a sequence of digit strings defined by the recursive formula:

• countAndSay(1) = "1"
• countAndSay(n) is the way you would “say” the digit string from countAndSay(n-1), which is then converted into a new digit string.

To determine how you “say” a digit string, split it into the minimal number of substrings such that each substring contains exactly one unique digit. Then for each substring, say the number of digits, then say the digit. Finally, concatenate every said digit.

For example, the saying of the digit string "3322251" is "23321511".

Given a positive integer n, return the nth term of the count-and-say sequence.

Example 1:

Input: n = 4
Output: "1211"
Explanation:
countAndSay(1) = "1"
countAndSay(2) = say "1" = one 1 = "11"
countAndSay(3) = say "11" = two 1s = "21"
countAndSay(4) = say "21" = one 2 + one 1 = "12" + "11" = "1211"

Solution Approach

This problem is best solved iteratively. We start with the base case, n=1, which is "1". Then, we generate the next term in the sequence based on the previous one, repeating this n-1 times.

To generate the next term from a given string s:

1. Initialize an empty string or string builder for the new term.
2. Iterate through the string s using a pointer or index, let’s call it i.
3. At each position i, count the number of consecutive occurrences of the character s[i]. Let’s say it appears count times.
4. Append the count and the character s[i] to our new string.
5. Advance the pointer i by count to move to the next new character.
6. Repeat until the entire string s is processed.
7. The newly constructed string is the next term in the sequence.

Let’s trace for n=4:

• n=1: res = "1"
• n=2: Generate from "1". We see one ‘1’. New string is "11". res = "11".
• n=3: Generate from "11". We see two ‘1’s. New string is "21". res = "21".
• n=4: Generate from "21". We see one ‘2’, then one ‘1’. New string is "12" + "11" = "1211". res = "1211".

This iterative approach builds the solution step-by-step.

Code

# Please provide your solution code.
# I will add it here.

Complexity Analysis

Let L be the average length of the strings generated.

• Time: We iterate from 1 to n. In each iteration, we scan the previously generated string to create the new one. This scan takes time proportional to the length of the string. The length of the strings can grow, but it’s not a simple exponential. The complexity is roughly O(n * L), where L is the length of the terms, which can be large.
• Space: We only need to store the previous term to generate the next one. So, the space complexity is O(L) to store the result string of the (n-1)th iteration.

Notes & Learnings

This problem tests careful string manipulation and iteration. The description can be a bit tricky to understand at first, so working through an example is key. Using a string builder or a list of characters to build the new string is more efficient in many languages (like Python or Java) than repeated string concatenation.