Count and Say

Count and Say

What is this problem about?

The Count and Say interview question asks you to generate the $n^{th}$ term in a specific mathematical sequence. The sequence starts with 1. Each subsequent term is a "saying" of the previous term. For example, to get the second term, you "say" the first term: "one 1", which becomes 11. To get the third term, you "say" the second term: "two 1s", which becomes 21.

Why is this asked in interviews?

Companies like Amazon, Microsoft, and Bloomberg ask the Count and Say coding problem to test your string manipulation and simulation skills. It's a "Medium" difficulty task that evaluates how well you can process a string character by character while keeping track of counts. It also tests your ability to use recursion or iteration to build the sequence step-by-step.

Algorithmic pattern used

This is a Simulation problem that uses a Two Pointers or Counter approach.

1. Start with the string s = "1".
2. To generate the next term, iterate through the current string s.
3. Count the consecutive occurrences of the same character.
4. Append the count followed by the character to a new string.
5. Repeat this process n-1 times.

Example explanation

n = 4

1. n=1: "1"
2. n=2: Say "1" -> "one 1" -> "11"
3. n=3: Say "11" -> "two 1s" -> "21"
4. n=4: Say "21" -> "one 2, one 1" -> "1211" Result for n=4 is "1211".

Common mistakes candidates make

• Inefficient String Building: Using string concatenation (s += ...) inside a loop in languages like Java or Python, which leads to O(N^2) complexity. Use StringBuilder or list joining instead.
• Off-by-one errors: Not correctly handling the last group of characters in the string.
• Recursive Depth: While recursion is intuitive, a very large n could theoretically cause a stack overflow. Iteration is generally safer.

Interview preparation tip

This problem is all about "Group By" logic. Practice the pattern of while (i < s.length() && s[i] == s[i+1]) to quickly count identical consecutive elements in an array or string.