## Similar Problems

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# Count And Say - Leetcode Solution

LeetCode: Count And Say Leetcode Solution

Difficulty: Medium

Topics: string

Problem Statement: 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 different digit string.

To determine how you "say" a digit string, split it into the minimal number of groups so that each group is a contiguous section all of the same character. Then for each group, say the number of characters, followed by the character. To convert the saying into a digit string, replace the counts with a number and concatenate every saying.

Example:

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

Approach: We can solve this problem by applying a recursive approach. We can start from n=1 and keep returning the resulting string by applying the rule of the count-and-say sequence. For a given n, we can recursively call the function to calculate the string countAndSay(n-1) which is given. To get the required string countAndSay(n), we can apply the following steps:

- Initialize the resulting string as an empty string.
- Initialize count as 1 and the starting character as the first character in countAndSay(n-1).
- Traverse the string countAndSay(n-1) from the second character to the end.
- If the current character is the same as the previous character, increment the count.
- If the current character is different from the previous character, append the count and the previous character to the resulting string and start a new count for the current character.
- After the traversal is complete, append the last count and character to the resulting string and return the resulting string.

For example, consider the following code for the countAndSay function:

```
def countAndSay(n: int) -> str:
if n == 1:
return "1"
res = ""
prev = countAndSay(n-1)
count = 1
prev_char = prev[0]
for i in range(1, len(prev)):
if prev[i] == prev_char:
count += 1
else:
res += str(count) + prev_char
count = 1
prev_char = prev[i]
res += str(count) + prev_char
return res
```

Time Complexity: Each recursive call performs a linear traversal of the previous string, so the time complexity of the countAndSay function is O(n*m), where n is the input value of n, and m is the length of the previous string. In the worst case, m = 2^(n-1) which is the length of the maximum possible string.

Space Complexity: Each recursive call requires a new string of length at most 2^(n-1). Therefore, the space complexity of the countAndSay function is O(2^n).

## Count And Say Solution Code

`1`