# Solution For Count And Say

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:

“`python

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).

## Step by Step Implementation For Count And Say

public class Solution { public String countAndSay(int n) { // Base case if (n == 1) { return "1"; } // Recursive case String prev = countAndSay(n - 1); StringBuilder sb = new StringBuilder(); // Iterate through the previous result string for (int i = 0; i < prev.length(); i++) { // Keep track of the number of consecutive occurrences int count = 1; // Check for consecutive occurrences while (i < prev.length() - 1 && prev.charAt(i) == prev.charAt(i + 1)) { count++; i++; } // Append the number of occurrences and the character to the string builder sb.append(count); sb.append(prev.charAt(i)); } // Convert the string builder to a string and return return sb.toString(); } }

def countAndSay(n): if n == 1: return "1" # Base case prev = countAndSay(n-1) # Recursive call count = 1 # Initialize count i = 1 res = "" # Initialize result while i < len(prev): if prev[i] == prev[i-1]: count += 1 else: # Append count and character res += str(count) res += prev[i-1] count = 1 i += 1 # If last character is same as # previous one if prev[i-1] == prev[i-2]: res += str(count) res += prev[i-1] return res

var countAndSay = function(n) { // your code goes here };

int countAndSay(int n) { // Base cases if (n == 1) return "1"; if (n == 2) return "11"; // Find n'th term by generating all terms from 3 to // n-1. Every term is generated using previous term string str = "11"; // Initialize previous term for (int i = 3; i<=n; i++) { // In below for loop, previous character is processed in // current iteration. That is why a dummy character is // added to make sure that loop runs one extra iteration. str += '$'; int len = str.length(); int cnt = 1; // Initialize count of matching chars string tmp = ""; // Initialize i'th term in series // Process previous term to find the next term for (int j = 1; j < len; j++) { // If current character does't match if (str[j] != str[j-1]) { // Append count of str[j-1] to temp tmp += cnt + '0'; // Append str[j-1] tmp += str[j-1]; // Reset count cnt = 1; } // If matches, then increment count of matching // characters else cnt++; } // Update str str = tmp; } return str; }

public class Solution { public string CountAndSay(int n) { // Base case if (n == 1) { return "1"; } // Use previous result to generate next result string prevResult = CountAndSay(n - 1); return GenerateNextResult(prevResult); } private string GenerateNextResult(string prevResult) { // Track current number and count char currNum = prevResult[0]; int currCount = 1; // Use StringBuilder to efficiently create result string StringBuilder result = new StringBuilder(); // Iterate through previous result string for (int i = 1; i < prevResult.Length; i++) { // If current number is different than previous number if (prevResult[i] != currNum) { // Append current number and count to result result.Append(currCount); result.Append(currNum); // Reset current number and count currNum = prevResult[i]; currCount = 1; } // Otherwise, increment count else { currCount++; } } // Append last number and count to result result.Append(currCount); result.Append(currNum); return result.ToString(); } }