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:

1. Initialize the resulting string as an empty string.
2. Initialize count as 1 and the starting character as the first character in countAndSay(n-1).
3. Traverse the string countAndSay(n-1) from the second character to the end.
4. If the current character is the same as the previous character, increment the count.
5. 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.
6. 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();
    }
}