Count And Say

                                            
class Solution {
public:
    string countAndSay(int n) {
        if (n == 1) return "1";
        
        string result = "1";
        for (int i = 2; i <= n; ++i) {
            string current = "";
            int count = 1;
            char prev = result[0];
            
            for (int j = 1; j < result.size(); ++j) {
                if (result[j] == prev) {
                    count++;
                } else {
                    current += to_string(count) + prev;
                    prev = result[j];
                    count = 1;
                }
            }
            current += to_string(count) + prev;
            result = current;
        }
        
        return result;
    }
};

                                            
class Solution {
    public String countAndSay(int n) {
        if (n == 1) {
            return "1";
        }
        
        String prev = countAndSay(n-1);
        StringBuilder sb = new StringBuilder();
        int count = 1;
        for (int i = 1; i < prev.length(); i++) {
            if (prev.charAt(i) == prev.charAt(i-1)) {
                count++;
            } else {
                sb.append(count);
                sb.append(prev.charAt(i-1));
                count = 1;
            }
        }
        sb.append(count);
        sb.append(prev.charAt(prev.length()-1));
        
        return sb.toString();
    }
}

                                            
class Solution(object):
    def countAndSay(self, n):
        if n == 1:
            return "1"
        
        prev = self.countAndSay(n-1)
        result = ""
        
        count = 1
        for i in range(1, len(prev)):
            if prev[i] == prev[i-1]:
                count += 1
            else:
                result += str(count) + prev[i-1]
                count = 1
        
        result += str(count) + prev[-1]
        
        return result

                                            
char *countAndSay(int n) {
    if (n == 1) {
        return "1";
    }
    
    char *prev = countAndSay(n - 1);
    char *result = (char *)malloc(2 * strlen(prev) + 1);
    
    int i = 0, count = 1, j = 0;
    while (prev[i] != '\0') {
        if (prev[i] == prev[i + 1]) {
            count++;
        } else {
            result[j++] = count + '0';
            result[j++] = prev[i];
            count = 1;
        }
        i++;
    }
    result[j] = '\0';
    
    return result;
}

                                            
public class Solution {
    public string CountAndSay(int n) {
        if (n == 1) {
            return "1";
        }
        
        string prev = CountAndSay(n - 1);
        StringBuilder result = new StringBuilder();
        int count = 1;
        
        for (int i = 0; i < prev.Length; i++) {
            if (i + 1 < prev.Length && prev[i] == prev[i + 1]) {
                count++;
            } else {
                result.Append(count);
                result.Append(prev[i]);
                count = 1;
            }
        }
        
        return result.ToString();
    }
}

                                            
/**
 * @param {number} n
 * @return {string}
 */
var countAndSay = function(n) {
    let result = "1";
    for (let i = 1; i < n; i++) {
        let temp = "";
        let count = 1;
        for (let j = 0; j < result.length; j++) {
            if (result[j] === result[j + 1]) {
                count++;
            } else {
                temp += count + result[j];
                count = 1;
            }
        }
        result = temp;
    }
    return result;
};

                                            
function countAndSay(n: number): string {
    if (n === 1) return "1";
    
    let result = "1";
    for (let i = 2; i <= n; i++) {
        let temp = "";
        let count = 1;
        for (let j = 1; j < result.length; j++) {
            if (result[j] === result[j - 1]) {
                count++;
            } else {
                temp += count.toString() + result[j - 1];
                count = 1;
            }
        }
        temp += count.toString() + result[result.length - 1];
        result = temp;
    }
    
    return result;
};

                                            
class Solution {
    /**
     * @param Integer $n
     * @return String
     */
    function countAndSay($n) {
        $result = "1";
        
        for ($i = 1; $i < $n; $i++) {
            $temp = "";
            $count = 1;
            
            for ($j = 0; $j < strlen($result); $j++) {
                if ($j + 1 < strlen($result) && $result[$j] == $result[$j + 1]) {
                    $count++;
                } else {
                    $temp .= $count . $result[$j];
                    $count = 1;
                }
            }
            
            $result = $temp;
        }
        
        return $result;
    }
}

                                            
class Solution {
    func countAndSay(_ n: Int) -> String {
        var result = "1"
        
        for _ in 1..<n {
            var current = ""
            var i = 0
            
            while i < result.count {
                var count = 1
                let currentChar = result[result.index(result.startIndex, offsetBy: i)]
                
                while i + 1 < result.count && result[result.index(result.startIndex, offsetBy: i)] == result[result.index(result.startIndex, offsetBy: i + 1)] {
                    i += 1
                    count += 1
                }
                
                current += "\(count)\(currentChar)"
                i += 1
            }
            
            result = current
        }
        
        return result
    }
}

                                            
class Solution {
    fun countAndSay(n: Int): String {
        var result = "1"
        for (x in 1 until n) {
            var current = StringBuilder()
            var i = 0
            while (i < result.length) {
                var count = 1
                var currentChar = result[i]
                while (i + 1 < result.length && result[i] == result[i + 1]) {
                    count++
                    i++
                }
                current.append(count)
                current.append(currentChar)
                i++
            }
            result = current.toString()
        }
        return result
    }
}

                                            
class Solution {
  String countAndSay(int n) {
    if (n == 1) {
      return '1';
    }
    
    String result = '1';
    for (int i = 2; i <= n; i++) {
      result = generateNext(result);
    }
    
    return result;
  }
  
  String generateNext(String s) {
    String result = '';
    int count = 1;
    for (int i = 1; i < s.length; i++) {
      if (s[i] == s[i - 1]) {
        count++;
      } else {
        result += '$count${s[i - 1]}';
        count = 1;
      }
    }
    result += '$count${s[s.length - 1]}';
    
    return result;
  }
}

                                            
func countAndSay(n int) string {
    if n == 1 {
        return "1"
    }
    
    prev := countAndSay(n - 1)
    result := ""
    count := 1
    
    for i := 0; i < len(prev); i++ {
        if i+1 < len(prev) && prev[i] == prev[i+1] {
            count++
        } else {
            result += strconv.Itoa(count) + string(prev[i])
            count = 1
        }
    }
    
    return result
}

                                            
# @param {Integer} n
# @return {String}
def count_and_say(n)
    return "1" if n == 1
    prev = count_and_say(n-1)
    result = ""
    count = 1
    (1..prev.length).each do |i|
        if prev[i] != prev[i-1]
            result += count.to_s + prev[i-1]
            count = 1
        else
            count += 1
        end
    end
    return result
end

                                            
object Solution {
    def countAndSay(n: Int): String = {
        if (n == 1) "1"
        else {
            val prev = countAndSay(n - 1)
            var result = ""
            var count = 1
            var i = 0
            while (i < prev.length) {
                if (i + 1 < prev.length && prev(i) == prev(i + 1)) {
                    count += 1
                } else {
                    result += count.toString + prev(i)
                    count = 1
                }
                i += 1
            }
            result
        }
    }
}

                                            
impl Solution {
    pub fn count_and_say(n: i32) -> String {
        if n == 1 {
            return "1".to_string();
        }
        
        let mut prev = "1".to_string();
        
        for _ in 2..=n {
            let mut new_str = String::new();
            let mut chars = prev.chars().collect::<Vec<char>>();
            
            let mut count = 1;
            let mut curr_char = chars[0];
            
            for j in 1..chars.len() {
                if chars[j] == curr_char {
                    count += 1;
                } else {
                    new_str.push_str(&count.to_string());
                    new_str.push(curr_char);
                    count = 1;
                    curr_char = chars[j];
                }
            }
            
            new_str.push_str(&count.to_string());
            new_str.push(curr_char);
            
            prev = new_str;
        }
        
        prev
    }
}

To solve this coding challenge, we need to implement the count-and-say sequence. The approach requires us to recursively build the sequence, with each step derived from the previous step using run-length encoding (RLE).

# Explanation:

Base Case : The sequence starts with "1" when
n

is 1.
Recursive Case : For any
n

,
countAndSay(n)

is derived from
countAndSay(n-1)

using RLE.
Run-Length Encoding (RLE) : It involves replacing consecutive identical characters with a string that comprises the count of those characters followed by the character itself.

Detailed Steps in Pseudocode:

Step 1 : Initialize the sequence with the base case, i.e., "1" for

n = 1

Step 2 : If

n

is greater than 1, perform the following steps recursively:

Get the previous sequence string which is
countAndSay(n-1)

.
Apply RLE to this string to generate the current sequence.

Step 3 : For each character in the previous sequence:

Count consecutive identical characters.
Formulate the new string by concatenating the count and the character.

Step 4 : Return the newly formed string.

Pseudocode:

Base Case :

                                            
FUNCTION countAndSay(n):
IF n == 1:
RETURN "1"
END IF

Comment: This handles the base case directly by returning "1" when

n

is 1.

Recursive Call :

                                            
FUNCTION countAndSay(n):
IF n == 1:
RETURN "1"
END IF

previous_string = countAndSay(n - 1)
result_string = ""

// Initialize count for the first character
count = 1
character_to_count = previous_string[0]

Comment: Initialize the previous string from the recursive call to

countAndSay(n-1)

, prepare an empty result string, and set up the initial count and character.

RLE loop :

                                            
    // Loop through the previous string starting from the second character
    FOR i FROM 1 TO LENGTH OF previous_string:
        // Check if the current character is the same as the previous character
        IF previous_string[i] == character_to_count:
            count += 1
        ELSE:
            // Add the current count and character to result_string
            result_string = result_string + str(count) + character_to_count
            
            // Reset count and character_to_count
            count = 1
            character_to_count = previous_string[i]
        END IF
    END FOR
    
    // After the loop, add the final count and character
    result_string = result_string + str(count) + character_to_count
    
    RETURN result_string

Comment: This loop handles both tracking consecutive characters and resetting counters when a different character is encountered. The final part ensures the last set of characters is added to the result string.

Complete Pseudocode:

                                            
FUNCTION countAndSay(n):
    // Base case
    IF n == 1:
        RETURN "1"
    END IF
    
    // Get the previous sequence
    previous_string = countAndSay(n - 1)
    result_string = ""
    
    // Initialize the count and character
    count = 1
    character_to_count = previous_string[0]
    
    // Loop through the previous string
    FOR i FROM 1 TO LENGTH OF previous_string:
        IF previous_string[i] == character_to_count:
            // Increment the count if the same character is found
            count += 1
        ELSE:
            // Update the result string with the count and character
            result_string = result_string + str(count) + character_to_count
            
            // Reset the count and character_to_count for the new character
            count = 1
            character_to_count = previous_string[i]
        END IF
    END FOR
    
    // Append the remaining count and character
    result_string = result_string + str(count) + character_to_count
    
    RETURN result_string
END FUNCTION

Detailed Process Walkthrough:

Initial Call
n=4

:

Calls
countAndSay(3)

Calls
countAndSay(2)

Calls
countAndSay(1)

Returns "1"

Generates "11" from "1"

Generates "21" from "11"

Generates "1211" from "21"

By breaking down the problem recursively and applying RLE to each string, we effectively generate the

n

-th element of the count-and-say sequence. This pseudocode can be translated into any programming language that supports basic string operations and recursion.