The count-and-say sequence is a sequence of digit strings defined by the following recursive formula:
For example:
Given a positive integer n, return the nth term of the count-and-say sequence.
Input: n = 4
Output: "1211"
Explanation:
countAndSay(1) = "1"
countAndSay(2) is described as "one 1" = "11"
countAndSay(3) is described as "two 1s" = "21"
countAndSay(4) is described as "one 2, then one 1" = "1211"
Input: n = 1
Output: "1"
Explanation:
This is the base case where countAndSay(1) is "1".
Input: n = 5
A simple way to solve the problem is to take the help of recursion which will come in handy while determining the (n-1)th term of the count-and-say sequence.
The nth string of the count-and-say sequence can be determined by counting the identical consecutive digits.
To solve this problem, a recursive approach will be followed:
#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
/* Recursive function to return the nth
term of the count-and-say sequence */
string countAndSay(int n) {
if(n == 1) return "1";
// Recursive call
string prev = countAndSay(n-1);
int len = prev.length();
// To store the answer
string ans = "";
// To count the frequency of identicals
int count = 1;
// Traverse the string
for(int i=1; i < len; i++) {
// If identicals are found, increment the counter
if(prev[i] == prev[i-1]) count++;
// Otherwise
else {
ans.push_back('0' + count); // Add frequency
ans.push_back(prev[i-1]); // Add the digit
count = 1; // Reset counter to 1
}
}
// Adding the frequency for the last digit and the last digit
ans.push_back('0' + count);
ans.push_back(prev[len-1]);
return ans; // Return the result
}
};
int main(){
int n = 4;
// Creating an instance of Solution class
Solution sol;
/* Recursive function call to return the
nth term of the count-and-say sequence */
string ans = sol.countAndSay(n);
// Output
cout << "The nth term of the count-and-say sequence is: " << ans << endl;
}import java.util.*;
class Solution {
/* Recursive function to return the nth
term of the count-and-say sequence */
public String countAndSay(int n) {
if (n == 1) return "1";
// Recursive call
String prev = countAndSay(n - 1);
int len = prev.length();
// To store the answer
String ans = "";
// To count the frequency of identicals
int count = 1;
// Traverse the string
for (int i = 1; i < len; i++) {
// If identicals are found, increment the counter
if (prev.charAt(i) == prev.charAt(i - 1)) count++;
// Otherwise
else {
ans += (char) ('0' + count); // Add frequency
ans += prev.charAt(i - 1); // Add the digit
count = 1; // Reset counter to 1
}
}
// Adding the frequency for the last digit and the last digit
ans += (char) ('0' + count);
ans += prev.charAt(len - 1);
return ans; // Return the result
}
}
class Main {
public static void main(String[] args) {
int n = 4;
// Creating an instance of Solution class
Solution sol = new Solution();
/* Recursive function call to return the
nth term of the count-and-say sequence */
String ans = sol.countAndSay(n);
// Output
System.out.println("The nth term of the count-and-say sequence is: " + ans);
}
}class Solution:
# Recursive function to return the nth
# term of the count-and-say sequence
def countAndSay(self, n: int) -> str:
if n == 1:
return "1"
# Recursive call
prev = self.countAndSay(n - 1)
length = len(prev)
# To store the answer
ans = ""
# To count the frequency of identicals
count = 1
# Traverse the string
for i in range(1, length):
# If identicals are found, increment the counter
if prev[i] == prev[i - 1]:
count += 1
# Otherwise
else:
ans += str(count) # Add frequency
ans += prev[i - 1] # Add the digit
count = 1 # Reset counter to 1
# Adding the frequency for the last digit and the last digit
ans += str(count)
ans += prev[-1]
return ans # Return the result
# Main function
if __name__ == "__main__":
n = 4
# Creating an instance of Solution class
sol = Solution()
# Recursive function call to return the nth term of the count-and-say sequence
ans = sol.countAndSay(n)
# Output
print(f"The nth term of the count-and-say sequence is: {ans}")class Solution {
/* Recursive function to return the nth
term of the count-and-say sequence */
countAndSay(n) {
if (n === 1) return "1";
// Recursive call
let prev = this.countAndSay(n - 1);
let len = prev.length;
// To store the answer
let ans = "";
// To count the frequency of identicals
let count = 1;
// Traverse the string
for (let i = 1; i < len; i++) {
// If identicals are found, increment the counter
if (prev[i] === prev[i - 1]) count++;
// Otherwise
else {
ans += (count).toString(); // Add frequency
ans += prev[i - 1]; // Add the digit
count = 1; // Reset counter to 1
}
}
// Adding the frequency for the last digit and the last digit
ans += (count).toString();
ans += prev[len - 1];
return ans; // Return the result
}
}
// Main function
const n = 4;
// Creating an instance of Solution class
const sol = new Solution();
/* Recursive function call to return the
nth term of the count-and-say sequence */
const ans = sol.countAndSay(n);
// Output
console.log(`The nth term of the count-and-say sequence is: ${ans}`);
Time Complexity: Can't be determined exactly
Because it will depend on the length of the string in each step which is uncertain.
Space Complexity: Can't be determined exactly
The recursive stack space will take O(n) space and during each step, the string generated will take a variable space.
Q: Why can’t we use recursion directly?
A: A naive recursive approach leads to redundant recomputation, causing stack overflow for large n. An iterative approach avoids this problem.
Q: Why do we process characters one-by-one instead of using regex or string splits?
A: The problem requires processing consecutive characters, which is best handled with a two-pointer approach instead of splitting.
Q: How would you modify the approach if we needed to generate only the k-th character of countAndSay(n)?
A: Instead of generating the full sequence, use on-the-fly processing to retrieve only the required k-th character, reducing memory usage.
Q: How does this compare to Fibonacci sequence generation?
A: Fibonacci grows linearly in O(n) using DP. Count-and-say grows exponentially in length but remains O(n * m) due to processing constraints.