grid47 Exploring patterns and algorithms
Nov 3, 2024
5 min read
Solution to LeetCode 38: Count and Say Problem
The Count-and-Say sequence starts with ‘1’. Each subsequent term is generated by describing the previous term in terms of the count of consecutive digits. Given a positive integer n, return the nth term of the Count-and-Say sequence.
Problem
Approach
Steps
Complexity
Input: A positive integer n representing the term of the Count-and-Say sequence to be returned.
Example: Input: n = 4
Constraints:
• 1 <= n <= 30
Output: Return the nth term of the Count-and-Say sequence.
Example: Output: "1211"
Constraints:
• Return a string representing the nth term.
Goal: The goal is to iteratively generate the nth term by describing the previous term using run-length encoding.
Steps:
• Start with '1'.
• For each subsequent term, describe the previous term using run-length encoding.
• For each group of consecutive digits in the previous term, note the count followed by the digit.
Goal: The input integer n will always be between 1 and 30.
Steps:
• 1 <= n <= 30
Assumptions:
• n is always a positive integer.
• The Count-and-Say sequence starts with '1'.
• Input: Input: n = 4
• Explanation: The 4th term in the Count-and-Say sequence is '1211', which is generated by describing the 3rd term '21' as 'one 2, one 1'.
• Input: Input: n = 2
• Explanation: The 2nd term in the Count-and-Say sequence is '11', generated by describing the first term '1' as 'one 1'.
Approach: Start from the first term of the sequence and iteratively generate the next terms using run-length encoding to describe the previous term.
Observations:
• The sequence can be generated iteratively by processing one term at a time.
• A simple way to describe each term is by counting consecutive characters and then concatenating the count and the character.
Steps:
• Start with the first term '1'.
• For each subsequent term, iterate through the previous term and count consecutive digits.
• Concatenate the count and the digit to form the next term.
• Repeat this process for n-1 iterations.
Empty Inputs:
• N/A, since n is always at least 1.
Large Inputs:
• The constraints (n <= 30) ensure that large inputs will not be problematic.
Special Values:
• Consider how the sequence behaves with the smallest input n = 1, which simply returns '1'.
Constraints:
• n will always be within the range [1, 30].
string countAndSay(int n) {
if (n ==1) return"1";
string tmp = countAndSay(n-1);
string ans ="";
for(int i =0; i < tmp.size(); ) {
char ch = tmp[i];
int cnt =0;
while(tmp[i] == ch && i < tmp.size()) {
cnt++;
i++;
}
ans.push_back(cnt+'0');
ans.push_back(tmp[i-1]);
}
return ans;
}
1 : Function Declaration
string countAndSay(int n) {
This line declares a function named `countAndSay` that takes an integer `n` as input and returns a string representing the nth term of the count-and-say sequence.
2 : Base Case
if (n ==1) return"1";
This line handles the base case where `n` is 1. The first term of the sequence is simply "1", so the function returns it.
3 : Recursive Call
string tmp = countAndSay(n-1);
This line recursively calls the `countAndSay` function to get the (n-1)th term of the sequence and stores it in the `tmp` string.
4 : String Initialization
string ans ="";
This line initializes an empty string `ans` to store the nth term of the sequence.
5 : Loop Iteration
for(int i =0; i < tmp.size(); ) {
This line starts a `for` loop to iterate over the characters of the `tmp` string.
6 : Variable Initialization
char ch = tmp[i];
int cnt =0;
These lines initialize two variables: `ch` to store the current character being processed and `cnt` to count the occurrences of that character.
7 : Loop Iteration and Condition
while(tmp[i] == ch && i < tmp.size()) {
This `while` loop continues as long as the current character `tmp[i]` is the same as the previous character `ch` and the index `i` is within the bounds of the `tmp` string.
8 : Counter Increment
cnt++;
This line increments the `cnt` variable to count the occurrences of the current character.
9 : Index Increment
i++;
This line increments the `i` index to move to the next character in the `tmp` string.
10 : String Manipulation
ans.push_back(cnt+'0');
ans.push_back(tmp[i-1]);
These lines append the count of the current character (`cnt`) and the character itself (`tmp[i-1]`) to the `ans` string. The `cnt+'0'` converts the integer count to a character representing the digit.
11 : Return
return ans;
After processing all characters in the `tmp` string, the function returns the `ans` string, which represents the nth term of the count-and-say sequence.
Best Case: O(n)
Average Case: O(n^2)
Worst Case: O(n^2)
Description: The time complexity is O(n^2) in the worst case since the number of digits in each term grows approximately exponentially.
Best Case: O(n)
Worst Case: O(n^2)
Description: Space complexity is O(n) for storing the current term, with additional space needed to store previous terms.
