The count-and-say sequence is a sequence of digit strings defined by the recursive formula:
countAndSay(1) = "1"countAndSay(n) is the run-length encoding of countAndSay(n - 1).Run-length encoding (RLE) is a string compression method that works by replacing consecutive identical characters (repeated 2 or more times) with the concatenation of the character and the number marking the count of the characters (length of the run). For example, to compress the string "3322251" we replace "33" with "23", replace "222" with "32", replace "5" with "15" and replace "1" with "11". Thus the compressed string becomes "23321511".
Given a positive integer n, return the nth element of the count-and-say sequence.
Example 1:
Input: n = 4
Output: "1211"
Explanation:
countAndSay(1) = "1" countAndSay(2) = RLE of "1" = "11" countAndSay(3) = RLE of "11" = "21" countAndSay(4) = RLE of "21" = "1211"
Example 2:
Input: n = 1
Output: "1"
Explanation:
This is the base case.
Constraints:
1 <= n <= 30The Count and Say sequence is generated by repeatedly describing the digits of the previous term. Starting from "1", each step reads the current string and groups consecutive identical digits. For every group, you append the count of digits followed by the digit itself. For example, reading "21" becomes "1211" (one 2, then one 1).
The most effective approach is iterative string construction. Begin with the base string and repeat the process n-1 times. During each iteration, traverse the current string, maintain a counter for consecutive characters, and build the next sequence using a string builder or dynamic string.
This problem mainly tests string traversal and grouping logic. Since each term depends on the previous one, the algorithm repeatedly scans and constructs strings. If L represents the maximum length of the generated term, the time complexity becomes O(n × L) because each iteration processes the full string. Space complexity is also O(L) for storing the current sequence.
| Approach | Time Complexity | Space Complexity |
|---|---|---|
| Iterative string construction (grouping consecutive digits) | O(n × L) | O(L) |
Amell Peralta
Use these hints if you're stuck. Try solving on your own first.
Create a helper function that maps an integer to pairs of its digits and their frequencies. For example, if you call this function with "223314444411", then it maps it to an array of pairs [[2,2], [3,2], [1,1], [4,5], [1, 2]].
Create another helper function that takes the array of pairs and creates a new integer. For example, if you call this function with [[2,2], [3,2], [1,1], [4,5], [1, 2]], it should create "22"+"23"+"11"+"54"+"21" = "2223115421".
Now, with the two helper functions, you can start with "1" and call the two functions alternatively n-1 times. The answer is the last integer you will obtain.
This approach involves constructing the sequence iteratively. We start from the base case, which is '1', and iteratively build up the strings by counting and saying the digits from the last constructed string.
Time Complexity: O(2^n) as the length of the string grows exponentially.
Space Complexity: O(2^n) for storing the string.
1#include <stdio.h>
2#include <stdlib.h>
3#include <string.h>
4
5char* countAndSay(int n) {
6 char* result = malloc(2);
7 strcpy(result, "1");
8 for (int i = 1; i < n; i++) {
9 int len = strlen(result);
10 char* temp = malloc(2 * len + 1);
11 int index = 0;
12 for (int j = 0; j < len;) {
13 char current_digit = result[j];
14 int count = 0;
15 while (j < len && result[j] == current_digit) {
16 count++;
17 j++;
18 }
19 index += sprintf(temp + index, "%d%c", count, current_digit);
20 }
21 free(result);
22 result = temp;
23 }
24 return result;
25}
26
27int main() {
28 int n = 4;
29 char* result = countAndSay(n);
30 printf("%s\n", result);
31 free(result);
32 return 0;
33}The solution initializes the result to '1'. For each iteration, it constructs the next sequence by reading the current result, counting contiguous digits, and forming a new string based on these counts. Memory is managed dynamically to accommodate changing string sizes.
The recursive approach defines the function countAndSay recursively by computing countAndSay(n-1), then generating the count-and-say encoding for it.
This method involves less memory usage on the function stack compared to an iterative approach but still leverages recursion for elegance.
Time Complexity: O(2^n) as strings grow exponentially.
Space Complexity: O(2^n) due to storing strings and O(n) recursion stack.
1
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Yes, Count and Say occasionally appears in coding interviews because it tests string manipulation and iteration logic. While not extremely common, it helps interviewers evaluate attention to detail and handling grouped characters.
A dynamic string builder or mutable string structure works best for this problem. It allows efficient concatenation while constructing the next sequence during each iteration of the algorithm.
The optimal approach is iterative string construction. Start with "1" and repeatedly read the previous string, grouping identical consecutive digits and appending the count followed by the digit. This efficiently generates each term until the nth sequence.
Although the logic is straightforward, candidates must carefully manage digit grouping and string construction. Handling consecutive counts correctly and avoiding off-by-one errors can make the implementation tricky.
The recursive C solution computes the sequence for n-1 and uses a helper function to calculate the next sequence by consecutively counting the digits in the current sequence.