Count And Say
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# LeetCode #38: Count And Say (C/C++).

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source: https://leetcode.com/problems/count-and-say/
C/C++ Solution to LeetCode problem 38. Count And Say.

## Problem

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 substrings such that each substring contains exactly one unique digit. Then for each substring, say the number of digits, then say the digit. Finally, concatenate every said digit.

For example, the saying and conversion for digit string "3322251":

Given a positive integer n, return the nth term of the count-and-say sequence.

## Examples

### Example 1:

Input: n = 1,
Output: “1”
Explanation: This is the base case.

### Example 2:

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”

## Constraints

• 1 <= n <= 30

## Solution

• We iterate through the string, if size == 0 then the string will be "1".
• We count how many times a character is repeated.
• Append to the final string the counter + the character.
• Call recursively the method, passing our last generated string.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 class Solution { private: string cAndS(int n, string cStr) { if (n==0) return cStr; string tmp = cStr.size() == 0 ? "1"s : ""s; for (int i=0; i<cStr.size(); i++) { int j=i+1; while(j<cStr.size() && cStr[j] == cStr[i]) j++; tmp += to_string(j-i) + cStr[i]; i = j-1; } return cAndS(n-1, tmp); } public: string countAndSay(int n) { return cAndS(n, ""); } };