Count and Say LeetCode Solution | Easy Approach

Minimum Cost to Merge Stones
Share:

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 groups so that each group is a contiguous section all of the same character. Then for each group, say the number of characters, then say the character. To convert the saying into a digit string, replace the counts with a number and concatenate every saying.

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.

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

Count and Say Solutions

Time: O(∣ans∣)
Space:  O(∣ans∣)

C++

class Solution {
 public:
  string countAndSay(int n) {
    string ans = "1";

    while (--n) {
      string next;
      for (int i = 0; i < ans.length(); ++i) {
        int count = 1;
        while (i + 1 < ans.length() && ans[i] == ans[i + 1]) {
          ++count;
          ++i;
        }
        next += to_string(count) + ans[i];
      }
      ans = move(next);
    }

    return ans;
  }
};

Java

 class Solution {
  public String countAndSay(int n) {
    StringBuilder sb = new StringBuilder("1");

    while (--n > 0) {
      StringBuilder next = new StringBuilder();
      for (int i = 0; i < sb.length(); ++i) {
        int count = 1;
        while (i + 1 < sb.length() && sb.charAt(i) == sb.charAt(i + 1)) {
          ++count;
          ++i;
        }
        next.append(count).append(sb.charAt(i));
      }
      sb = next;
    }

    return sb.toString();
  }
}

Python

class Solution:
  def countAndSay(self, n: int) -> str:
    ans = '1'

    for _ in range(n - 1):
      nxt = ''
      i = 0
      while i < len(ans):
        count = 1
        while i + 1 < len(ans) and ans[i] == ans[i + 1]:
          count += 1
          i += 1
        nxt += str(count) + ans[i]
        i += 1
      ans = nxt

    return ans

Watch Tutorial

Checkout more Solutions here

Leave a Comment

Your email address will not be published. Required fields are marked *

x