Count and Say Leetcode Solution

In this post, you will know how to solve the Count and Say Leetcode Solution problem of Leetcode. This Leetcode problem is done in many programming languages like C++, Java, and Python.

One more thing to add, don’t directly look for the solutions, first try to solve the problems of Leetcode by yourself. If you find any difficulty after trying several times, then you can look for solutions.

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.

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 Leetcode Solutions in 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

Count and Say Leetcode Solutions in CPP

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;
  }
};

Count and Say Leetcode Solutions in 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();
  }
}

Note: This problem Count and Say is generated by Leetcode but the solution is provided by BrokenProgrammers. This tutorial is only for Educational and Learning purposes.

NEXT: Combination Sum Leetcode Solution