Count and Say LeetCode (Look and Say Sequence)

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"

 NOTE:

• 1 <= n <= 30

This problem is popular in LeetCode and GeeksForGeeks A collection of hundreds of interview questions and solutions are available in our blog at Interview Question Solutions

Solution:

Source: Java

	/*
	* To change this license header, choose License Headers in Project Properties.
	* To change this template file, choose Tools | Templates
	* and open the template in the editor.
	*/
	package com.alg.ccup;
	/**
	*
	* The count-and-say sequence is the sequence of integers beginning as follows:
	* 1, 11, 21, 1211, 111221, ...
	*
	* 1 is read off as "one 1" or 11. 11 is read off as "two 1s" or 21. 21 is read
	* off as "one 2, then one 1" or 1211. Given an integer n, generate the nth
	* sequence.
	*
	* Note: The sequence of integers will be represented as a string.
	*
	* Base case: n = 0 print "1" for n = 1, look at previous string and write
	* number of times a digit is seen and the digit itself. In this case, digit 1
	* is seen 1 time in a row... so print "1 1" for n = 2, digit 1 is seen two
	* times in a row, so print "2 1" for n = 3, digit 2 is seen 1 time and then
	* digit 1 is seen 1 so print "1 2 1 1" for n = 4 you will print "1 1 1 2 2 1"
	*
	*
	* @author rbaral
	*/
	public class CountAndSay {
	public static String countAndSay(int n) {
	String resultString = "1";
	//base case
	if (n <= 1) {
	return "1";
	} else {
	int count = 1;
	for (int i = 1; i < n; i++) {
	StringBuilder lastString = new StringBuilder();
	for (int j = 1; j < resultString.length(); j++) {
	if (resultString.charAt(j) == resultString.charAt(j - 1)) {// a repeat character
	count++;
	} else {//new character
	lastString.append(count);
	lastString.append(resultString.charAt(j - 1));
	count = 1;
	}
	}
	lastString.append(count);
	lastString.append(resultString.charAt(resultString.length() - 1));
	resultString = lastString.toString();
	count = 1;
	}
	}
	return resultString;
	}
	public static void main(String args[]) {
	System.out.println(countAndSay(6));
	}
	}

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