Count and Say - String - Easy - LeetCode
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Count and Say - String - Easy - LeetCode

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  • 1The count-and-say sequence starts with '1' and generates subsequent terms by describing the previous term's digits.
  • 2Each term is constructed by counting consecutive digits and expressing them as 'count followed by digit'.
  • 3The time complexity for generating the nth term is O(n*m), where n is the term number and m is the string length.

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Key Insight
AskGif

"The count-and-say sequence starts with '1' and generates subsequent terms by describing the previous term's digits."

Count and Say - String - Easy - LeetCode

The count-and-say sequence is the sequence of integers with the first five terms as following:

1. 1

2. 11

3. 21

4. 1211

5. 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 where 1 ≤ n ≤ 30, generate the nth term of the count-and-say sequence. You can do so recursively, in other words from the previous member read off the digits, counting the number of digits in groups of the same digit.

Note: Each term of the sequence of integers will be represented as a string.

Example 1:

Input: 1

Output: "1"

Explanation: This is the base case.

Example 2:

Input: 4

Output: "1211"

Explanation: For n = 3 the term was "21" in which we have two groups "2" and "1", "2" can be read as "12" which means frequency = 1 and value = 2, the same way "1" is read as "11", so the answer is the concatenation of "12" and "11" which is "1211".

Solution:

using System;
using System.Collections.Generic;
using System.Text;

namespace LeetCode.AskGif.Easy.String
{
 public class CountAndSaySoln
 {
 public string CountAndSay(int n)
 {
 var strPrev = "1";
 
 for (int i = 1; i < n; i++)
 {
 var current = new StringBuilder();
 if (strPrev.Length == 1)
 {
 current.Append("1");
 current.Append(strPrev);
 strPrev = current.ToString();
 continue;
 }

 int count = 1;
 for (int j = 1; j <= strPrev.Length; j++)
 {
 if (j < strPrev.Length && strPrev[j - 1] == strPrev[j])
 {
 count++;
 }
 else
 {
 current.Append(count.ToString());
 current.Append(strPrev[j - 1]);
 count = 1;
 }
 }
 strPrev = current.ToString();
 }

 return strPrev;
 }
 }
}

Time Complexity: O(n*m) where n is the number while m is the length of the generated string.

Space Complexity: O(n) For StringBuilder

Unit Tests:

using LeetCode.AskGif.Easy.String;
using Microsoft.VisualStudio.TestTools.UnitTesting;
using System;
using System.Collections.Generic;
using System.Text;

namespace CodingUnitTest.Easy.String
{
 [TestClass]
 public class CountAndSaySolnTests
 {
 [TestMethod]
 public void CountAndSaySoln_First()
 {
 var n = 1;
 var output = "1";
 var res = new CountAndSaySoln().CountAndSay(n);

 Assert.AreEqual(res, output);
 }

 [TestMethod]
 public void CountAndSaySoln_Second()
 {
 var n = 4;
 var output = "1211";
 var res = new CountAndSaySoln().CountAndSay(n);

 Assert.AreEqual(res, output);
 }

 [TestMethod]
 public void CountAndSaySoln_Third()
 {
 var n = 5;
 var output = "111221";
 var res = new CountAndSaySoln().CountAndSay(n);

 Assert.AreEqual(res, output);
 }

 [TestMethod]
 public void CountAndSaySoln_fourth()
 {
 var n = 6;
 var output = "312211";
 var res = new CountAndSaySoln().CountAndSay(n);

 Assert.AreEqual(res, output);
 }
 }
}

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AskGif

Published on 4 June 2020 · 2 min read · 411 words

Part of AskGif Blog · coding

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