Algorithm
Problem Name: 38. Count and Say
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 fromcountAndSay(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
Code Examples
#1 Code Example with C Programming
Code -
C Programming
char* countAndSay(int n) {
char *s, *d, *x;
int i, c;
if (!n) return NULL;
#define BUF_SIZE 100 * 1024
s = malloc(BUF_SIZE);
x = d = malloc(BUF_SIZE);
s[0] = '1';
s[1] = 0;
while (n > 1) {
c = 1;
for (i = 0; s[i]; i ++) {
if (s[i] != s[i+1]) {
sprintf(x, "%d%c", c, s[i]);
x += strlen(x);
c = 1;
} else {
c ++;
}
}
x = s;
s = d;
d = x;
n --;
}
free(d);
return s;
}
Input
Output
#2 Code Example with C++ Programming
Code -
C++ Programming
class Solution {
public:
string countAndSay(int n) {
if(n == 1) return "1";
string s = countAndSay(n - 1);
string res = "";
for(int i = 0; i < s.size(); i++){
int count = 1;
while(i < s.size() - 1 && s[i] == s[i + 1]) i++, count++;
res.append(to_string(count) + s[i]>;
}
return res;
}
};
Input
Output
#3 Code Example with Java Programming
Code -
Java Programming
class Solution {
public String countAndSay(int n) {
String s = "1";
for (int i = 1; i < n; i++) {
StringBuilder sb = new StringBuilder();
for (int j = 1, count = 1; j < = s.length(); j++) {
if (j == s.length() || s.charAt(j - 1) != s.charAt(j)) {
sb.append(count);
sb.append(s.charAt(j - 1));
count = 1;
} else {
count++;
}
}
s = sb.toString();
}
return s;
}
}
Input
Output
#4 Code Example with Javascript Programming
Code -
Javascript Programming
const countAndSay = function(n) {
let str = '1'
for (let i = 2; i < = n; i++) {
let tempStr = ''
let count = 0
for (let j = 0, m = str.length; j < m; j++) {
const char = str.charAt(j)
count += 1
if (char !== str.charAt(j + 1)) {
tempStr += count + char
count = 0
}
}
str = tempStr
}
return str
}
Input
Output
#5 Code Example with Python Programming
Code -
Python Programming
class Solution:
def countAndSay(self, n):
curr = "1"
for i in range(n - 1):
tmp, cnt = "", 1
for j, c in enumerate(curr):
if j > 0 and curr[j - 1] == c:
cnt += 1
elif j > 0:
tmp += str(cnt) + curr[j - 1]
cnt = 1
if j == len(curr) - 1:
tmp += str(cnt) + curr[j]
curr = tmp
return curr
Input
Output
#6 Code Example with C# Programming
Code -
C# Programming
using System.Text;
namespace LeetCode
{
public class _038_CountAndSay
{
public string CountAndSay(int n)
{
var result = "1";
char currentCh;
int i, j, startNum;
var builder = new StringBuilder();
for (i = 1; i < n; i++)
{
currentCh = result[0];
startNum = 0;
for (j = 1; j < result.Length; j++)
{
if (currentCh != result[j])
{
builder.Append(j - startNum);
builder.Append(currentCh);
currentCh = result[j];
startNum = j;
}
}
builder.Append(j - startNum);
builder.Append(currentCh);
result = builder.ToString();
builder.Clear();
}
return result;
}
}
}
Input
Output