Solution For Count And Say
Problem Statement: 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, followed by the character. To convert the saying into a digit string, replace the counts with a number and concatenate every saying.
Example:
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”
Approach: We can solve this problem by applying a recursive approach. We can start from n=1 and keep returning the resulting string by applying the rule of the count-and-say sequence. For a given n, we can recursively call the function to calculate the string countAndSay(n-1) which is given. To get the required string countAndSay(n), we can apply the following steps:
- Initialize the resulting string as an empty string.
- Initialize count as 1 and the starting character as the first character in countAndSay(n-1).
- Traverse the string countAndSay(n-1) from the second character to the end.
- If the current character is the same as the previous character, increment the count.
- If the current character is different from the previous character, append the count and the previous character to the resulting string and start a new count for the current character.
- After the traversal is complete, append the last count and character to the resulting string and return the resulting string.
For example, consider the following code for the countAndSay function:
“`python
def countAndSay(n: int) -> str:
if n == 1:
return “1”
res = ""
prev = countAndSay(n-1)
count = 1
prev_char = prev[0]
for i in range(1, len(prev)):
if prev[i] == prev_char:
count += 1
else:
res += str(count) + prev_char
count = 1
prev_char = prev[i]
res += str(count) + prev_char
return res
“`
Time Complexity: Each recursive call performs a linear traversal of the previous string, so the time complexity of the countAndSay function is O(n*m), where n is the input value of n, and m is the length of the previous string. In the worst case, m = 2^(n-1) which is the length of the maximum possible string.
Space Complexity: Each recursive call requires a new string of length at most 2^(n-1). Therefore, the space complexity of the countAndSay function is O(2^n).
Step by Step Implementation For Count And Say
public class Solution {
public String countAndSay(int n) {
// Base case
if (n == 1) {
return "1";
}
// Recursive case
String prev = countAndSay(n - 1);
StringBuilder sb = new StringBuilder();
// Iterate through the previous result string
for (int i = 0; i < prev.length(); i++) {
// Keep track of the number of consecutive occurrences
int count = 1;
// Check for consecutive occurrences
while (i < prev.length() - 1 && prev.charAt(i) == prev.charAt(i + 1)) {
count++;
i++;
}
// Append the number of occurrences and the character to the string builder
sb.append(count);
sb.append(prev.charAt(i));
}
// Convert the string builder to a string and return
return sb.toString();
}
}def countAndSay(n): if n == 1: return "1" # Base case prev = countAndSay(n-1) # Recursive call count = 1 # Initialize count i = 1 res = "" # Initialize result while i < len(prev): if prev[i] == prev[i-1]: count += 1 else: # Append count and character res += str(count) res += prev[i-1] count = 1 i += 1 # If last character is same as # previous one if prev[i-1] == prev[i-2]: res += str(count) res += prev[i-1] return res
var countAndSay = function(n) {
// your code goes here
};int countAndSay(int n)
{
// Base cases
if (n == 1) return "1";
if (n == 2) return "11";
// Find n'th term by generating all terms from 3 to
// n-1. Every term is generated using previous term
string str = "11"; // Initialize previous term
for (int i = 3; i<=n; i++)
{
// In below for loop, previous character is processed in
// current iteration. That is why a dummy character is
// added to make sure that loop runs one extra iteration.
str += '$';
int len = str.length();
int cnt = 1; // Initialize count of matching chars
string tmp = ""; // Initialize i'th term in series
// Process previous term to find the next term
for (int j = 1; j < len; j++)
{
// If current character does't match
if (str[j] != str[j-1])
{
// Append count of str[j-1] to temp
tmp += cnt + '0';
// Append str[j-1]
tmp += str[j-1];
// Reset count
cnt = 1;
}
// If matches, then increment count of matching
// characters
else cnt++;
}
// Update str
str = tmp;
}
return str;
}public class Solution {
public string CountAndSay(int n) {
// Base case
if (n == 1) {
return "1";
}
// Use previous result to generate next result
string prevResult = CountAndSay(n - 1);
return GenerateNextResult(prevResult);
}
private string GenerateNextResult(string prevResult) {
// Track current number and count
char currNum = prevResult[0];
int currCount = 1;
// Use StringBuilder to efficiently create result string
StringBuilder result = new StringBuilder();
// Iterate through previous result string
for (int i = 1; i < prevResult.Length; i++) {
// If current number is different than previous number
if (prevResult[i] != currNum) {
// Append current number and count to result
result.Append(currCount);
result.Append(currNum);
// Reset current number and count
currNum = prevResult[i];
currCount = 1;
}
// Otherwise, increment count
else {
currCount++;
}
}
// Append last number and count to result
result.Append(currCount);
result.Append(currNum);
return result.ToString();
}
}