Solving Leetcode Interviews in Seconds with AI: Minimum Cost Good Caption
Introduction
In this blog post, we will explore how to solve the LeetCode problem "3441" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.
Problem Statement
You are given a string caption of length n. A good caption is a string where every character appears in groups of at least 3 consecutive occurrences. For example: "aaabbb" and "aaaaccc" are good captions. "aabbb" and "ccccd" are not good captions. You can perform the following operation any number of times: Choose an index i (where 0 <= i < n) and change the character at that index to either: The character immediately before it in the alphabet (if caption[i] != 'a'). The character immediately after it in the alphabet (if caption[i] != 'z'). Your task is to convert the given caption into a good caption using the minimum number of operations, and return it. If there are multiple possible good captions, return the lexicographically smallest one among them. If it is impossible to create a good caption, return an empty string "". Example 1: Input: caption = "cdcd" Output: "cccc" Explanation: It can be shown that the given caption cannot be transformed into a good caption with fewer than 2 operations. The possible good captions that can be created using exactly 2 operations are: "dddd": Change caption[0] and caption[2] to their next character 'd'. "cccc": Change caption[1] and caption[3] to their previous character 'c'. Since "cccc" is lexicographically smaller than "dddd", return "cccc". Example 2: Input: caption = "aca" Output: "aaa" Explanation: It can be proven that the given caption requires at least 2 operations to be transformed into a good caption. The only good caption that can be obtained with exactly 2 operations is as follows: Operation 1: Change caption[1] to 'b'. caption = "aba". Operation 2: Change caption[1] to 'a'. caption = "aaa". Thus, return "aaa". Example 3: Input: caption = "bc" Output: "" Explanation: It can be shown that the given caption cannot be converted to a good caption by using any number of operations. Constraints: 1 <= caption.length <= 5 * 104 caption consists only of lowercase English letters.
Explanation
Here's a breakdown of the solution and the code:
High-Level Approach:
- Iterate through the caption, identifying consecutive groups of the same character.
- If a group has fewer than 3 characters, attempt to extend it by modifying characters around it. Prioritize creating the lexicographically smallest good caption by decrementing characters whenever possible.
- If at any point, forming a good caption becomes impossible (e.g., a single character 'b' surrounded by 'a' and 'c'), return an empty string.
Complexity:
- Runtime Complexity: O(n), where n is the length of the caption.
- Storage Complexity: O(n) to store the modified caption as a list of characters.
def solve():
caption = input()
n = len(caption)
caption_list = list(caption)
i = 0
while i < n:
j = i
while j < n and caption_list[i] == caption_list[j]:
j += 1
length = j - i
if length >= 3:
i = j
continue
if length == 1:
if i == 0 and j == n:
return ""
elif i == 0:
if ord(caption_list[j]) - ord(caption_list[i]) > 1:
return ""
else:
caption_list[i] = chr(ord(caption_list[j]))
caption_list[i+1] = chr(ord(caption_list[j]))
elif j == n:
if ord(caption_list[i]) - ord(caption_list[i-1]) > 1:
return ""
else:
caption_list[i] = chr(ord(caption_list[i-1]))
caption_list[i+1] = chr(ord(caption_list[i-1]))
else:
if ord(caption_list[j]) - ord(caption_list[i]) <=1 and ord(caption_list[i]) - ord(caption_list[i-1]) >1:
caption_list[i] = chr(ord(caption_list[j]))
caption_list[i+1] = chr(ord(caption_list[j]))
elif ord(caption_list[j]) - ord(caption_list[i]) > 1 and ord(caption_list[i]) - ord(caption_list[i-1]) <=1:
caption_list[i] = chr(ord(caption_list[i-1]))
caption_list[i+1] = chr(ord(caption_list[i-1]))
elif ord(caption_list[j]) - ord(caption_list[i]) <=1 and ord(caption_list[i]) - ord(caption_list[i-1]) <=1:
if ord(caption_list[i-1]) < ord(caption_list[j]):
caption_list[i] = chr(ord(caption_list[i-1]))
caption_list[i+1] = chr(ord(caption_list[i-1]))
else:
caption_list[i] = chr(ord(caption_list[j]))
caption_list[i+1] = chr(ord(caption_list[j]))
else:
return ""
elif length == 2:
if i == 0:
caption_list[i+2] = chr(ord(caption_list[i]))
elif j == n:
caption_list[i-1] = chr(ord(caption_list[i]))
else:
if ord(caption_list[i]) - ord(caption_list[i-1]) <=1:
caption_list[i+2] = chr(ord(caption_list[i]))
caption_list[i-1] = chr(ord(caption_list[i]))
elif ord(caption_list[j]) - ord(caption_list[i]) <=1:
caption_list[i+2] = chr(ord(caption_list[i]))
caption_list[i-1] = chr(ord(caption_list[i]))
else:
return ""
else:
return ""
i = 0
while i < n-2:
if caption_list[i] != caption_list[i+1] or caption_list[i+1] != caption_list[i+2]:
return ""
i += 1
return "".join(caption_list)
return "".join(caption_list) if len(set(caption_list)) <= 1 and len(caption_list) >= 3 else ""
# Code
```python
def solve(): caption = input()
n = len(caption)
caption_list = list(caption)
i = 0
while i < n:
j = i
while j < n and caption_list[i] == caption_list[j]:
j += 1
length = j - i
if length >= 3:
i = j
continue
if length == 1:
if i == 0 and j == n:
return ""
elif i == 0:
if ord(caption_list[j]) - ord(caption_list[i]) > 1:
return ""
else:
caption_list[i] = chr(ord(caption_list[j]))
caption_list[i+1] = chr(ord(caption_list[j]))
elif j == n:
if ord(caption_list[i]) - ord(caption_list[i-1]) > 1:
return ""
else:
caption_list[i] = chr(ord(caption_list[i-1]))
caption_list[i+1] = chr(ord(caption_list[i-1]))
else:
if ord(caption_list[j]) - ord(caption_list[i]) <=1 and ord(caption_list[i]) - ord(caption_list[i-1]) >1:
caption_list[i] = chr(ord(caption_list[j]))
caption_list[i+1] = chr(ord(caption_list[j]))
elif ord(caption_list[j]) - ord(caption_list[i]) > 1 and ord(caption_list[i]) - ord(caption_list[i-1]) <=1:
caption_list[i] = chr(ord(caption_list[i-1]))
caption_list[i+1] = chr(ord(caption_list[i-1]))
elif ord(caption_list[j]) - ord(caption_list[i]) <=1 and ord(caption_list[i]) - ord(caption_list[i-1]) <=1:
if ord(caption_list[i-1]) < ord(caption_list[j]):
caption_list[i] = chr(ord(caption_list[i-1]))
caption_list[i+1] = chr(ord(caption_list[i-1]))
else:
caption_list[i] = chr(ord(caption_list[j]))
caption_list[i+1] = chr(ord(caption_list[j]))
else:
return ""
elif length == 2:
if i == 0:
caption_list[i+2] = chr(ord(caption_list[i]))
elif j == n:
caption_list[i-1] = chr(ord(caption_list[i]))
else:
if ord(caption_list[i]) - ord(caption_list[i-1]) <=1:
caption_list[i+2] = chr(ord(caption_list[i]))
caption_list[i-1] = chr(ord(caption_list[i]))
elif ord(caption_list[j]) - ord(caption_list[i]) <=1:
caption_list[i+2] = chr(ord(caption_list[i]))
caption_list[i-1] = chr(ord(caption_list[i]))
else:
return ""
else:
return ""
i = 0
while i < n-2:
if caption_list[i] != caption_list[i+1] or caption_list[i+1] != caption_list[i+2]:
return ""
i += 1
return "".join(caption_list)
return "".join(caption_list) if len(set(caption_list)) <= 1 and len(caption_list) >= 3 else ""
print(solve())