How do you solve Minimum Cost Good Caption on LeetCode?

Minimum Cost Good Caption (LeetCode #3441) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: A good caption requires characters to be grouped in threes.

What is the brute force solution for Minimum Cost Good Caption?

The brute force approach for Minimum Cost Good Caption is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking every possible combination of characters at each index to see how many operations are needed to transform the string into a good caption. This method is straightforward but inefficient, as it explores many unnecessary paths. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Cost Good Caption?

Minimum Cost Good Caption uses the following concepts: String, Dynamic Programming. The recommended patterns to study are: Dynamic Programming, Greedy Algorithms.

What are the interview tips for Minimum Cost Good Caption?

Always clarify the problem requirements before jumping into coding. Think about edge cases and how your solution handles them. Discuss your thought process and reasoning while coding.

What are common mistakes when solving Minimum Cost Good Caption?

Failing to consider lexicographical order when making changes. Not handling edge cases where it's impossible to create a good caption.

#3441

Minimum Cost Good Caption

Hard

StringDynamic ProgrammingDynamic ProgrammingGreedy Algorithms

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal approach uses dynamic programming to track the minimum operations needed to convert each character into a good caption while ensuring the result is lexicographically smallest. By systematically checking groups of characters, we can minimize changes effectively.

⚙️

Algorithm

3 steps

1Step 1: Initialize a DP array to store the minimum operations needed for each character.
2Step 2: Iterate through the string and for each character, check the previous two characters to determine if they can form a good caption.
3Step 3: Update the DP array based on the minimum operations required to convert the current character, ensuring to choose the lexicographically smallest option.

solution.py13 lines

1def min_cost_good_caption(caption):
2    n = len(caption)
3    dp = [[float('inf')] * 26 for _ in range(n)]
4    for i in range(26):
5        dp[0][i] = 0 if chr(i + 97) == caption[0] else 1
6    for i in range(1, n):
7        for j in range(26):
8            for k in range(max(0, j - 1), min(25, j + 1) + 1):
9                dp[i][j] = min(dp[i][j], dp[i - 1][k] + (0 if chr(j + 97) == caption[i] else 1))
10    min_ops = min(dp[n - 1])
11    if min_ops == float('inf'):
12        return ''
13    return ''.join(chr(i + 97) for i in range(26) if dp[n - 1][i] == min_ops)

ℹ

Complexity note: This complexity is due to the single pass through the string and the limited number of character checks (26 letters), making it efficient.

1A good caption requires characters to be grouped in threes.
2Dynamic programming can efficiently track minimum operations.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.