How do you solve Maximum Deletions on a String on LeetCode?

Maximum Deletions on a String (LeetCode #2430) 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: Dynamic programming can significantly reduce redundant calculations.

What is the brute force solution for Maximum Deletions on a String?

The brute force approach for Maximum Deletions on a String is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible prefix of the string to see if it can be removed based on the given conditions. This method is straightforward but inefficient, as it explores all combinations. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Maximum Deletions on a String?

Clearly explain your thought process and approach to the interviewer. Consider edge cases, such as very short strings or strings with no deletable prefixes. Practice explaining your code as you write it to simulate the interview environment.

What are common mistakes when solving Maximum Deletions on a String?

Not considering all possible prefix lengths for deletion. Overlooking the base case for the dp array.

#2430

Maximum Deletions on a String

Hard

StringDynamic ProgrammingRolling HashString MatchingHash FunctionDynamic ProgrammingString Matching

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(n)

The optimal solution leverages dynamic programming to efficiently calculate the maximum deletions by storing results of subproblems. This avoids redundant calculations and significantly reduces the time complexity.

⚙️

Algorithm

3 steps

1Step 1: Initialize a dp array of size n (length of s) to store the maximum deletions for each prefix.
2Step 2: Iterate through each character in the string, checking for valid deletions based on the prefix match condition.
3Step 3: For each valid deletion found, update the dp array to reflect the maximum deletions possible.

solution.py10 lines

1def maxDeletions(s):
2    n = len(s)
3    dp = [0] * n
4    for i in range(n):
5        dp[i] = dp[i - 1] + 1 if i > 0 else 1
6        for j in range(1, (i + 1) // 2 + 1):
7            if s[:j] == s[j:j + j]:
8                dp[i] = max(dp[i], dp[i - j] + 1)
9    return dp[-1]
10

ℹ

Complexity note: The time complexity is O(n) because we only traverse the string once, and the space complexity is O(n) due to the dp array storing results for each prefix.

1Dynamic programming can significantly reduce redundant calculations.
2Understanding prefix matching is crucial for optimizing deletions.

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