What is the time complexity of Maximum Number of Non-overlapping Palindrome Substrings?

The optimal solution for Maximum Number of Non-overlapping Palindrome Substrings runs in O(n²) time and uses O(n²) space. The time complexity is O(n²) due to the nested loops used to fill the dp array, but it is more efficient than brute force as it avoids redundant palindrome checks.

What is the brute force solution for Maximum Number of Non-overlapping Palindrome Substrings?

The brute force approach for Maximum Number of Non-overlapping Palindrome Substrings is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking every possible substring of the string to see if it is a palindrome and meets the length requirement. This is simple but inefficient, as it may involve a lot of unnecessary checks. The optimal Optimal Solution approach improves this to O(n²).

What algorithm or data structure is used to solve Maximum Number of Non-overlapping Palindrome Substrings?

Maximum Number of Non-overlapping Palindrome Substrings uses the following concepts: Two Pointers, String, Dynamic Programming, Greedy. The recommended patterns to study are: Dynamic Programming, Two Pointers.

What are the interview tips for Maximum Number of Non-overlapping Palindrome Substrings?

Clearly explain your thought process and how you approach identifying palindromes. Discuss the trade-offs between brute force and optimized solutions. Practice dynamic programming problems to get comfortable with state transitions.

What are common mistakes when solving Maximum Number of Non-overlapping Palindrome Substrings?

Failing to check for overlapping substrings correctly. Not initializing the dp array properly, leading to incorrect palindrome checks.

#2472

Maximum Number of Non-overlapping Palindrome Substrings

Hard

Two PointersStringDynamic ProgrammingGreedyDynamic ProgrammingTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n²)Space O(n²)

The optimal approach uses dynamic programming to efficiently track palindromic substrings and their counts. This reduces the need for repeated checks and allows us to build on previously computed results.

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Algorithm

4 steps

1Step 1: Create a 2D array dp where dp[i][j] is true if the substring s[i...j] is a palindrome.
2Step 2: Fill the dp array using a bottom-up approach, checking substrings of increasing lengths.
3Step 3: Create a 1D array count to store the maximum number of non-overlapping palindromic substrings ending at each index.
4Step 4: For each index, check if a valid palindrome ends there and update the count array accordingly.

solution.py19 lines

1# Full working Python code
2
3def max_palindromic_substrings(s, k):
4    n = len(s)
5    dp = [[False] * n for _ in range(n)]
6    count = [0] * n
7
8    for length in range(1, n + 1):
9        for i in range(n - length + 1):
10            j = i + length - 1
11            if s[i] == s[j] and (length <= 2 or dp[i + 1][j - 1]):
12                dp[i][j] = True
13                if length >= k:
14                    count[j] = max(count[j], (count[i - 1] if i > 0 else 0) + 1)
15
16    return count[-1]
17
18# Example usage
19print(max_palindromic_substrings('abaccdbbd', 3))  # Output: 2

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Complexity note: The time complexity is O(n²) due to the nested loops used to fill the dp array, but it is more efficient than brute force as it avoids redundant palindrome checks.

1Identifying palindromic substrings is crucial; use dynamic programming for efficiency.
2Non-overlapping substrings require careful tracking of indices to avoid counting overlaps.

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