How do you solve Can Make Palindrome from Substring on LeetCode?

Can Make Palindrome from Substring (LeetCode #1177) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n + m) time complexity and O(n) space complexity. Key insight: Understanding character frequencies is crucial for determining palindrome possibilities.

What is the brute force solution for Can Make Palindrome from Substring?

The brute force approach for Can Make Palindrome from Substring is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach checks each query by extracting the substring, counting character frequencies, and determining if it can be rearranged into a palindrome. This method is straightforward but inefficient for larger strings due to repeated calculations. The optimal Optimal Solution approach improves this to O(n + m).

What algorithm or data structure is used to solve Can Make Palindrome from Substring?

Can Make Palindrome from Substring uses the following concepts: Array, Hash Table, String, Bit Manipulation, Prefix Sum. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Can Make Palindrome from Substring?

Always clarify the constraints and edge cases with the interviewer. Discuss your thought process and potential optimizations as you work through the problem. Practice implementing prefix sums and frequency counting as they are common in string manipulation problems.

What are common mistakes when solving Can Make Palindrome from Substring?

Not considering the case where k is zero and expecting odd frequencies to be handled. Failing to account for the rearrangement of characters when counting odd frequencies.

#1177

Can Make Palindrome from Substring

Medium

ArrayHash TableStringBit ManipulationPrefix SumHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n + m)Space O(n)

The optimal solution uses a prefix frequency array to efficiently calculate character frequencies for any substring in constant time. This reduces the need to repeatedly count characters, making the solution much faster.

⚙️

Algorithm

4 steps

1Step 1: Create a prefix frequency array where each entry at index i contains the frequency of each character from the start of the string up to index i.
2Step 2: For each query, use the prefix frequency array to quickly get the character frequencies for the substring defined by the query.
3Step 3: Count the number of characters with odd frequencies in the substring.
4Step 4: Determine if the number of odd frequencies is less than or equal to k (the number of allowed replacements).

solution.py16 lines

1def canMakePaliQueries(s, queries):
2    n = len(s)
3    prefix = [[0] * 26 for _ in range(n + 1)]
4    for i in range(n):
5        for j in range(26):
6            prefix[i + 1][j] = prefix[i][j]
7        prefix[i + 1][ord(s[i]) - ord('a')] += 1
8
9    result = []
10    for left, right, k in queries:
11        odd_count = 0
12        for j in range(26):
13            if (prefix[right + 1][j] - prefix[left][j]) % 2 != 0:
14                odd_count += 1
15        result.append(odd_count // 2 <= k)
16    return result

ℹ

Complexity note: The time complexity is O(n + m), where n is the length of the string and m is the number of queries. The prefix frequency array allows us to compute character frequencies in constant time for each query.

1Understanding character frequencies is crucial for determining palindrome possibilities.
2Using prefix sums can significantly optimize the frequency counting process.

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