How do you solve Construct K Palindrome Strings on LeetCode?

Construct K Palindrome Strings (LeetCode #1400) 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: If the length of the string is less than k, it's impossible to form k non-empty strings.

What is the brute force solution for Construct K Palindrome Strings?

The brute force approach for Construct K Palindrome Strings is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks every possible way to partition the string into k parts and verifies if each part can form a palindrome. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Construct K Palindrome Strings?

Construct K Palindrome Strings uses the following concepts: Hash Table, String, Greedy, Counting. The recommended patterns to study are: Hash Map, Counting.

What are the interview tips for Construct K Palindrome Strings?

Always check edge cases, such as when k is greater than the length of the string. Explain your thought process clearly, especially when transitioning from brute force to optimal solutions. Practice counting character frequencies and understanding their implications for palindrome properties.

What are common mistakes when solving Construct K Palindrome Strings?

Failing to check if the string length is less than k before proceeding. Misunderstanding how to count odd frequencies and their impact on palindrome formation.

#1400

Construct K Palindrome Strings

Medium

Hash TableStringGreedyCountingHash MapCounting

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 solution focuses on counting character frequencies and determining the minimum number of palindromes that can be formed based on odd character counts. This is efficient and avoids unnecessary partitioning.

⚙️

Algorithm

3 steps

1Step 1: Count the frequency of each character in the string.
2Step 2: Count how many characters have odd frequencies.
3Step 3: If the number of odd frequencies is greater than k, return false; otherwise, return true.

solution.py9 lines

1# Full working Python code
2from collections import Counter
3
4def can_construct_k_palindromes(s, k):
5    if len(s) < k:
6        return False
7    freq = Counter(s)
8    odd_count = sum(1 for count in freq.values() if count % 2 != 0)
9    return odd_count <= k

ℹ

Complexity note: The complexity is linear because we only traverse the string once to count frequencies and then check the counts, leading to efficient performance.

1If the length of the string is less than k, it's impossible to form k non-empty strings.
2The number of characters with odd frequencies determines the minimum number of palindromes that can be formed.

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