How do you solve Largest Palindromic Number on LeetCode?

Largest Palindromic Number (LeetCode #2384) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: To form a palindrome, digits must be paired symmetrically.

What is the brute force solution for Largest Palindromic Number?

The brute force approach for Largest Palindromic Number is Brute Force, with O(n²) time complexity and O(n!) space complexity. The brute-force approach involves generating all possible permutations of the digits and checking each one to see if it forms a palindrome. This is straightforward but inefficient, especially for larger strings. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Largest Palindromic Number?

Largest Palindromic Number uses the following concepts: Hash Table, String, Greedy, Counting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Largest Palindromic Number?

Always clarify the requirements regarding leading zeros. Discuss your thought process and any edge cases you consider. Practice explaining your solution clearly, as communication is key in interviews.

What are common mistakes when solving Largest Palindromic Number?

Not considering leading zeros in the final output. Overlooking the need for a center digit in odd-length palindromes.

#2384

Largest Palindromic Number

Medium

Hash TableStringGreedyCountingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(1)

The optimal approach counts the frequency of each digit and constructs the largest palindromic number by placing pairs of digits symmetrically around a potential center digit. This is efficient and avoids unnecessary permutations.

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Algorithm

4 steps

1Step 1: Count the frequency of each digit in the string.
2Step 2: Construct the first half of the palindrome using pairs of digits, starting from the largest digit.
3Step 3: If any digit has an odd count, it can be used as the center of the palindrome.
4Step 4: Combine the first half, the center (if any), and the reverse of the first half to form the palindrome.

solution.py16 lines

1# Full working Python code
2from collections import Counter
3
4def largest_palindromic(num):
5    count = Counter(num)
6    first_half = []
7    center = ''
8    for digit in sorted(count.keys(), reverse=True):
9        pairs = count[digit] // 2
10        first_half.append(digit * pairs)
11        if count[digit] % 2 == 1 and center == '':
12            center = digit
13    first_half_str = ''.join(first_half)
14    if first_half_str == '' and center == '0':
15        return '0'
16    return first_half_str + center + first_half_str[::-1]

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Complexity note: The time complexity is O(n) because we traverse the string once to count digits. The space complexity is O(1) since we only use a fixed-size array for digit counts.

1To form a palindrome, digits must be paired symmetrically.
2The largest digits should be prioritized to maximize the palindrome's value.

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