How do you solve Find the Count of Good Integers on LeetCode?

Find the Count of Good Integers (LeetCode #3272) 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: A number can only be rearranged into a palindrome if at most one digit has an odd frequency.

What is the brute force solution for Find the Count of Good Integers?

The brute force approach for Find the Count of Good Integers is Brute Force, with O(n²) time complexity and O(1) space complexity. We can generate all n-digit numbers and check if they can be rearranged into a k-palindromic integer. This is straightforward but inefficient as it involves checking many combinations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Find the Count of Good Integers?

Find the Count of Good Integers uses the following concepts: Hash Table, Math, Combinatorics, Enumeration. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Find the Count of Good Integers?

Always clarify the constraints and edge cases before diving into coding. Discuss your thought process and approach before writing code to ensure clarity. Test your solution with small examples to validate correctness.

What are common mistakes when solving Find the Count of Good Integers?

Failing to account for leading zeros when generating numbers. Not checking the frequency of digits correctly to determine if a palindrome can be formed.

#3272

Find the Count of Good Integers

Hard

Hash TableMathCombinatoricsEnumerationHash MapArray

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)

Instead of checking each number, we can directly calculate how many digit combinations can form a k-palindromic integer based on their frequency. This avoids unnecessary checks.

⚙️

Algorithm

3 steps

1Step 1: Generate all possible digit frequencies for n digits ensuring the first digit is non-zero.
2Step 2: For each frequency, check if it can form a palindrome (at most one digit can have an odd frequency).
3Step 3: Count how many of these frequencies are divisible by k.

solution.py17 lines

1from itertools import product
2
3def countGoodIntegers(n, k):
4    count = 0
5    for digits in product(range(10), repeat=n):
6        if digits[0] == 0:
7            continue
8        freq = [0] * 10
9        for d in digits:
10            freq[d] += 1
11        if canFormPalindrome(freq) and sum(d * f for d, f in enumerate(freq)) % k == 0:
12            count += 1
13    return count
14
15def canFormPalindrome(freq):
16    odd_count = sum(1 for f in freq if f % 2 != 0)
17    return odd_count <= 1

ℹ

Complexity note: The time complexity is O(n) because we efficiently calculate the digit frequencies and check conditions without generating all numbers explicitly.

1A number can only be rearranged into a palindrome if at most one digit has an odd frequency.
2Divisibility by k must be checked after confirming the number can form a palindrome.

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