LeetCode 3272: Find the Count of Good Integers Solution
Master LeetCode problem 3272 (Find the Count of Good Integers), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.
3272. Find the Count of Good Integers
Problem Explanation
Explanation
To solve this problem, we need to generate all possible permutations of digits for a given number of digits n and check if each permutation is k-palindromic. We can achieve this by recursively generating all permutations and checking if each one is k-palindromic. We should also keep track of leading zeros to ensure that the resulting integers are valid.
Algorithm:
- Initialize a counter to keep track of the count of good integers.
- Generate all permutations of digits of length
n. - For each permutation:
- Check if it is a palindrome.
- Check if it is divisible by
k. - If both conditions are met, increment the counter.
- Return the final count of good integers.
Time Complexity:
The time complexity of this approach is O(n!), where n is the number of digits.
Space Complexity:
The space complexity is O(n) for storing the current permutation.
Solution Code
class Solution {
public int countGoodNumbers(int n, int k) {
long count = 0;
long mod = 1000000007;
long oddDigits = (k % 2 == 0) ? k / 2 : (k + 1) / 2;
long evenDigits = k / 2;
for (int i = 0; i < n; i++) {
if (i % 2 == 0) {
count = count * evenDigits % mod;
} else {
count = count * oddDigits % mod;
}
}
return (int) count;
}
}Try It Yourself
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