LeetCode 2638: Count the Number of K-Free Subsets Solution
Master LeetCode problem 2638 (Count the Number of K-Free Subsets), a medium 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.
2638. Count the Number of K-Free Subsets
Problem Explanation
Explanation:
To solve this problem, we can use dynamic programming. We can iterate over all possible subsets of the given array and count the number of subsets that are k-free, i.e., they contain no two elements that are divisible by k.
For each subset, we can check if it is k-free by iterating over all pairs of elements in the subset and verifying if their sum is not divisible by k. If the subset is k-free, we increment the count.
To optimize the solution, we can use bitmasking to represent subsets. We can use a bitmask to represent whether an element is included in the subset or not. By iterating over all possible bitmasks, we can generate all subsets efficiently.
At each step, we can check if the sum of elements in the subset is not divisible by k. If it is not divisible, we increment the count.
Time Complexity: The time complexity of this solution is O(2^n) where n is the number of elements in the given array. Since we are iterating over all possible subsets, the time complexity is exponential.
Space Complexity: The space complexity of this solution is O(n) for storing the input array elements and the bitmask for subsets.
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Solution Code
class Solution {
public int countKFreeSubsets(int[] nums, int k) {
int n = nums.length;
int count = 0;
for (int mask = 0; mask < (1 << n); mask++) {
boolean kFree = true;
for (int i = 0; i < n; i++) {
if ((mask & (1 << i)) != 0) {
for (int j = i + 1; j < n; j++) {
if ((mask & (1 << j)) != 0 && (nums[i] + nums[j]) % k == 0) {
kFree = false;
break;
}
}
}
}
if (kFree) {
count++;
}
}
return count;
}
}Try It Yourself
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