LeetCode 2044: Count Number of Maximum Bitwise-OR Subsets Solution

Master LeetCode problem 2044 (Count Number of Maximum Bitwise-OR 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.

2044. Count Number of Maximum Bitwise-OR Subsets

Medium

Array Backtracking Bit Manipulation Enumeration

Problem Explanation

Explanation

To solve this problem, we can iterate through all possible subsets of the given array nums. For each subset, we calculate its bitwise OR value and keep track of the maximum bitwise OR value seen so far. Additionally, we count the number of subsets that have this maximum bitwise OR value.

We can use bitwise operations to efficiently calculate the bitwise OR value of a subset. We start by initializing the maximum OR value and the count of subsets with this value to 0. Then, we iterate through all possible subsets using bit manipulation techniques.

Algorithm:

Initialize maxOr to 0 and maxOrCount to 0.
Iterate i from 0 to 2^n - 1, where n is the length of nums.
For each i, iterate through the elements of nums and calculate the bitwise OR value of the subset based on the bits of i.
If the calculated bitwise OR value is greater than maxOr, update maxOr and reset maxOrCount to 1.
If the calculated bitwise OR value is equal to maxOr, increment maxOrCount.
Return maxOrCount.

Time Complexity: O(2^n * n), where n is the length of nums. Space Complexity: O(1)

Solution Code

class Solution {
    public int countMaxOrSubsets(int[] nums) {
        int maxOr = 0, maxOrCount = 0;
        for (int i = 0; i < (1 << nums.length); i++) {
            int currentOr = 0;
            for (int j = 0; j < nums.length; j++) {
                if ((i & (1 << j)) > 0) {
                    currentOr |= nums[j];
                }
            }
            if (currentOr > maxOr) {
                maxOr = currentOr;
                maxOrCount = 1;
            } else if (currentOr == maxOr) {
                maxOrCount++;
            }
        }
        return maxOrCount;
    }
}

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