Leetcode 2044: Count Number of Maximum Bitwise-OR Subsets
Given an integer array
nums, find the maximum bitwise OR that can be achieved by any subset of nums and count how many different non-empty subsets yield this maximum value. A subset is defined as any combination of elements from the array, and subsets are considered different if their elements are selected from different indices.Problem
Approach
Steps
Complexity
Input: The input is an integer array `nums` representing the elements from which subsets can be formed. The goal is to determine the maximum bitwise OR and count subsets achieving it.
Example: Input: nums = [4, 2, 6]
Constraints:
• 1 <= nums.length <= 16
• 1 <= nums[i] <= 10^5
Output: The output is an integer representing the number of subsets that achieve the maximum bitwise OR value.
Example: Output: 3
Constraints:
• The output is always a non-negative integer.
Goal: Determine the maximum bitwise OR value that can be obtained from any subset of `nums`, and count the subsets achieving this maximum.
Steps:
• Iterate through all subsets of `nums` using a bitmask.
• Calculate the bitwise OR for each subset.
• Track the maximum bitwise OR value and count the subsets that achieve this value.
Goal: The solution must handle up to 2^16 subsets efficiently while calculating the OR values and counting subsets.
Steps:
• The number of subsets is limited by 2^n, where n is the length of the array.
• Bitwise OR operations must handle integers up to 10^5.
Assumptions:
• The input array is non-empty.
• All elements in the array are positive integers.
• Input: Input: nums = [4, 1, 7]
• Explanation: The maximum possible bitwise OR is 7. There are 4 subsets achieving this: [7], [4, 7], [1, 7], and [4, 1, 7]. Thus, the output is 4.
• Input: Input: nums = [1, 1, 1]
• Explanation: The maximum possible bitwise OR is 1. All subsets are non-empty and achieve this value. Total subsets: 2^3 - 1 = 7. Output is 7.
Approach: Use dynamic programming to count subsets with a specific OR value while iterating through the array. This ensures efficient handling of overlapping subsets with similar OR results.
Observations:
• The maximum bitwise OR value is achieved by including specific elements in the subset.
• The problem can be solved efficiently by tracking the OR values dynamically instead of recalculating for each subset.
• Iterate through the elements of `nums`, updating the count of subsets for each OR value dynamically.
Steps:
• Initialize an array `dp` to store the count of subsets for each OR value.
• Iterate through `nums`, updating `dp` for all OR values it can combine with.
• Track the maximum OR value and count subsets achieving it.
Empty Inputs:
• No empty input cases exist since nums is guaranteed to be non-empty.
Large Inputs:
• Handle cases where nums contains 16 elements, leading to 2^16 subsets.
Special Values:
• All elements in nums are the same, e.g., nums = [7, 7, 7].
• The array nums contains powers of 2, e.g., nums = [1, 2, 4, 8].
Constraints:
• Maximum bitwise OR value must be calculated for all combinations efficiently.
int countMaxOrSubsets(vector<int>& nums) {
int mx = 0;
int dp[1 << 17] = {1};
dp[0] = 1;
for(int a : nums) {
for(int i = mx; i >= 0; i--) {
dp[i | a] += dp[i];
}
mx |= a;
}
return dp[mx];
}
1 : Function Declaration
int countMaxOrSubsets(vector<int>& nums) {
This is the function header that defines the `countMaxOrSubsets` function, which takes a vector of integers `nums` as input.
2 : Variable Initialization
int mx = 0;
The variable `mx` is initialized to 0. It will keep track of the maximum bitwise OR of all elements encountered so far.
3 : Array Initialization
int dp[1 << 17] = {1};
The array `dp` is initialized with size `1 << 17` (a large size to cover all possible OR results) and the first element set to 1.
4 : Array Initialization
dp[0] = 1;
Sets the base case for the dynamic programming array, indicating that there is one way to form a subset with a bitwise OR of 0.
5 : Loop Start
for(int a : nums) {
This loop iterates through each element `a` in the input vector `nums`.
6 : Nested Loop Start
for(int i = mx; i >= 0; i--) {
This nested loop iterates backwards from the current maximum OR value `mx` to 0, ensuring all subsets are considered.
7 : Dynamic Programming Update
dp[i | a] += dp[i];
This updates the dynamic programming array by adding the number of subsets that result in the OR value `i | a` to the current subset count.
8 : Nested Loop End
}
End of the inner loop where all possible OR values with the current element `a` are processed.
9 : Update Maximum OR
mx |= a;
Updates `mx` to reflect the maximum OR value encountered so far, combining it with the current element `a`.
10 : Return Statement
return dp[mx];
Returns the value in `dp[mx]`, which represents the number of subsets with the maximum OR value.
Best Case: O(n * max_or_values) when elements overlap minimally.
Average Case: O(n * max_or_values).
Worst Case: O(n * max_or_values) for highly overlapping subsets.
Description: The time complexity depends on the number of OR values dynamically tracked during the iterations.
Best Case: O(max_or_values).
Worst Case: O(max_or_values) for storing DP results.
Description: Space is used for the DP array storing counts for each OR value.
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