2044. Count Number of Maximum Bitwise-OR Subsets

Description

Given an integer array nums, find the maximum possible bitwise OR of a subset of nums and return the number of different non-empty subsets with the maximum bitwise OR.

An array a is a subset of an array b if a can be obtained from b by deleting some (possibly zero) elements of b. Two subsets are considered different if the indices of the elements chosen are different.

The bitwise OR of an array a is equal to a[0] OR a[1] OR ... OR a[a.length - 1] (0-indexed).

 

Example 1:

Input: nums = [3,1]
Output: 2
Explanation: The maximum possible bitwise OR of a subset is 3. There are 2 subsets with a bitwise OR of 3:
- [3]
- [3,1]

Example 2:

Input: nums = [2,2,2]
Output: 7
Explanation: All non-empty subsets of [2,2,2] have a bitwise OR of 2. There are 23 - 1 = 7 total subsets.

Example 3:

Input: nums = [3,2,1,5]
Output: 6
Explanation: The maximum possible bitwise OR of a subset is 7. There are 6 subsets with a bitwise OR of 7:
- [3,5]
- [3,1,5]
- [3,2,5]
- [3,2,1,5]
- [2,5]
- [2,1,5]

 

Constraints:

• 1 <= nums.length <= 16
• 1 <= nums[i] <= 105

 

Solutions

Solution: Backtracking

• Time complexity: O(n²)
• Space complexity: O(n²)

 

JavaScript

/**
 * @param {number[]} nums
 * @return {number}
 */
const countMaxOrSubsets = function (nums) {
  const n = nums.length;
  const target = nums.reduce((result, num) => result | num);
  let result = 0;

  const findTargetSubset = (index, xor) => {
    if (index >= n) {
      result += xor === target ? 1 : 0;
      return;
    }

    findTargetSubset(index + 1, xor);
    findTargetSubset(index + 1, xor | nums[index]);
  };

  findTargetSubset(0, 0);

  return result;
};