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Count Number of Maximum Bitwise-OR Subsets

Tags : leetcode, cpp, medium

Problem Statement - link #

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).

Examples #

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 #

Solutions #


class Solution {
public:
    int countMaxOrSubsets(vector<int>& nums) {
      
      unordered_map<int, int> map; 
        int ans = 0;
      
      for(int i=0; i<pow(2,nums.size()); i++){
        
        int _or = 0;
        
        for(int j=0; j<nums.size(); j++){
          if((i & (1<<j))!=0)
          _or = _or | nums[j];
        }
        
        if(_or >= ans){
          if(map.find(_or)==map.end())
            map[_or] = 1;
          else
            map[_or]++;
          ans = max(ans, _or);          
        }
        
      }
      
      return map[ans];
    }
};