Today I have done Bitwise ORs of Subarrays on leetcode and leetcode's November LeetCoding Challenge with cpp.

Bitwise ORs of Subarrays

Description

We have an array A of non-negative integers.

For every (contiguous) subarray B = [A[i], A[i+1], ..., A[j]] (with i <= j), we take the bitwise OR of all the elements in B, obtaining a result A[i] | A[i+1] | ... | A[j].

Return the number of possible results. (Results that occur more than once are only counted once in the final answer.)

Example 1:

Input: [0]
Output: 1
Explanation: 
There is only one possible result: 0.

Example 2:

Input: [1,1,2]
Output: 3
Explanation: 
The possible subarrays are [1], [1], [2], [1, 1], [1, 2], [1, 1, 2].
These yield the results 1, 1, 2, 1, 3, 3.
There are 3 unique values, so the answer is 3.

Example 3:

Input: [1,2,4]
Output: 6
Explanation: 
The possible results are 1, 2, 3, 4, 6, and 7.

Note:

1. 1 <= A.length <= 50000
2. 0 <= A[i] <= 10^9

Solution

nothing to say

class Solution {
public:
    int subarrayBitwiseORs(vector<int>& A) {
        unordered_set<int> cur[2];
        unordered_set<int> answer;
        cur[0].insert(0);
        for(int i = 0; i < A.size(); ++i) {
            cur[1].clear();
            for(auto num : cur[0]) {
                cur[1].insert(num|A[i]);
            }
            cur[1].insert(A[i]);
            cur[0] = cur[1];
            answer.merge(cur[1]);
        }
        return answer.size();
    }
};

November LeetCoding Challenge 6

Description

Find the Smallest Divisor Given a Threshold

Given an array of integers nums and an integer threshold, we will choose a positive integer divisor and divide all the array by it and sum the result of the division. Find the smallest divisor such that the result mentioned above is less than or equal to threshold.

Each result of division is rounded to the nearest integer greater than or equal to that element. (For example: 7/3 = 3 and 10/2 = 5).

It is guaranteed that there will be an answer.

Example 1:

Input: nums = [1,2,5,9], threshold = 6
Output: 5
Explanation: We can get a sum to 17 (1+2+5+9) if the divisor is 1. 
If the divisor is 4 we can get a sum to 7 (1+1+2+3) and if the divisor is 5 the sum will be 5 (1+1+1+2). 

Example 2:

Input: nums = [2,3,5,7,11], threshold = 11
Output: 3

Example 3:

Input: nums = [19], threshold = 5
Output: 4

Constraints:

• 1 <= nums.length <= 5 * 10^4
• 1 <= nums[i] <= 10^6
• nums.length <= threshold <= 10^6

Solution

parity

class Solution {
  bool satisfied(vector<int> &nums, int threshold, double divisor) {
    int sum = 0;
    for(auto n : nums) {
      sum += ceil(n/divisor);
    }
    return sum <= threshold;
  }
public:
  int smallestDivisor(vector<int> &nums, int threshold) {
    int start = 1, end = INT_MAX-1;
    while(start < end) {
      int mid = (end + start) / 2;
      if(satisfied(nums, threshold, mid)) end = mid;
      else start = mid + 1; 
    }
    return start;
  }
};