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package g0801_0900.s0898_bitwise_ors_of_subarrays;

// #Medium #Array #Dynamic_Programming #Bit_Manipulation
// #2022_03_28_Time_151_ms_(97.74%)_Space_71.9_MB_(85.31%)

import java.util.HashSet;
import java.util.Set;

/**
 * 898 - Bitwise ORs of Subarrays\.
 *
 * Medium
 *
 * We have an array `arr` of non-negative integers.
 *
 * For every (contiguous) subarray `sub = [arr[i], arr[i + 1], ..., arr[j]]` (with `i <= j`), we take the bitwise OR of all the elements in `sub`, obtaining a result `arr[i] | arr[i + 1] | ... | arr[j]`.
 *
 * Return the number of possible results. Results that occur more than once are only counted once in the final answer
 *
 * **Example 1:**
 *
 * **Input:** arr = [0]
 *
 * **Output:** 1
 *
 * **Explanation:** There is only one possible result: 0. 
 *
 * **Example 2:**
 *
 * **Input:** arr = [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:** arr = [1,2,4]
 *
 * **Output:** 6
 *
 * **Explanation:** The possible results are 1, 2, 3, 4, 6, and 7. 
 *
 * **Constraints:**
 *
 * *   1 <= nums.length <= 5 * 10⁴
 * *   0 <= nums[i] <= 10⁹
**/
public class Solution {
    public int subarrayBitwiseORs(int[] arr) {
        Set set = new HashSet<>();
        for (int i = 0; i < arr.length; i++) {
            set.add(arr[i]);
            for (int j = i - 1; j >= 0; j--) {
                if ((arr[i] | arr[j]) == arr[j]) {
                    break;
                }
                arr[j] |= arr[i];
                set.add(arr[j]);
            }
        }
        return set.size();
    }
}