898. Bitwise ORs of Subarrays
Explanation
To solve this problem, we can iterate through the array and keep track of the possible bitwise ORs of all the subarrays ending at each index. We can use a set to store the unique values of bitwise ORs.
- Initialize a set to store unique bitwise OR results.
- Initialize a set to store the bitwise OR results that end at each index.
- Iterate through the array:
- For each element, calculate the bitwise OR with each value in the set of bitwise ORs ending at the previous index and add the result to the set of bitwise ORs ending at the current index.
- Add the element itself to the set of bitwise ORs ending at the current index.
- Add all values in the set of bitwise ORs ending at the current index to the main set of unique bitwise ORs.
- Return the size of the set of unique bitwise ORs.
Time complexity: O(n * log(max(arr))) Space complexity: O(n)
class Solution {
public int subarrayBitwiseORs(int[] arr) {
Set<Integer> uniqueORs = new HashSet<>();
Set<Integer> curORs = new HashSet<>();
for (int num : arr) {
Set<Integer> newCurORs = new HashSet<>();
for (int prev : curORs) {
newCurORs.add(num | prev);
}
newCurORs.add(num);
curORs = newCurORs;
uniqueORs.addAll(curORs);
}
return uniqueORs.size();
}
}Code Editor (Testing phase)
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