left and right to 0, which will represent the window's boundaries.current_and to 0, which will store the cumulative AND of the current window.max_length to 1, which will store the maximum length of a nice subarray found so far.right over the elements of nums.
a. For each element, perform a bitwise AND with current_and.
b. If the result is non-zero, it means there is a conflict, and we need to shrink the window from the left.
c. While there is a conflict, use XOR to remove the leftmost element's bits from current_and and increment left.current_and by performing a bitwise OR with the current element.max_length with the maximum of its current value and the size of the current window (right - left + 1).max_length as the result.