From the most significant bit to the least significant bit, maintain the bits that will not be included in the final answer in a variable <code>mask</code>.

For a fixed bit, add it to <code>mask</code> then check if there exists some sequence of <code>k</code> operations such that <code>mask & answer == 0 </code> where <code>answer</code> is the bitwise-or of the remaining elements of <code>nums</code>. If there is no such sequence of operations, remove the current bit from <code>mask</code>. How can we perform this check?

Let <code>x</code> be the bitwise-and of all elements of <code>nums</code>. If <code>x AND mask != 0</code>, there is no sequence of operations that satisfies the condition in the previous hint. This is because even if we perform this operation <code>n - 1</code> times on the array, we will end up with <code>x</code> as the final element.

Otherwise, there exists at least one such sequence. It is sufficient to check if the number of operations in such a sequence is less than <code>k</code>. Let’s calculate the minimum number of operations in such a sequence.

Iterate over the array from left to right, if <code>nums[i] & mask != 0</code>, apply the operation on index <code>i</code>.

After iterating over all elements, let <code>x</code> be the bitwise-and of all elements of <code>nums</code>. If <code>x == 0</code>, then we have found the minimum number of operations. Otherwise, It can be proven that we need exactly one more operation so that <code>x == 0</code>.

The condition in the second hint is satisfied if and only if the minimum number of operations is less than or equal to <code>k</code>.