<p>You are given a <strong>0-indexed</strong> integer array <code>nums</code> of length <code>n</code> and an integer <code>k</code>. In an operation, you can choose an element and multiply it by <code>2</code>.</p>

<p>Return <em>the maximum possible value of </em><code>nums[0] | nums[1] | ... | nums[n - 1]</code> <em>that can be obtained after applying the operation on nums at most </em><code>k</code><em> times</em>.</p>

<p>Note that <code>a | b</code> denotes the <strong>bitwise or</strong> between two integers <code>a</code> and <code>b</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [12,9], k = 1
<strong>Output:</strong> 30
<strong>Explanation:</strong> If we apply the operation to index 1, our new array nums will be equal to [12,18]. Thus, we return the bitwise or of 12 and 18, which is 30.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [8,1,2], k = 2
<strong>Output:</strong> 35
<strong>Explanation:</strong> If we apply the operation twice on index 0, we yield a new array of [32,1,2]. Thus, we return 32|1|2 = 35.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
	<li><code>1 &lt;= k &lt;= 15</code></li>
</ul>

class Solution:
    def maximumOr(self, nums: List[int], k: int) -> int:
        # print(98, 2, 1 << 2, 98)
        mx_save = max(nums)
        mx = mx_save
        depth = 0
        while mx:
            mx = mx >> 1
            depth +=1
        candidates = []
        for i in range(len(nums)):
            if nums[i] >= (1 << (depth-1)):
                candidates.append(i)
        top_candidates = []
        if k >= depth:
            mx_index = nums.index(mx_save)
            res = 0
            for i in range(len(nums)):
                if i==mx_index:
                    res |= nums[i] << k
                else:
                    res |= nums[i]
            return res
        # print([(nums[i], nums[i] >> (depth-k)) for i in candidates])
        shortened = [nums[i] >> (depth-k) for i in candidates]
        # print("shortneed", shortened)
        if not shortened:
            mx_index = nums.index(mx_save)
            res = 0
            for i in range(len(nums)):
                if i==mx_index:
                    res |= nums[i] << k
                else:
                    res |= nums[i]
            return res
        short_mx = max(shortened)
        top_candidates = [candidates[i] for i in range(len(candidates)) if shortened[i] == short_mx]
        # print("top_candidates", [nums[i] for i in top_candidates])
        overall_max = 0
        for top_candidate in top_candidates:
            res = 0
            for i in range(len(nums)):
                if i==top_candidate:
                    res |= nums[i] << k
                else:
                    res |= nums[i]
            overall_max = max(overall_max, res)
        return overall_max

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