<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:
        bigInd = []
        maxlen = 0
        n = len(nums)
        for i in range(n):
            currlen = 0
            currNum = nums[i]
            while currNum > 0:
                currNum = currNum >> 1
                currlen += 1
            maxlen = max(maxlen, currlen)
        for i in range(n):
            if nums[i] >> (maxlen - 1):
                bigInd.append(i)
        
        maxSum = 0
        prefsum = [0] * n
        for i in range(1, n):
            prefsum[i] = prefsum[i-1] | nums[i-1]
        suffsum = [0] * n
        for i in range(n-2, -1, -1):
            suffsum[i] = suffsum[i+1] | nums[i+1]
        maxSum = 0
        for i in range(len(bigInd)):
            currElem = nums[bigInd[i]] << k
            currSum = prefsum[bigInd[i]] | suffsum[bigInd[i]] | currElem
            maxSum = max(maxSum, currSum)


        return maxSum
917