Given an array A[] of length N, the task is to find the minimum possible value of bitwise OR of the elements after removing at most K elements.
Examples:
Input: A = {1, 10, 9, 4, 5, 16, 8}, K = 3
Output: 11
Explanation: Remove 4, 5, 16 for the given array.
The remaining elements are {1, 10, 9, 5}.
The OR of these elements are 11 which is the minimum possible.Input: A = {1, 2, 4, 8}, K = 1
Output: 7
Explanation: Remove 8 form the Array, Minimum OR= 1  2  4 = 7
Approach: This problem can be solved using bit manipulation on the basis of the following idea:
Try to remove the highest MSB elements first, then the elements having 2nd highest MSB and so on until K elements are removed.
Follow the illustration given below for a better understanding.
Illustration :
Consider A[] = [1, 10, 9, 4, 5, 16, 8 ]
 Binary representation of the elements are:
1 – 00001
10 – 01010
9 – 01001
4 – 00100
5 – 00101
16 – 10000
8 – 01000 The positions of the MSB of every elements are:
1 > 0, 10 > 3, 9 > 3, 4 > 2, 5 > 2, 16 > 4, 8 > 3 Therefore count of elements having MSB at ith position are as given below:
setBitCount = [1, 0, 2, 3, 1 ]
MSB position 0, 1, 2, 3, 4 Traverse array and check if current setBitCount is less than k, then remove all elements with current Bit as FirstSetBit.
For, K = 3, traverse array:
=>When i = 4: setBitCount[i] = 1, which is less than K so remove 16, and now K = 2.
=>When i = 3: K < setBitCount[i] (i.e 3). So don’t remove it.
=>When i = 2: setBitCount[i] = 2 ≤ K, so remove 4 and 5. Now, K =0. Calculate OR for all remaining elements i.e (1, 5, 9, 10) = 11
Follow the steps mentioned below to implement the above observation:
 Create a hash array.
 Traverse the given array and increment the MSB Index of every element into the hash array.
 Now, traverse the hash array from MSB and in each iteration:
 Check if it is possible to remove all elements having the ith bit as MSB i.e. setBitCount ≤ K.
 If setBitCount ≤ K, don’t calculate OR for that elements and, just decrease the K.
 if not, calculate the OR for elements with ith bit as MSB.
 Return Calculate OR after removing K elements.
Below is the implementation of the above approach:
C++

Time Complexity: O(N* log(N))
Auxiliary Space: O(N)