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Minimum bitwise OR after removing at most K elements from given Array

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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++

  

#include <bits/stdc++.h>

using namespace std;

  

int minimumOR(vector<int> A, int N, int K)

{

    vector<int> SetBitCount(32, 0);

    int OR = 0, FirstSetBit = 0;

  

    

    sort(A.rbegin(), A.rend());

    for (int i = 0; i < N; i++) {

        FirstSetBit = log2(A[i]) + 1;

        SetBitCount[FirstSetBit]++;

    }

  

    

    for (int i = 31, j = 0; i >= 0; i--) {

        if (SetBitCount[i] <= K) {

            K -= SetBitCount[i];

            j += SetBitCount[i];

        }

        else {

            for (int x = j;

                 j < x + SetBitCount[i]; j++)

                OR = OR | A[j];

        }

    }

  

    return OR;

}

  

int main()

{

    int N = 7;

    vector<int> A = { 1, 10, 9, 4, 5, 16, 8 };

    int K = 3;

  

    cout << minimumOR(A, N, K);

    return 0;

}

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

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