Size of the smallest subset with maximum Bitwise OR

Difficulty Level : Medium
Last Updated : 06 Oct, 2020

Given an array of positive integers. The task is to find the size of the smallest subset such that the Bitwise OR of that set is Maximum possible.

Examples:

Input : arr[] = {5, 1, 3, 4, 2}
Output : 2
7 is the maximum value possible of OR, 
5|2 = 7 and 5|3 = 7

Input : arr[] = {2, 6, 2, 8, 4, 5}
Output : 3
15 is the maximum value of OR and set
elements are 8, 6, 5

Source: Sprinklr on Campus Internship

Doing bitwise OR of a number with some value does not decrease its value. It either keeps the value same or increases. If we take a closer look at the problem we can notice that the maximum OR value that we can get is by doing bitwise OR of all array elements. But this includes all elements and here want to know the smallest subset. So we do the following.
I) Find bitwise OR of all array elements. This is the OR we are looking for.
II) Now we need to find the smallest subset with this bitwise OR. This problem is similar to the subset-sum problem, We can solve it in two ways :

We recursively generate all subsets and return the smallest size with given OR
We use Dynamic Programming to solve the problem. This solution is going to be very similar to Maximum size subset with given sum

The time complexity of the recursive solution is O(2ⁿ) and time complexity of the Dynamic Programming solution is O(OR * n) where OR is OR of all array elements and n is the size of the input array.

Using Method 1 : (recursively generating all subsets and return the smallest size with given OR)

C++

// CPP Code for above approach
#include <bits/stdc++.h>
using namespace std;
 
// Compute bitwise or of all elements 
// in array of size sz
int OR(int data[], int sz)
{
    int mOR = 0;
      for (int i = 0; i < sz; ++i) {
        mOR |= data[i];
    }
   
      return mOR;
}
 
// Recursively calculating the size of 
// minimum subset with maximum or
void minSubset(int data[], int sz, int pos, int len,
               int subset[], int req, int& ans)
{
     
    // checking the conditions for maxOR in the
    // resultant subset and setting len strictly 
    // above 0
    if (pos == sz && OR(subset, len) == req && 
                                        len > 0) 
    {
      
        ans = min(len, ans);
    }
    else if (pos < sz) 
    {
        
        // Try the current element in the subset.
        subset[len] = data[pos];
        minSubset(data, sz, pos + 1, len + 1, subset,
                                            req, ans);
 
        // Skip the current element.
        minSubset(data, sz, pos + 1, len, subset, 
                                            req, ans);
    }
}
 
// Driver code
int main()
{
    int data[] = { 5, 1, 3, 4, 2 };
    int sz = sizeof(data) / sizeof(0);
    int req = OR(data, sz);
    int ans = sz;
 
    // Creating a temporary subset with 
    // all elements 0
    int subset[sz];
     
    // Function Call
    minSubset(data, sz, 0, 0, subset, req, ans);
    cout << ans;
}
 
// Code contibuted by Adhiraj Singh

Java

// Java Program for above approach
import java.io.*;
import java.util.*;
 
class Solution 
{
   
  // Compute bitwise or of all elements 
  // in array of size sz
  private static int OR(int[] arr) 
  {
    int mOR = 0;
    for (int i = 0; i < arr.length; ++i) 
    {
      mOR |= arr[i];
    }
    return mOR;
  }
   
  // Recursively calculating the size of 
  // minimum subset with maximum or
  private static int maxSubset(int[] arr, int i, 
                int curOr, int curSize, int maxOr) 
  {
       
    // If i is arr.length
    if (i == arr.length) 
    {
       
      // If curOr is equal to maxOr
      if (curOr == maxOr) 
      {
          return curSize;
      } 
       
      // Return arr.length
      else
      {
          return arr.length;
      }
    }
     
    // Try the current element in the subset
    int take = maxSubset(arr, i + 1, curOr | 
                          arr[i], curSize + 1, maxOr);
     
    // Skip the current element
    int notTake = maxSubset(arr, i + 1, curOr, 
                                      curSize, maxOr);
     
     
    // Return minimum of take and notTake
    return Math.min(take, notTake);
  }
   
  // Driver Code
  public static void main(String[] args) 
  {
    int[] data = {5, 1, 3, 4, 2};
     
    int maxOr = OR(data);
     
    // Function Call
    int maxSubsetSize = maxSubset(data, 0, 0, 0, maxOr);
    System.out.println(maxSubsetSize);
  }
}
 
// Code contibuted by Abdelaziz EROUI

Output

Time complexity: O(2ⁿ)

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