Given an array A of non-negative integers, where
Examples:
Input: A = [1, 2] Output: 3 Explanation: The possible subarrays are [1], [2], [1, 2]. These Bitwise OR of subarrays are 1, 2, 3. There are 3 distinct values, so the answer is 3. Input: A = [1, 2, 4] Output: 6 Explanation: The possible distinct values are 1, 2, 3, 4, 6, and 7.
Approach: The Naive approach is to generate all possible subarrays and take bitwise OR of all elements in the subarray. Store each result in set and return length of the set.
Efficient Approach: We can make the above approach better. The Naive approach is to calculate all possible result where, res(i, j) = A[i] | A[i+1] | … | A[j]. However we can speed this up by taking note of the fact that res(i, j+1) = res(i, j) | A[j+1]. At the kth step, say we have all of the res(i, k) in some set pre. Then we can find the next pre set (for k -> k+1) by using res(i, k+1) = res(i, k) | A[k+1].
However, the number of unique values in this set pre is atmost 32, since the list res(k, k), res(k-1, k), res(k-2, k), … is monotone increasing, and any subsequent values that are different from previous must have more 1’s in it’s binary representation which can have maximum of 32 ones.
Below is the implementation of above approach.
#include <iostream> #include <vector> #include <set> using namespace std;
// function to return count of distinct bitwise OR int subarrayBitwiseOR(vector< int >& A) {
// res contains distinct values
set< int > res;
set< int > pre;
pre.insert(0);
for ( int x : A) {
set< int > temp;
for ( int y : pre) {
temp.insert(x | y);
}
temp.insert(x);
pre = temp;
res.insert(pre.begin(), pre.end());
}
return res.size();
} // Driver program int main() {
vector< int > A = {1, 2, 4};
// print required answer
cout << subarrayBitwiseOR(A) << endl;
return 0;
} |
import java.util.*;
class Main {
// function to return count of distinct bitwise OR
public static int subarrayBitwiseOR( int [] A) {
// res contains distinct values
Set<Integer> res = new HashSet<>();
Set<Integer> pre = new HashSet<>();
pre.add( 0 );
for ( int x : A) {
Set<Integer> temp = new HashSet<>();
for ( int y : pre) {
temp.add(x | y);
}
temp.add(x);
pre = temp;
res.addAll(pre);
}
return res.size();
}
// Driver program
public static void main(String[] args) {
int [] A = { 1 , 2 , 4 };
// print required answer
System.out.println(subarrayBitwiseOR(A));
}
} |
# Python implementation of the above approach # function to return count of distinct bitwise OR def subarrayBitwiseOR(A):
# res contains distinct values
res = set ()
pre = { 0 }
for x in A:
pre = {x | y for y in pre} | {x}
res | = pre
return len (res)
# Driver program A = [ 1 , 2 , 4 ]
# print required answer print (subarrayBitwiseOR(A))
# This code is written by # Sanjit_Prasad |
using System;
using System.Collections.Generic;
class MainClass {
// function to return count of distinct bitwise OR
public static int subarrayBitwiseOR( int [] A) {
// res contains distinct values
HashSet< int > res = new HashSet< int >();
HashSet< int > pre = new HashSet< int >();
pre.Add(0);
foreach ( int x in A) {
HashSet< int > temp = new HashSet< int >();
foreach ( int y in pre) {
temp.Add(x | y);
}
temp.Add(x);
pre = temp;
res.UnionWith(pre);
}
return res.Count;
}
// Driver program
public static void Main( string [] args) {
int [] A = {1, 2, 4};
// print required answer
Console.WriteLine(subarrayBitwiseOR(A));
}
} |
// JavaScript implementation of the above approach // function to return count of distinct bitwise OR function subarrayBitwiseOR(A) {
// res contains distinct values
let res = new Set();
let pre = new Set([0]);
for (let x of A) {
let temp = new Set();
for (let y of pre) {
temp.add(x | y);
}
temp.add(x);
res = new Set([...res, ...temp]);
pre = new Set(temp);
}
return res.size;
} // Driver program let A = [1, 2, 4]; // print required answer console.log(subarrayBitwiseOR(A)); |
Output
6
Time Complexity: O(N*log(K)), where N is the length of A, and K is the maximum size of elements in A.
C++ implementation of the above approach.
// C++ implementation of the above approach #include <bits/stdc++.h> using namespace std;
// function to calculate count of // distinct bitwise OR of all // subarrays. int distintBitwiseOR( int arr[], int n)
{ unordered_set< int > ans, prev;
for ( int i = 0; i < n; i++) {
unordered_set< int > ne;
for ( auto x : prev)
ne.insert(arr[i] | x);
ne.insert(arr[i]);
for ( auto x : ne)
ans.insert(x);
prev = ne;
}
return ans.size();
} // Driver Code int main()
{ int n = 3;
int arr[] = { 1, 2, 4 };
cout << distintBitwiseOR(arr, n);
return 0;
} |
// Java implementation of the above approach import java.io.*;
import java.util.*;
class GFG
{ // function to calculate count of
// distinct bitwise OR of all // subarrays. static int distintBitwiseOR( int arr[], int n)
{
HashSet<Integer>ans = new HashSet<>();
HashSet<Integer>prev = new HashSet<>();
for ( int i = 0 ; i < n; i++)
{
HashSet<Integer>ne = new HashSet<>();
ne.add(arr[i]);
for ( int x :prev)
{
ne.add(arr[i]|x);
}
for ( int x :ne)
{
ans.add(x);
}
prev = ne;
}
return ans.size();
}
// Driver code
public static void main (String[] args) {
int n = 3 ;
int arr[] = { 1 , 2 , 4 };
System.out.println(distintBitwiseOR(arr, n));
}
} // This code is contributed by iramkhalid24. |
# Python implementation of the above approach # function to calculate count of # distinct bitwise OR of all # subarrays. def distintBitwiseOR(arr,n):
ans,prev = set (), set ()
for i in range (n):
ne = set ()
for x in prev:
ne.add(arr[i] | x)
ne.add(arr[i])
for x in ne:
ans.add(x)
prev = ne
return len (ans)
# Driver Code n = 3
arr = [ 1 , 2 , 4 ]
print (distintBitwiseOR(arr, n))
# This code is written by Shinjanpatra |
// C# implementation of the above approach using System;
using System.Collections.Generic;
public class GFG
{ // function to calculate count of
// distinct bitwise OR of all
// subarrays.
static int distintBitwiseOR( int [] arr, int n)
{
HashSet< int > ans = new HashSet< int >();
HashSet< int > prev = new HashSet< int >();
for ( int i = 0; i < n; i++) {
HashSet< int > ne = new HashSet< int >();
ne.Add(arr[i]);
foreach ( var x in prev) { ne.Add(arr[i] | x); }
foreach ( var x in ne) { ans.Add(x); }
prev = ne;
}
return ans.Count;
}
// Driver code
public static void Main( string [] args)
{
int n = 3;
int [] arr = { 1, 2, 4 };
// Function call
Console.WriteLine(distintBitwiseOR(arr, n));
}
} // This code is contributed by phasing17 |
<script> // JavaScript implementation of the above approach // function to calculate count of // distinct bitwise OR of all // subarrays. function distintBitwiseOR(arr,n)
{ let ans = new Set(), prev = new Set();
for (let i = 0; i < n; i++) {
let ne = new Set();
for (let x of prev)
ne.add(arr[i] | x);
ne.add(arr[i]);
for (let x of ne)
ans.add(x);
prev = ne;
}
return ans.size;
} // Driver Code let n = 3; let arr = [ 1, 2, 4 ]; document.write(distintBitwiseOR(arr, n)); // This code is written by Shinjanpatra </script> |
Output
6
Time Complexity: O(N*K) where N is the length of A, and K is the maximum size of elements in A.
Auxiliary Space: O(K)