Hard
Find Subarray With Bitwise OR Closest to K — C++
Full explanation · Time O(nlogr) · Space O(logr)
// Time: O(nlogr), r = max(nums)
// Space: O(logr)
// freq table, two pointers, sliding window, lc1521
class Solution {
public:
int minimumDifference(vector<int>& nums, int k) {
BitCount count(bit_length(ranges::max(nums)));
int result = numeric_limits<int>::max();
for (int left = 0, right = 0; right < size(nums); ++right) {
count += nums[right];
while (left <= right) {
const auto& f = count.bitOr();
result = min(result, abs(f - k));
if (f <= k) {
break;
}
count -= nums[left++];
}
}
return result;
}
private:
int bit_length(int x) {
return (x ? std::__lg(x) : -1) + 1;
}
class BitCount {
public:
BitCount(int n)
: l_(0)
, n_(n)
, count_(n) {
}
int bitOr() const {
int num = 0;
for (int i = 0; i < n_; ++i) {
if (count_[i]) {
num |= 1 << i;
}
}
return num;
}
void operator+=(int num) {
++l_;
for (int i = 0; i < n_; ++i) {
if (num & (1 << i)) {
++count_[i];
}
}
}
void operator-=(int num) {
--l_;
for (int i = 0; i < n_; ++i) {
if (num & (1 << i)) {
--count_[i];
}
}
}
private:
int l_;
int n_;
vector<int> count_;
};
};
// Time: O(nlogr), r = max(nums)
// Space: O(logr)
// freq table, two pointers, lc1521
class Solution2 {
public:
int minimumDifference(vector<int>& nums, int k) {
int result = numeric_limits<int>::max();
unordered_set<int> dp; // at most O(logr) dp states
for (const auto& x : nums) {
unordered_set<int> new_dp = {x};
for (const auto& f: dp) {
new_dp.emplace(f | x);
}
for (const auto& f : new_dp) {
result = min(result, abs(f - k));
}
dp = move(new_dp);
}
return result;
}
};