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Problem solution · C++

Find Subarray with Bitwise or Closest to K

Find Subarray with Bitwise or Closest to K: a C++ solution using binary search. Learn the idea, check the complexity, and read the full code, with credit to Kamyu LeetCode Solutions.

Technique
Binary search
Source
Kamyu LeetCode Solutions
Length
94 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Binary search

For Find Subarray with Bitwise or Closest to K, the implementation exploits a monotonic condition to discard half of the remaining search space after every check.

  1. Identify the ordered answer range or sorted search domain.
  2. Write a predicate whose truth changes only once.
  3. Move the appropriate boundary after each midpoint check and return the final feasible position.

Code notes

  • 94 lines of C++ from the credited upstream file find-subarray-with-bitwise-or-closest-to-k.cpp.
  • The implementation visibly relies on sequence storage, hash lookup.
  • 8 loop blocks detected.

Complexity

Multiply the logarithmic number of midpoint checks by the cost of one predicate evaluation.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from Kamyu LeetCode Solutions by kamyu104 and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeFind Subarray with Bitwise or Closest to K · C++C++
Use this to learn the idea, then write your own version.
// Time:  O(nlogr), r = max(nums)// Space: O(logr) // freq table, two pointers, sliding window, lc1521class 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, lc1521class 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;    }};

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