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Problem solution · C++

Maximum Subarray Xor with Bounded Range

Maximum Subarray Xor with Bounded Range: a C++ solution using sliding window or two pointers. Learn the idea, check the complexity, and read the full code, with credit to Kamyu LeetCode Solutions.

Technique
Sliding window or two pointers
Source
Kamyu LeetCode Solutions
Length
135 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

Sliding window or two pointers

For Maximum Subarray Xor with Bounded Range, the implementation maintains a moving interval and updates only the information that enters or leaves the window.

  1. Choose the invariant that makes a window valid or useful.
  2. Advance the right boundary and add the new element.
  3. Move the left boundary only as needed while maintaining the invariant and updating the answer.

Code notes

  • 135 lines of C++ from the credited upstream file maximum-subarray-xor-with-bounded-range.cpp.
  • The implementation visibly relies on sequence storage, hash lookup, work queue.
  • 13 loop blocks detected.

Complexity

Confirm that neither pointer moves backwards; if so, the scan is usually linear apart from the window’s data-structure operations.

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 codeMaximum Subarray Xor with Bounded Range · C++C++
Use this to learn the idea, then write your own version.
// Time:  O(nlogr), r = max(max(nums), 1)// Space: O(n) // two pointers, mono deque, bitmasks, prefix sum, hash tableclass Solution {public:    int maxXor(vector<int>& nums, int k) {        vector<int> lookup(size(nums), -1);        deque<int> max_dq, min_dq;        for (int right = 0, left = 0; right < size(nums); ++right) {            while (!empty(max_dq) && nums[max_dq.back()] <= nums[right]) {                max_dq.pop_back();            }            max_dq.emplace_back(right);            while (!empty(min_dq) && nums[min_dq.back()] >= nums[right]) {                min_dq.pop_back();            }            min_dq.emplace_back(right);            while (nums[max_dq[0]] - nums[min_dq[0]] > k) {                if (!empty(max_dq) && max_dq[0] == left) {                    max_dq.pop_front();                }                if (!empty(min_dq) && min_dq[0] == left) {                    min_dq.pop_front();                }                ++left;            }            lookup[right] = left;        }        int result = 0;        const uint32_t mx = max(ranges::max(nums), 1);        for (int i = bit_width(mx) - 1; i >= 0; --i) {            unordered_map<int, int> lookup2;            lookup2[0] = 0;            for (int right = 0, prefix = 0; right < size(nums); ++right) {                prefix ^= nums[right] >> i;                if (lookup2.count(((result >> i) | 1) ^ prefix) && lookup2[((result >> i) | 1) ^ prefix] >= lookup[right]) {                    result |= 1 << i;                    break;                }                lookup2[prefix] = right + 1;            }        }        return result;    }}; // Time:  O(nlogr), r = max(max(nums), 1)// Space: O(n + t)// two pointers, mono deque, bitmasks, prefix sum, trieclass Solution2 {private:    class Trie {    public:        Trie(int bit_length)         : bit_length_(bit_length)         , nodes_() {             new_node();         }         void add(int num, int diff) {            int curr = 0;            for (int i = bit_length_ - 1; i >= 0; --i) {                const auto& x = (num >> i) & 1;                if (nodes_[curr][x] == -1) {                    nodes_[curr][x] = new_node();                }                curr = nodes_[curr][x];                cnts_[curr] += diff;            }        }         int query(int prefix) {            int result = 0, curr = 0;            for (int i = bit_length_ - 1; i >= 0 && curr != -1; --i) {                const auto& x = (prefix >> i) & 1;                if (nodes_[curr][x ^ 1] != -1 && cnts_[nodes_[curr][x ^ 1]]) {                    result |= 1 << i;                    curr = nodes_[curr][x ^ 1];                } else {                    curr = nodes_[curr][x];                }            }            return result;        }     private:        int new_node() {            nodes_.push_back(array<int, 2>{-1, -1});            cnts_.emplace_back(0);            return size(nodes_) - 1;        }         const int bit_length_;        vector<array<int, 2>> nodes_;        vector<int> cnts_;    }; public:    int maxXor(vector<int>& nums, int k) {        int result = 0;        vector<int> prefix(size(nums) + 1);        for (int i = 0; i < size(nums); ++i) {            prefix[i + 1] = prefix[i] ^ nums[i];        }        const uint32_t mx = max(ranges::max(nums), 1);        Trie trie(bit_width(mx));        trie.add(prefix[0], +1);        deque<int> max_dq, min_dq;        for (int right = 0, left = 0; right < size(nums); ++right) {            while (!empty(max_dq) && nums[max_dq.back()] <= nums[right]) {                max_dq.pop_back();            }            max_dq.emplace_back(right);            while (!empty(min_dq) && nums[min_dq.back()] >= nums[right]) {                min_dq.pop_back();            }            min_dq.emplace_back(right);            while (nums[max_dq[0]] - nums[min_dq[0]] > k) {                trie.add(prefix[left], -1);                if (!empty(max_dq) && max_dq[0] == left) {                    max_dq.pop_front();                }                if (!empty(min_dq) && min_dq[0] == left) {                    min_dq.pop_front();                }                ++left;            }            result = max(result, trie.query(prefix[right + 1]));            trie.add(prefix[right + 1], +1);        }        return result;    }};

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