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3845. Maximum Subarray XOR with Bounded Range
Description
You are given a non-negative integer array nums and an integer k.
You must select a subarray of nums such that the difference between its maximum and minimum elements is at most k. The value of this subarray is the bitwise XOR of all elements in the subarray.
Return an integer denoting the maximum possible value of the selected subarray.
Example 1:
Input: nums = [5,4,5,6], k = 2
Output: 7
Explanation:
- Select the subarray
[5, 4, 5, 6]. - The difference between its maximum and minimum elements is
6 - 4 = 2 <= k. - The value is
4 XOR 5 XOR 6 = 7.
Example 2:
Input: nums = [5,4,5,6], k = 1
Output: 6
Explanation:
- Select the subarray
[5, 4, 5, 6]. - The difference between its maximum and minimum elements is
6 - 6 = 0 <= k. - The value is 6.
Constraints:
1 <= nums.length <= 4 * 1040 <= nums[i] < 2150 <= k < 215
Solutions
Solution 1
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class Solution { // Trie node for storing prefix XOR values in binary form class TrieNode { TrieNode[] children = new TrieNode[2]; // 0 and 1 branches int count = 0; // number of prefix values passing through this node } TrieNode root = new TrieNode(); // Insert or remove a prefix XOR value from the trie void updateTrie(int value, int delta) { TrieNode current = root; for (int bit = 14; bit >= 0; bit--) { int currentBit = (value >> bit) & 1; if (current.children[currentBit] == null) { current.children[currentBit] = new TrieNode(); } current = current.children[currentBit]; current.count += delta; } } // Find maximum XOR of given value with any value currently in the trie int getMaxXor(int value) { TrieNode current = root; int maxXor = 0; for (int bit = 14; bit >= 0; bit--) { int currentBit = (value >> bit) & 1; int oppositeBit = 1 - currentBit; if (current.children[oppositeBit] != null && current.children[oppositeBit].count > 0) { maxXor |= (1 << bit); current = current.children[oppositeBit]; } else { current = current.children[currentBit]; } } return maxXor; } public int maxXor(int[] nums, int limit) { int length = nums.length; // Prefix XOR array int[] prefixXor = new int[length + 1]; for (int i = 0; i < length; i++) { prefixXor[i + 1] = prefixXor[i] ^ nums[i]; } // Monotonic queues to maintain max and min in sliding window Deque<Integer> maxDeque = new ArrayDeque<>(); Deque<Integer> minDeque = new ArrayDeque<>(); int left = 0; int result = 0; updateTrie(prefixXor[0], 1); for (int right = 0; right < length; right++) { // Maintain decreasing deque for maximum while (!maxDeque.isEmpty() && nums[maxDeque.peekLast()] <= nums[right]) { maxDeque.pollLast(); } // Maintain increasing deque for minimum while (!minDeque.isEmpty() && nums[minDeque.peekLast()] >= nums[right]) { minDeque.pollLast(); } maxDeque.addLast(right); minDeque.addLast(right); // Shrink window if max - min exceeds limit while (nums[maxDeque.peekFirst()] - nums[minDeque.peekFirst()] > limit) { if (maxDeque.peekFirst() == left) { maxDeque.pollFirst(); } if (minDeque.peekFirst() == left) { minDeque.pollFirst(); } updateTrie(prefixXor[left], -1); left++; } result = Math.max(result, getMaxXor(prefixXor[right + 1])); updateTrie(prefixXor[right + 1], 1); } return result; } } -
class Solution { public: int maxXor(vector<int>& nums, int k) { root = new TrieNode(); int n = nums.size(); vector<int> prefix(n + 1); for (int i = 0; i < n; ++i) { prefix[i + 1] = prefix[i] ^ nums[i]; } deque<int> maxq, minq; int left = 0, ans = 0; update(prefix[0], 1); for (int right = 0; right < n; ++right) { while (!maxq.empty() && nums[maxq.back()] <= nums[right]) { maxq.pop_back(); } while (!minq.empty() && nums[minq.back()] >= nums[right]) { minq.pop_back(); } maxq.push_back(right); minq.push_back(right); while (nums[maxq.front()] - nums[minq.front()] > k) { if (maxq.front() == left) { maxq.pop_front(); } if (minq.front() == left) { minq.pop_front(); } update(prefix[left], -1); ++left; } ans = max(ans, getMaxXor(prefix[right + 1])); update(prefix[right + 1], 1); } return ans; } private: struct TrieNode { TrieNode* children[2]{}; int count = 0; }; TrieNode* root; void update(int value, int delta) { TrieNode* cur = root; for (int bit = 14; bit >= 0; --bit) { int b = (value >> bit) & 1; if (!cur->children[b]) { cur->children[b] = new TrieNode(); } cur = cur->children[b]; cur->count += delta; } } int getMaxXor(int value) { TrieNode* cur = root; int ans = 0; for (int bit = 14; bit >= 0; --bit) { int b = (value >> bit) & 1; int opp = 1 - b; if (cur->children[opp] && cur->children[opp]->count > 0) { ans |= 1 << bit; cur = cur->children[opp]; } else { cur = cur->children[b]; } } return ans; } }; -
class TrieNode: __slots__ = ("children", "count") def __init__(self): self.children = [None, None] self.count = 0 class Solution: def maxXor(self, nums: List[int], k: int) -> int: root = TrieNode() def update(value: int, delta: int) -> None: cur = root for bit in range(14, -1, -1): b = (value >> bit) & 1 if cur.children[b] is None: cur.children[b] = TrieNode() cur = cur.children[b] cur.count += delta def get_max_xor(value: int) -> int: cur = root ans = 0 for bit in range(14, -1, -1): b = (value >> bit) & 1 opp = 1 - b if cur.children[opp] is not None and cur.children[opp].count > 0: ans |= 1 << bit cur = cur.children[opp] else: cur = cur.children[b] return ans n = len(nums) prefix = [0] * (n + 1) for i, x in enumerate(nums): prefix[i + 1] = prefix[i] ^ x maxq, minq = deque(), deque() left = 0 ans = 0 update(prefix[0], 1) for right, x in enumerate(nums): while maxq and nums[maxq[-1]] <= x: maxq.pop() while minq and nums[minq[-1]] >= x: minq.pop() maxq.append(right) minq.append(right) while nums[maxq[0]] - nums[minq[0]] > k: if maxq[0] == left: maxq.popleft() if minq[0] == left: minq.popleft() update(prefix[left], -1) left += 1 ans = max(ans, get_max_xor(prefix[right + 1])) update(prefix[right + 1], 1) return ans -
type xorTrieNode struct { children [2]*xorTrieNode count int } func updateTrie(root *xorTrieNode, value, delta int) { cur := root for bit := 14; bit >= 0; bit-- { b := (value >> bit) & 1 if cur.children[b] == nil { cur.children[b] = &xorTrieNode{} } cur = cur.children[b] cur.count += delta } } func getMaxXor(root *xorTrieNode, value int) int { cur := root ans := 0 for bit := 14; bit >= 0; bit-- { b := (value >> bit) & 1 opp := 1 - b if cur.children[opp] != nil && cur.children[opp].count > 0 { ans |= 1 << bit cur = cur.children[opp] } else { cur = cur.children[b] } } return ans } func maxXor(nums []int, k int) int { n := len(nums) prefix := make([]int, n+1) for i, x := range nums { prefix[i+1] = prefix[i] ^ x } root := &xorTrieNode{} maxq, minq := []int{}, []int{} left, ans := 0, 0 updateTrie(root, prefix[0], 1) for right, x := range nums { for len(maxq) > 0 && nums[maxq[len(maxq)-1]] <= x { maxq = maxq[:len(maxq)-1] } for len(minq) > 0 && nums[minq[len(minq)-1]] >= x { minq = minq[:len(minq)-1] } maxq = append(maxq, right) minq = append(minq, right) for nums[maxq[0]]-nums[minq[0]] > k { if maxq[0] == left { maxq = maxq[1:] } if minq[0] == left { minq = minq[1:] } updateTrie(root, prefix[left], -1) left++ } ans = max(ans, getMaxXor(root, prefix[right+1])) updateTrie(root, prefix[right+1], 1) } return ans }