花花酱 LeetCode 1707. Maximum XOR With an Element From Array

zxi — Sun, 27 Dec 2020 21:54:20 +0000

You are given an array nums consisting of non-negative integers. You are also given a queries array, where queries[i] = [x_i, m_i].

The answer to the i^th query is the maximum bitwise XOR value of x_i and any element of nums that does not exceed m_i. In other words, the answer is max(nums[j] XOR x_i) for all j such that nums[j] <= m_i. If all elements in nums are larger than m_i, then the answer is -1.

Return an integer array answer where answer.length == queries.length and answer[i] is the answer to the i^th query.

Example 1:

Input: nums = [0,1,2,3,4], queries = [[3,1],[1,3],[5,6]]
Output: [3,3,7]
Explanation:
1) 0 and 1 are the only two integers not greater than 1. 0 XOR 3 = 3 and 1 XOR 3 = 2. The larger of the two is 3.
2) 1 XOR 2 = 3.
3) 5 XOR 2 = 7.

Example 2:

Input: nums = [5,2,4,6,6,3], queries = [[12,4],[8,1],[6,3]]
Output: [15,-1,5]

Constraints:

1 <= nums.length, queries.length <= 10⁵
queries[i].length == 2
0 <= nums[j], x_i, m_i <= 10⁹

Solution: Trie on the fly

We can build the trie on the fly by sorting nums in ascending order and queries by its limit, insert nums into the trie up the limit.

Time complexity: O(nlogn + QlogQ)
Space complexity: O(n)

C++

// Author: Huahua
class Trie {
public:
  Trie(): children{nullptr, nullptr} {}  
  Trie* children[2];
};
class Solution {
public:
  vector maximizeXor(vector& nums, vector>& queries) {
    const int n = nums.size();
    sort(begin(nums), end(nums));    
    
    const int Q = queries.size();
    for (int i = 0; i < Q; ++i)
      queries[i].push_back(i);
    sort(begin(queries), end(queries), [](const auto& q1, const auto& q2) {
      return q1[1] < q2[1];
    });
    
    Trie root;    
    auto insert = [&](int num) {
      Trie* node = &root; 
      for (int i = 31; i >= 0; --i) {
        int bit = (num >> i) & 1;
        if (!node->children[bit])
          node->children[bit] = new Trie();
        node = node->children[bit];
      }
    };
        
    auto query = [&](int num) {
      Trie* node = &root;
      int sum = 0;
      for (int i = 31; i >= 0; --i) {
        if (!node) return -1;
        int bit = (num >> i) & 1;
        if (node->children[1 - bit]) {
          sum |= (1 << i);
          node = node->children[1 - bit];
        } else {
          node = node->children[bit];
        }
      }
      return sum;
    };
    
    vector ans(Q);
    int i = 0;
    for (const auto& q : queries) {
      while (i < n && nums[i] <= q[1]) insert(nums[i++]);
      ans[q[2]] = query(q[0]);
    }  
    return ans;
  }
};

The post 花花酱 LeetCode 1707. Maximum XOR With an Element From Array appeared first on Huahua's Tech Road.

花花酱 LeetCode 421. Maximum XOR of Two Numbers in an Array

zxi — Sun, 27 Dec 2020 21:31:59 +0000

Given an integer array nums, return the maximum result of nums[i] XOR nums[j], where 0 ≤ i ≤ j < n.

Follow up: Could you do this in O(n) runtime?

Example 1:

Input: nums = [3,10,5,25,2,8]
Output: 28
Explanation: The maximum result is 5 XOR 25 = 28.

Example 2:

Input: nums = [0]
Output: 0

Example 3:

Input: nums = [2,4]
Output: 6

Example 4:

Input: nums = [8,10,2]
Output: 10

Example 5:

Input: nums = [14,70,53,83,49,91,36,80,92,51,66,70]
Output: 127

Constraints:

1 <= nums.length <= 2 * 10⁴
0 <= nums[i] <= 2³¹ - 1

Solution: Trie

Time complexity: O(31*2*n)
Space complexity: O(31*2*n)

C++

// Author: Huahua
class Trie {
public:
  Trie(): children(2) {}  
  vector> children;
};
class Solution {
public:
  int findMaximumXOR(vector& nums) {
    Trie root;
    
    // Insert a number into the trie.
    auto insert = [&](Trie* node, int num) {
      for (int i = 31; i >= 0; --i) {
        int bit = (num >> i) & 1;
        if (!node->children[bit])
          node->children[bit] = std::make_unique();
        node = node->children[bit].get();
      }
    };
  
    // Find max xor sum of num and another element in the array.
    auto query = [&](Trie* node, int num) {
      int sum = 0;
      for (int i = 31; i >= 0; --i) {
        int bit = (num >> i) & 1;
        if (node->children[1 - bit]) {
          sum += (1 << i);
          node = node->children[1 - bit].get();
        } else {
          node = node->children[bit].get();
        }
      }
      return sum;
    };
    
    // Insert all numbers.
    for (int x : nums) 
      insert(&root, x);
    
    int ans = 0;
    // For each number find the maximum xor sum.
    for (int x : nums) 
      ans = max(ans, query(&root, x));
    return ans;
  }
};

The post 花花酱 LeetCode 421. Maximum XOR of Two Numbers in an Array appeared first on Huahua's Tech Road.

Trie Archives - Huahua's Tech Road

花花酱 LeetCode 1707. Maximum XOR With an Element From Array

Solution: Trie on the fly

C++

花花酱 LeetCode 421. Maximum XOR of Two Numbers in an Array

Solution: Trie

C++