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421. Maximum XOR of Two Numbers in an Array

Description

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

 

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

 

Constraints:

  • 1 <= nums.length <= 2 * 105
  • 0 <= nums[i] <= 231 - 1

Solutions

Solution 1

Thinking

Trying every pair is \(O(n^2)\) and \(n\) can be \(2\times 10^5\). A large XOR wants the opposite bit as early as possible.

Insert each number into a binary trie from the high bit. When querying \(x\), prefer the child with the flipped bit so that bit of the answer becomes \(1\). Thirty-one bits cover the value range.

Each query walks the trie once and is equivalent to choosing the best partner among the inserted numbers.

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class Trie:
    __slots__ = ("children",)

    def __init__(self):
        self.children: List[Trie | None] = [None, None]

    def insert(self, x: int):
        node = self
        for i in range(30, -1, -1):
            v = x >> i & 1
            if node.children[v] is None:
                node.children[v] = Trie()
            node = node.children[v]

    def search(self, x: int) -> int:
        node = self
        ans = 0
        for i in range(30, -1, -1):
            v = x >> i & 1
            if node.children[v ^ 1]:
                ans |= 1 << i
                node = node.children[v ^ 1]
            else:
                node = node.children[v]
        return ans


class Solution:
    def findMaximumXOR(self, nums: List[int]) -> int:
        trie = Trie()
        for x in nums:
            trie.insert(x)
        return max(trie.search(x) for x in nums)

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class Trie {
    private Trie[] children = new Trie[2];

    public Trie() {
    }

    public void insert(int x) {
        Trie node = this;
        for (int i = 30; i >= 0; --i) {
            int v = x >> i & 1;
            if (node.children[v] == null) {
                node.children[v] = new Trie();
            }
            node = node.children[v];
        }
    }

    public int search(int x) {
        Trie node = this;
        int ans = 0;
        for (int i = 30; i >= 0; --i) {
            int v = x >> i & 1;
            if (node.children[v ^ 1] != null) {
                ans |= 1 << i;
                node = node.children[v ^ 1];
            } else {
                node = node.children[v];
            }
        }
        return ans;
    }
}

class Solution {
    public int findMaximumXOR(int[] nums) {
        Trie trie = new Trie();
        int ans = 0;
        for (int x : nums) {
            trie.insert(x);
            ans = Math.max(ans, trie.search(x));
        }
        return ans;
    }
}

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class Trie {
public:
    Trie* children[2];

    Trie()
        : children{nullptr, nullptr} {}

    void insert(int x) {
        Trie* node = this;
        for (int i = 30; ~i; --i) {
            int v = x >> i & 1;
            if (!node->children[v]) {
                node->children[v] = new Trie();
            }
            node = node->children[v];
        }
    }

    int search(int x) {
        Trie* node = this;
        int ans = 0;
        for (int i = 30; ~i; --i) {
            int v = x >> i & 1;
            if (node->children[v ^ 1]) {
                ans |= 1 << i;
                node = node->children[v ^ 1];
            } else {
                node = node->children[v];
            }
        }
        return ans;
    }
};

class Solution {
public:
    int findMaximumXOR(vector<int>& nums) {
        Trie* trie = new Trie();
        int ans = 0;
        for (int x : nums) {
            trie->insert(x);
            ans = max(ans, trie->search(x));
        }
        return ans;
    }
};

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type Trie struct {
    children [2]*Trie
}

func newTrie() *Trie {
    return &Trie{}
}

func (t *Trie) insert(x int) {
    node := t
    for i := 30; i >= 0; i-- {
        v := x >> i & 1
        if node.children[v] == nil {
            node.children[v] = newTrie()
        }
        node = node.children[v]
    }
}

func (t *Trie) search(x int) int {
    node := t
    ans := 0
    for i := 30; i >= 0; i-- {
        v := x >> i & 1
        if node.children[v^1] != nil {
            ans |= 1 << i
            node = node.children[v^1]
        } else {
            node = node.children[v]
        }
    }
    return ans
}

func findMaximumXOR(nums []int) (ans int) {
    trie := newTrie()
    for _, x := range nums {
        trie.insert(x)
        ans = max(ans, trie.search(x))
    }
    return ans
}

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struct Trie {
    children: [Option<Box<Trie>>; 2],
}

impl Trie {
    fn new() -> Trie {
        Trie {
            children: [None, None],
        }
    }

    fn insert(&mut self, x: i32) {
        let mut node = self;
        for i in (0..=30).rev() {
            let v = ((x >> i) & 1) as usize;
            if node.children[v].is_none() {
                node.children[v] = Some(Box::new(Trie::new()));
            }
            node = node.children[v].as_mut().unwrap();
        }
    }

    fn search(&self, x: i32) -> i32 {
        let mut node = self;
        let mut ans = 0;
        for i in (0..=30).rev() {
            let v = ((x >> i) & 1) as usize;
            if let Some(child) = &node.children[v ^ 1] {
                ans |= 1 << i;
                node = child.as_ref();
            } else {
                node = node.children[v].as_ref().unwrap();
            }
        }
        ans
    }
}

impl Solution {
    pub fn find_maximum_xor(nums: Vec<i32>) -> i32 {
        let mut trie = Trie::new();
        let mut ans = 0;
        for &x in nums.iter() {
            trie.insert(x);
            ans = ans.max(trie.search(x));
        }
        ans
    }
}

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