Maximum XOR With Element From Array

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Steps

setup

Sort nums and queries by m_i

We start by sorting the nums array and the queries by their m_i values to process queries in ascending order of constraints.

💡 Sorting queries by m_i allows incremental insertion of nums elements into the trie, avoiding repeated work.

Line:

nums.sort()
queries = sorted([(x, m, i) for i, (x, m) in enumerate(queries)], key=lambda x: x[1])

💡 Sorting prepares the data for efficient incremental processing.

setup

Prepare sorted queries with indices

We create a sorted list of queries by m_i: [(3,1,0), (1,3,1), (5,6,2)] where each tuple is (x, m, original_index).

💡 Tracking original indices lets us store answers in correct order after processing sorted queries.

Line:

queries = sorted([(x, m, i) for i, (x, m) in enumerate(queries)], key=lambda x: x[1])

💡 Sorting queries by m_i and keeping original indices is essential for offline query processing.

insert

Process first query (3,1,0): Insert nums ≤ 1 into trie

For the first query with m=1, we insert nums elements ≤1 into the trie: nums[0]=0 and nums[1]=1.

💡 Inserting only eligible nums elements keeps the trie minimal and relevant for the current query.

Line:

while idx < n and nums[idx] <= m:
    trie.insert(nums[idx])
    idx += 1

💡 Incremental insertion avoids reprocessing nums for each query.

compare

Query trie for max XOR with x=3

We query the trie to find the maximum XOR of 3 with inserted nums (0 and 1). The max XOR is 3 (3^0=3).

💡 Querying the trie finds the best candidate efficiently without checking all nums explicitly.

Line:

res[i] = trie.maxXor(x)

💡 Trie structure enables bitwise greedy search for max XOR.

insert

Process second query (1,3,1): Insert nums ≤ 3 into trie

For the second query with m=3, we insert nums elements ≤3 that are not yet inserted: nums[2]=2 and nums[3]=3.

💡 We continue incremental insertion from where we left off to avoid redundant work.

Line:

while idx < n and nums[idx] <= m:
    trie.insert(nums[idx])
    idx += 1

💡 Incremental insertion ensures trie always contains all nums ≤ current m.

compare

Query trie for max XOR with x=1

We query the trie to find the maximum XOR of 1 with inserted nums (0,1,2,3). The max XOR is 3 (1^2=3).

💡 The trie quickly finds the best XOR partner without checking all elements.

Line:

res[i] = trie.maxXor(x)

💡 Trie traversal picks bits to maximize XOR at each level.

insert

Process third query (5,6,2): Insert nums ≤ 6 into trie

For the third query with m=6, we insert remaining nums elements ≤6: nums[4]=4.

💡 We insert only the new eligible element to keep trie updated for this query.

Line:

while idx < n and nums[idx] <= m:
    trie.insert(nums[idx])
    idx += 1

💡 Trie now contains all nums elements for the largest m in queries.

compare

Query trie for max XOR with x=5

We query the trie to find the maximum XOR of 5 with all inserted nums (0,1,2,3,4). The max XOR is 7 (5^2=7).

💡 The trie efficiently finds the best XOR partner among all nums.

Line:

res[i] = trie.maxXor(x)

💡 Trie traversal maximizes XOR bit by bit.

reconstruct

All queries processed, results stored in original order

We have computed max XOR results for all queries and stored them in the results array in original query order.

💡 Storing results by original indices ensures output matches input query order.

Line:

return res

💡 Offline query processing allows efficient batch answering.

reconstruct

Final state: Output the results

The final results array [3,3,7] corresponds to the maximum XOR values for each query in input order.

💡 This final step confirms the algorithm's correctness and completes the process.

Line:

print(maximizeXor(nums, queries))

💡 The algorithm efficiently answers all queries using sorting, trie insertion, and querying.

class TrieNode:
    def __init__(self):
        self.children = [None, None]

class Trie:
    def __init__(self):
        self.root = TrieNode()

    def insert(self, num):
        node = self.root
        for i in range(31, -1, -1):  # STEP 3,5,7 insert bits
            bit = (num >> i) & 1
            if not node.children[bit]:
                node.children[bit] = TrieNode()
            node = node.children[bit]

    def maxXor(self, num):
        node = self.root
        if not node.children[0] and not node.children[1]:
            return -1
        max_xor = 0
        for i in range(31, -1, -1):  # STEP 4,6,8 query bits
            bit = (num >> i) & 1
            toggled_bit = 1 - bit
            if node.children[toggled_bit]:
                max_xor |= (1 << i)
                node = node.children[toggled_bit]
            else:
                node = node.children[bit]
        return max_xor

def maximizeXor(nums, queries):
    nums.sort()  # STEP 1
    queries = sorted([(x, m, i) for i, (x, m) in enumerate(queries)], key=lambda x: x[1])  # STEP 2
    trie = Trie()
    res = [-1] * len(queries)
    idx = 0
    n = len(nums)
    for x, m, i in queries:  # STEP 3-8
        while idx < n and nums[idx] <= m:
            trie.insert(nums[idx])  # STEP 3,5,7
            idx += 1
        res[i] = trie.maxXor(x)  # STEP 4,6,8
    return res  # STEP 9

if __name__ == '__main__':
    nums = [0,1,2,3,4]
    queries = [[3,1],[1,3],[5,6]]
    print(maximizeXor(nums, queries))  # STEP 10

📊

Maximum XOR With Element From Array - Watch the Algorithm Execute, Step by Step

Watching the algorithm step through sorting, insertion, and querying reveals how offline queries and tries optimize what would otherwise be a costly brute force approach.

setup

query

0

0

1

1

2

2

3

3

4

4

Result: "[-1, -1, -1]"

setup

query

(3,1,0)

0

(1,3,1)

1

(5,6,2)

2

Result: "[-1, -1, -1]"

insert

query

0

0

1

1

idx

2

2

3

3

4

4

Result: "[-1, -1, -1]"

compare

query

0

0

1

1

idx

2

2

3

3

4

4

Result: "[3, -1, -1]"

insert

0

0

query

1

1

2

2

3

3

idx

4

4

Result: "[3, -1, -1]"

compare

0

0

query

1

1

2

2

3

3

idx

4

4

Result: "[3, 3, -1]"

insert

0

0

1

1

query

2

2

3

3

4

4

Result: "[3, 3, -1]"

compare

0

0

1

1

query

2

2

3

3

4

4

Result: "[3, 3, 7]"

reconstruct

0

0

1

1

2

2

3

3

4

4

Result: "[3, 3, 7]"

output

0

0

1

1

2

2

3

3

4

4

Result: "[3, 3, 7]"

Key Takeaways

✓ Sorting queries by their m_i values enables incremental insertion of nums into the trie, avoiding repeated work.

This insight is hard to see from code alone because the benefit of offline queries and sorting is conceptual and not explicit in a single line.

✓ The bitwise trie allows efficient bit-level traversal to find the maximum XOR partner without brute force comparisons.

Understanding how the trie encodes bits and how toggling bits maximizes XOR is easier to grasp visually than from code.

✓ Incremental insertion of nums elements as m_i increases ensures the trie always contains exactly the elements relevant for the current query.

Seeing pointer 'idx' move forward and insert elements step-by-step clarifies why the algorithm is efficient.

Practice

1. Consider the following Python code snippet implementing the Trie + DFS approach for palindrome pairs. Given the input words = ["bat", "tab", "cat"], what is the final output of palindromePairs(words)?

easy

A. [[0, 1], [1, 0]]

B. [[1, 0], [0, 1]]

C. [[0, 1], [1, 0], [2, 2]]

D. []

Solution

1. Step 1: Trace add_word for "bat" and "tab"

   Words are added reversed: "tab" for "bat", "bat" for "tab". Palindrome suffixes recorded accordingly.

2. Step 2: Trace search for "bat" (index 0)

   Matches "tab" reversed in trie, finds palindrome suffixes leading to pair [0,1]. Similarly for "tab" (index 1) finds [1,0]. "cat" has no pairs.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Output matches expected palindrome pairs [OK]

Hint: Pairs are symmetric for "bat" and "tab" only [OK]

Common Mistakes:

• Confusing order of pairs
• Including self-pairs
• Missing palindrome suffix checks

2. You are given a dictionary of root words and a sentence. For each word in the sentence, you want to replace it with the shortest root word from the dictionary that is a prefix of the word. Which algorithmic approach guarantees the most efficient and correct solution for this problem?

easy

A. Use dynamic programming to find the shortest prefix for each word by building solutions from smaller subproblems.

B. Check each word against all root words in the dictionary by brute force prefix matching.

C. Sort the root words and for each word in the sentence, perform binary search to find the closest prefix root.

D. Build a trie from the root words and for each word in the sentence, search for the shortest prefix in the trie.

Solution

1. Step 1: Understand the problem constraints

   The problem requires replacing words with the shortest root prefix efficiently, which suggests a prefix-based data structure.

2. Step 2: Evaluate approaches

   Brute force and sorting with binary search are less efficient due to repeated prefix checks. Dynamic programming is not suitable as the problem is not about overlapping subproblems but prefix matching. A trie efficiently stores prefixes and allows quick shortest prefix search.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Trie enables O(m*k) prefix search, optimal for this problem [OK]

Hint: Trie is best for prefix-based replacement [OK]

Common Mistakes:

• Thinking DP applies here
• Using brute force without pruning

3. Consider the following code snippet implementing the optimal trie-based solution for shortest unique prefixes. Given the input words = ["dog", "dove", "duck"], what is the output of shortest_unique_prefix(words)?

easy

A. ["do", "dov", "du"]

B. ["d", "dov", "du"]

C. ["dog", "dove", "duck"]

D. ["dog", "dov", "du"]

Solution

1. Step 1: Trace insertion frequencies

   Insert "dog": d(1), o(1), g(1); "dove": d(2), o(2), v(1), e(1); "duck": d(3), u(1), c(1), k(1).

2. Step 2: Find unique prefixes

   For "dog": prefix "dog" (g freq=1) unique; for "dove": prefix "dov" (v freq=1) unique; for "duck": prefix "du" (u freq=1) unique.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Prefixes match first node with freq=1 [OK]

Hint: Unique prefix ends where node freq equals 1 [OK]

Common Mistakes:

• Stopping prefix too early
• Assuming first letter always unique

4. What is the time complexity of the optimal trie-based solution to replace words in a sentence, where m is the number of words in the sentence and k is the average length of each word, assuming the trie is already built from n root words with average length l?

medium

A. O(n * l + m * k) because building the trie takes O(n * l) and searching each word takes O(k)

B. O(m * k) because each word is searched character by character in the trie until a prefix is found or word ends

C. O(m * k * l) because for each word, the trie traversal depends on both word and root lengths

D. O(m * n * l) because each word is compared against all roots in the trie

Solution

1. Step 1: Identify trie construction cost

   Building the trie from n roots each of length l takes O(n * l) time.

2. Step 2: Identify search cost per word

   Searching each of the m words of length k in the trie takes O(k) time per word.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Sum of build and search times matches O(n*l + m*k) [OK]

Hint: Trie build and search costs add up [OK]

Common Mistakes:

• Assuming O(m*n*l) like brute force
• Multiplying k and l unnecessarily

5. Suppose the problem is modified so that root words can be reused multiple times to replace multiple prefixes in a single word (e.g., 'catcat' could be replaced by 'catcat' if 'cat' is a root). Which modification to the trie-based approach correctly handles this variant?

hard

A. Use the existing shortest prefix search once per word; no change needed as reuse is irrelevant.

B. After finding the shortest prefix, continue searching the remaining suffix recursively for more prefixes to replace.

C. Modify search to find the longest prefix instead of shortest, then recursively replace the suffix with roots.

D. Replace the entire word only if it exactly matches a root; otherwise, leave it unchanged.

Solution

1. Step 1: Understand reuse requirement

   Words can be replaced by concatenations of multiple root prefixes, so partial replacements must continue recursively.

2. Step 2: Modify search logic

   After finding a shortest prefix, recursively apply the same prefix search on the remaining suffix to replace all possible prefixes.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Recursive prefix replacement handles multiple root reuse correctly [OK]

Hint: Recursively replace suffixes after shortest prefix [OK]

Common Mistakes:

• Stopping after first prefix
• Searching for longest prefix only