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. You need to transform one string into another using the minimum number of insertions, deletions, and substitutions. Which algorithmic approach guarantees finding the optimal solution efficiently?

easy

A. Greedy algorithm that replaces characters whenever they differ

B. Pure brute force recursion without memoization

C. Dynamic Programming that builds a table of subproblem solutions based on prefixes

D. Divide and conquer by splitting strings into halves and solving independently

Solution

1. Step 1: Understand problem requires minimum edits

   The problem asks for minimum insertions, deletions, and substitutions to convert one string to another, which is a classic DP problem.

2. Step 2: Identify approach guaranteeing optimal substructure

   Dynamic Programming builds solutions for prefixes and combines them, ensuring optimality and polynomial time.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   DP table approach is standard for edit distance [OK]

Hint: Edit distance requires DP over prefixes [OK]

Common Mistakes:

• Thinking greedy or divide-and-conquer works optimally

2. What is the worst-case time complexity of the put operation in the brute force LFU cache implementation that scans all keys to find the least frequently used item for eviction?

medium

A. O(1) -- constant time due to direct access

B. O(n) -- linear time scanning all keys for eviction

C. O(log n) -- due to sorting keys by frequency

D. O(n^2) -- nested scanning for frequency and recency

Solution

1. Step 1: Identify eviction process

   Eviction requires scanning all keys to find the minimum frequency, which takes O(n) time where n is the number of keys.

2. Step 2: Confirm no sorting or nested loops

   The code uses min() and list comprehensions but does not sort or nest loops over keys, so complexity is O(n), not O(log n) or O(n^2).

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Eviction scanning all keys -> O(n) time [OK]

Hint: Eviction scans all keys linearly -> O(n) time [OK]

Common Mistakes:

• Assuming O(1) due to hashmap
• Confusing min() with sorting
• Overestimating complexity as O(n^2)

3. The following code attempts to solve the Russian Doll Envelopes problem optimally. Identify the line containing the subtle bug that causes incorrect results on some inputs.

medium

A. Line 4: Sorting envelopes with key (x[0], x[1]) ascending both

B. Line 6: Extracting heights from envelopes

C. Line 8: Using bisect_left to find insertion index

D. Line 11: Replacing tails element at found index

Solution

1. Step 1: Examine sorting key

   Sorting by (width asc, height asc) allows envelopes with same width and increasing height, which breaks strict nesting rules.

2. Step 2: Correct sorting key

   Sorting by (width asc, height desc) prevents envelopes with equal width but increasing height from being counted incorrectly.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Changing sorting to (x[0], -x[1]) fixes the bug [OK]

Hint: Sort height descending for equal widths to avoid counting duplicates [OK]

Common Mistakes:

• Sorting both width and height ascending
• Using <= instead of < in comparisons
• Not handling equal widths properly

4. What is the time complexity of the divide and conquer approach to the Skyline Problem when given n buildings, and why might a common misconception about the complexity be incorrect?

medium

A. O(n log n) because the problem is divided into halves and merged efficiently

B. O(n log n) time but O(n²) space due to storing all intermediate skylines

C. O(n log n) but with additional O(n) space overhead for recursion stack making it O(n log n) space

D. O(n²) because each merge step compares all points in both skylines

Solution

1. Step 1: Identify divide and conquer structure

   The buildings are split into halves recursively, leading to log n levels.

2. Step 2: Analyze merge cost per level

   Each merge step processes all points in the two skylines combined, which sums to O(n) per level.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Divide and conquer with linear merge per level -> O(n log n) time [OK]

Hint: Divide and conquer merges cost O(n) per level over log n levels [OK]

Common Mistakes:

• Assuming merge is O(n²) due to nested loops
• Confusing recursion stack space with time complexity
• Thinking space grows quadratically due to intermediate skylines

5. Suppose you want to find the median of two sorted arrays where elements can be repeated infinitely (i.e., arrays represent multisets with unlimited duplicates). Which modification to the optimal binary search partition approach is necessary to handle this scenario correctly?

hard

A. Modify the partition condition to handle infinite duplicates by counting occurrences explicitly.

B. No modification needed; the existing binary search partition approach works as is.

C. Switch to a two-pointer partial merge approach to handle duplicates efficiently.

D. Use a frequency map and binary search on value range instead of indices.

Solution

1. Step 1: Understand infinite duplicates impact

   Infinite duplicates mean the arrays represent multisets with repeated elements beyond fixed length, so index-based partitioning is insufficient.

2. Step 2: Modify partition condition

   The binary search partition must be adapted to count occurrences and handle duplicates explicitly to find correct median position.

3. Step 3: Why other options fail

   No modification ignores duplicates; two-pointer merge loses logarithmic efficiency; frequency map changes problem nature.

4. Final Answer:

   Option A -> Option A

5. Quick Check:

   Handling infinite duplicates requires counting occurrences in partition [OK]

Hint: Infinite duplicates require counting occurrences in partition [OK]

Common Mistakes:

• Assuming no change needed for duplicates
• Switching to merge loses optimal time
• Using frequency map changes problem scope