Count Pairs With XOR in a Range

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Steps

setup

Initialize Trie Root

Create the root node of the trie with no children and zero count. This node will be the starting point for all insertions and queries.

💡 The root node is essential as the entry point for representing all numbers in binary form within the trie.

Line:

root = TrieNode()

💡 The trie starts empty, ready to build paths for numbers bit by bit.

insert

Insert Number 1 - Bit 20 (Most Significant Bit)

Start inserting number 1 into the trie from the highest bit (bit 20). Extract bit 20 (which is 0) and create a child node for bit 0 since it does not exist yet.

💡 Inserting bits from high to low builds the binary representation path for the number in the trie.

Line:

bit = (num >> i) & 1
if not node.children[bit]:
  node.children[bit] = TrieNode()
node = node.children[bit]
node.count += 1

💡 Each bit leads to a branch; creating a new node means this bit pattern is new in the trie.

insert

Insert Number 1 - Bits 19 to 1

Continue inserting bits 19 down to 1 for number 1. All these bits are 0, so create child nodes for bit 0 along the path, incrementing counts.

💡 Most bits are zero for small numbers, so the trie path extends mostly along 0-child nodes.

Line:

for i in range(20, -1, -1):
  bit = (num >> i) & 1
  if not node.children[bit]:
    node.children[bit] = TrieNode()
  node = node.children[bit]
  node.count += 1

💡 The trie path for 1 is a chain of 0 bits except the last bit.

compare

Query Trie for Number 4 with Limit 7 (high+1)

Query how many previously inserted numbers have XOR with 4 less than 7. Start from bit 20, comparing bits of 4 and limit 7 to decide traversal and counting.

💡 Querying uses bitwise logic to count how many paths in the trie satisfy XOR < limit at each bit.

Line:

bit_num = (num >> i) & 1
bit_lim = (limit >> i) & 1
if bit_lim == 1:
  if node.children[bit_num]:
    count += node.children[bit_num].count
  node = node.children[1 - bit_num]
else:
  node = node.children[bit_num]

💡 When limit bit is 0, only paths matching num's bit continue; no counts added here.

compare

Query Trie for Number 4 - Bit 19

At bit 19, bit_lim=0, bit_num=0; continue traversal to child matching bit_num=0 without adding to count.

💡 Continuing traversal when limit bit is 0 narrows down valid paths for XOR < limit.

Line:

if bit_lim == 1:
  ...
else:
  node = node.children[bit_num]

💡 No counts added; path must match num's bit to stay under limit.

compare

Query Trie for Number 4 - Bit 18

At bit 18, bit_lim=0, bit_num=0; continue traversal to child 0 without adding count.

💡 Each bit with limit bit 0 forces traversal to matching bit child to keep XOR < limit.

Line:

else:
  node = node.children[bit_num]

💡 Traversal narrows possible XOR pairs stepwise.

compare

Query Trie for Number 4 - Bit 2 (Limit Bit 1)

At bit 2, bit_lim=1, bit_num=1; add count of child matching bit_num=1, then traverse child 0 (1-bit flipped).

💡 When limit bit is 1, counts from one child are added, and traversal continues on the other child to count valid pairs.

Line:

if bit_lim == 1:
  if node.children[bit_num]:
    count += node.children[bit_num].count
  node = node.children[1 - bit_num]

💡 This step accumulates counts of numbers with XOR bits less than limit at this position.

insert

Insert Number 4 into Trie

Insert number 4 bit by bit from bit 20 down to bit 0, creating new nodes or traversing existing ones, incrementing counts.

💡 Inserting each number updates the trie structure and counts to reflect all numbers inserted so far.

Line:

for i in range(20, -1, -1):
  bit = (num >> i) & 1
  if not node.children[bit]:
    node.children[bit] = TrieNode()
  node = node.children[bit]
  node.count += 1

💡 Trie grows and counts update to represent all inserted numbers.

traverse

Count Pairs Less Than high+1 (7) Computed

After querying and inserting all numbers, the count of pairs with XOR less than 7 is computed.

💡 This count includes all pairs with XOR strictly less than 7.

Line:

countHigh = countPairsLessThan(nums, high + 1)

💡 This partial count is the first part of the final answer calculation.

traverse

Count Pairs Less Than low (2) Computed

Similarly, count pairs with XOR less than 2 by querying the trie with limit 2.

💡 This count excludes pairs with XOR equal or greater than 2.

Line:

countLow = countPairsLessThan(nums, low)

💡 This partial count will be subtracted from the previous to get pairs in [low, high].

reconstruct

Calculate Final Answer: countHigh - countLow

Subtract countLow from countHigh to get the number of pairs with XOR in the range [2, 6].

💡 This subtraction isolates pairs whose XOR lies within the desired range.

Line:

return countPairsLessThan(nums, high + 1) - countPairsLessThan(nums, low)

💡 The final answer is the count of pairs with XOR in [low, high].

class TrieNode:
    def __init__(self):
        self.children = [None, None]
        self.count = 0

class Solution:
    def countPairsLessThan(self, nums, limit):
        root = TrieNode()  # STEP 1
        def insert(num):
            node = root
            for i in range(20, -1, -1):  # STEP 2-3
                bit = (num >> i) & 1
                if not node.children[bit]:
                    node.children[bit] = TrieNode()
                node = node.children[bit]
                node.count += 1

        def query(num, limit):
            node = root
            count = 0
            for i in range(20, -1, -1):  # STEP 4-7
                if not node:
                    break
                bit_num = (num >> i) & 1
                bit_lim = (limit >> i) & 1
                if bit_lim == 1:
                    if node.children[bit_num]:
                        count += node.children[bit_num].count
                    node = node.children[1 - bit_num]
                else:
                    node = node.children[bit_num]
            return count

        result = 0
        for num in nums:  # STEP 8
            result += query(num, limit)
            insert(num)
        return result

    def countPairs(self, nums, low, high):  # STEP 9-11
        return self.countPairsLessThan(nums, high + 1) - self.countPairsLessThan(nums, low)

if __name__ == '__main__':
    nums = [1,4,2,7]
    low, high = 2, 6
    sol = Solution()
    print(sol.countPairs(nums, low, high))  # Output: 6

📊

Count Pairs With XOR in a Range - Watch the Algorithm Execute, Step by Step

Watching the trie construction and query process step-by-step reveals how bitwise properties enable efficient counting of XOR pairs, which is difficult to grasp from code alone.

insert

insert

1

0

insert

1

000000000000000000001

query

4

0

query

4

00

query

4

00

query

4

001

insert

4

0000000000000000000011

query

query

countPairs

Search: 6

Key Takeaways

✓ The trie efficiently encodes all numbers bitwise, enabling fast queries to count pairs with XOR below a limit.

This insight is hard to see from code alone because the bitwise trie structure and counting logic are abstract without visualization.

✓ Querying the trie uses bitwise comparisons of the number and limit bits to decide which branches to count and traverse.

Understanding how limit bits guide traversal and counting clarifies why the algorithm counts pairs correctly.

✓ The final answer is computed by subtracting counts of pairs with XOR less than low from those less than high+1.

This subtraction isolates pairs within the desired XOR range, a key step that is subtle without seeing intermediate counts.

Practice

1. Given the following code snippet for MapSum, after inserting keys "apple" with value 3 and "app" with value 2, what is the value of node.sum at the trie node corresponding to 'p' (the second character) after the second insert?

easy

A. 3

B. 5

C. 2

D. 0

Solution

1. Step 1: Trace insert of "apple" with value 3

   Delta = 3 - 0 = 3. Root.sum += 3 -> 3. For each character 'a','p','p','l','e', node.sum += 3. After first insert, node at second 'p' has sum 3.

2. Step 2: Trace insert of "app" with value 2

   Delta = 2 - 0 = 2. Root.sum += 2 -> 5. For 'a','p','p', node.sum += 2. Node at second 'p' sum was 3, now 3 + 2 = 5.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Sum at second 'p' after both inserts is 5 [OK]

Hint: Sum accumulates delta on each insert per prefix node [OK]

Common Mistakes:

• Forgetting to add delta on second insert
• Confusing first and second 'p' nodes
• Using old value instead of delta

2. What is the time complexity of the optimal trie-based solution for finding shortest unique prefixes for n words each of maximum length m?

medium

A. O(n^2 * m) because each prefix is compared with all other words

B. O(n * m) because each word is inserted and traversed once in the trie

C. O(n * m^2) due to nested prefix checks during insertion

D. O(n * log n * m) because sorting is involved before trie insertion

Solution

1. Step 1: Identify operations per word

   Each word of length m is inserted into the trie once, updating frequency counts along the path.

2. Step 2: Analyze total complexity

   Insertion and prefix search both take O(m) per word, so total is O(n * m). No sorting or nested loops over words occur.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Trie insertion and search scale linearly with total characters [OK]

Hint: Trie operations scale with total characters, not pairwise comparisons [OK]

Common Mistakes:

• Confusing trie insertion with pairwise prefix checks
• Assuming sorting is required

3. Suppose the problem changes: you want to find the maximum XOR of two numbers in an array, but now numbers can be negative 32-bit integers. Which modification to the bitwise trie approach is necessary to handle negative numbers correctly?

hard

A. Use two's complement 32-bit representation and treat bits as unsigned during trie insertion and traversal

B. Ignore sign bit and only build trie from absolute values of numbers

C. Use 64-bit representation to store numbers in trie to avoid sign issues

D. Preprocess numbers by adding a large constant to make all numbers positive before building trie

Solution

1. Step 1: Understand representation of negative numbers

   Negative integers in 32-bit are stored in two's complement form, which can be treated as unsigned bit patterns for trie operations.

2. Step 2: Modify trie insertion and traversal

   Insert and traverse bits as unsigned 32-bit integers, including the sign bit, to correctly handle negative numbers.

3. Step 3: Avoid incorrect approaches

   Using 64-bit or adding constants changes problem semantics; ignoring sign bit loses information.

4. Final Answer:

   Option A -> Option A

5. Quick Check:

   Two's complement allows uniform bitwise trie handling of negative numbers [OK]

Hint: Two's complement bits can be treated as unsigned [OK]

Common Mistakes:

• Ignoring sign bit or converting to absolute values

4. Suppose the problem changes: now the stream can contain repeated queries of the same letter, and you want to support queries that allow reusing letters multiple times to form words (e.g., words can be formed by concatenating repeated suffixes). Which modification to the optimal StreamChecker approach correctly supports this reuse without degrading query time significantly?

hard

A. Keep the same reversed trie and stream buffer, but on each query, perform a DFS from the root to find all possible suffix matches allowing repeated traversal of nodes.

B. Modify the trie to store counts of repeated letters and update the stream buffer to track counts, then traverse accordingly.

C. Increase the stream buffer size to the total length of all words to capture all repeated suffixes.

D. Use a suffix automaton or Aho-Corasick automaton instead of a trie to handle repeated patterns efficiently in queries.

Solution

1. Step 1: Understand the reuse requirement

   Allowing repeated letters and concatenations means suffix matching alone is insufficient; need to handle overlapping patterns efficiently.

2. Step 2: Evaluate data structures

   DFS on trie is expensive and can degrade query time. Counting repeated letters is complex and error-prone. Increasing buffer size is inefficient. Suffix automaton or Aho-Corasick automaton efficiently handle repeated patterns and multiple matches.

3. Step 3: Choose optimal solution

   Using suffix or Aho-Corasick automaton supports repeated pattern matching with efficient query time.

4. Final Answer:

   Option D -> Option D

5. Quick Check:

   Suffix automaton handles repeated substrings efficiently [OK]

Hint: Suffix automaton handles repeated patterns efficiently [OK]

Common Mistakes:

• Trying DFS on trie causing exponential time
• Increasing buffer size unboundedly
• Tracking counts in trie nodes incorrectly

5. Suppose the Word Break problem is modified so that dictionary words can be reused unlimited times in the segmentation. Which modification to the algorithm is necessary to correctly handle this variant?

hard

A. Modify the inner loop to continue checking all prefixes even after dp[i] is set to True

B. Change dp array to count number of ways instead of boolean flags

C. Remove the dp array and use pure recursion with memoization only

D. No change needed; the original DP with trie approach already supports reuse

Solution

1. Step 1: Understand reuse impact on DP

   When reuse is allowed, multiple segmentations can overlap. The DP must consider all possible prefixes, not stop at first True.

2. Step 2: Modify inner loop behavior

   Breaking early when dp[i] is True prevents exploring all segmentations. Removing the break allows full exploration, correctly handling reuse.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Allowing all prefixes ensures reuse is handled [OK]

Hint: Breaking early blocks multiple reuse paths [OK]

Common Mistakes:

• Assuming original DP supports reuse without change
• Changing dp to count ways unnecessarily
• Removing dp and relying on recursion only