Count Pairs With XOR in a Range

📋

Count Pairs With XOR in a Range

Edge cases:

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IDE

def countPairs(nums: list[int], low: int, high: int) -> int:public int countPairs(int[] nums, int low, int high)int countPairs(vector<int>& nums, int low, int high)function countPairs(nums, low, high)

def countPairs(nums, low, high):
    # Write your solution here
    pass

class Solution {
    public int countPairs(int[] nums, int low, int high) {
        // Write your solution here
        return 0;
    }
}

#include <vector>
using namespace std;

int countPairs(vector<int>& nums, int low, int high) {
    // Write your solution here
    return 0;
}

function countPairs(nums, low, high) {
    // Write your solution here
}

Common Bugs to Avoid

Wrong: 0Fails to count any pairs, possibly returns 0 for all inputs.✅ Ensure nested loops or trie counting logic is implemented to count pairs properly.

Wrong: Incorrect count missing pairs with XOR exactly equal to low or highUsing strict inequalities instead of inclusive range checks.✅ Change condition to 'if low <= xor_val <= high' instead of strict inequalities.

Wrong: TLEUsing brute force O(n^2) approach on large inputs.✅ Implement bit trie based counting with O(n*log M) complexity.

Wrong: Wrong count on empty or single-element arraysNot handling base cases where no pairs exist.✅ Return 0 immediately if nums length < 2.

Wrong: Incorrect count with duplicates or all zerosNot counting all pairs when XOR is zero and range includes zero.✅ Count all pairs using n*(n-1)/2 when XOR=0 is in range.

✓

Test Cases

Try testing your solution on empty and single-element inputs to handle base cases.

Consider tricky inputs that break greedy or partial counting approaches.

Optimize your solution to use bit trie for large inputs to avoid TLE.

t1_01basic

Input{"nums":[1,4,2,7],"low":2,"high":6}

Expected6

⏱ Performance - must finish in 2000ms

Pairs with XOR in [2,6] are (1,4)=5, (1,2)=3, (4,2)=6, (4,7)=3, (2,7)=5, (1,7)=6

💡 Consider all pairs and their XOR values.

💡 Count pairs where XOR is between low and high inclusive.

💡 Use nested loops or a trie-based approach to count efficiently.

Why it failed: Incorrect count likely due to missing pairs or wrong XOR range check. Fix by ensuring XOR is checked with 'if low <= xor_val <= high'.

✓ Correctly counts all pairs with XOR in the given range.

t1_02basic

Input{"nums":[3,8,5,6],"low":1,"high":4}

Expected1

⏱ Performance - must finish in 2000ms

Pairs with XOR in [1,4]: (5,6)=3 only.

💡 Try enumerating all pairs and their XOR values.

💡 Check which XOR values fall within the range [low, high].

💡 Count only pairs where XOR >= low and XOR <= high.

Why it failed: Wrong count due to incorrect XOR range filtering or pair enumeration. Fix by enumerating all pairs and applying correct range check.

✓ Correctly counts pairs with XOR in the specified range.

t2_01edge

Input{"nums":[],"low":0,"high":0}

Expected0

⏱ Performance - must finish in 2000ms

Empty array has no pairs, so count is 0.

💡 Consider what happens when nums is empty.

💡 No pairs exist if array length is less than 2.

💡 Return 0 immediately if nums length < 2.

Why it failed: Fails on empty input by not handling zero-length array. Fix by returning 0 if nums length < 2.

✓ Correctly handles empty input with zero pairs.

t2_02edge

Input{"nums":[5],"low":0,"high":10}

Expected0

⏱ Performance - must finish in 2000ms

Single element array has no pairs, so count is 0.

💡 Check behavior when only one element is present.

💡 Pairs require two distinct indices i < j.

💡 Return 0 if nums length < 2.

Why it failed: Fails single element case by not returning 0. Fix by checking nums length before counting pairs.

✓ Correctly returns 0 for single element input.

t2_03edge

Input{"nums":[0,0,0,0],"low":0,"high":0}

Expected6

⏱ Performance - must finish in 2000ms

All pairs XOR to 0, which is within [0,0]. Number of pairs = C(4,2) = 6.

💡 All elements zero means XOR is always zero.

💡 Count all pairs since XOR=0 is within range.

💡 Use formula n*(n-1)/2 for total pairs.

Why it failed: Fails to count pairs when XOR=0 and range is [0,0]. Fix by including pairs with XOR exactly equal to low and high.

✓ Correctly counts all pairs with XOR zero in range [0,0].

t2_04edge

Input{"nums":[1000000,999999,500000],"low":0,"high":1000000}

Expected3

⏱ Performance - must finish in 2000ms

All pairs XOR values are within [0,1000000], so count all pairs: C(3,2)=3.

💡 Test maximum element values near 10^6.

💡 Ensure bit trie handles large values correctly.

💡 Count all pairs if XOR in range for all pairs.

Why it failed: Fails due to incorrect handling of large values or bit length. Fix by using consistent bit length and correct XOR checks.

✓ Correctly counts pairs with large values.

t3_01corner

Input{"nums":[1,2,3,4,5],"low":0,"high":1}

Expected2

⏱ Performance - must finish in 2000ms

Pairs with XOR in [0,1]: (2,3)=1 and (4,5)=1.

💡 Check if your solution handles XOR range boundaries correctly.

💡 Beware of greedy approaches that pick pairs without full enumeration.

💡 Ensure counting pairs with XOR exactly equal to low or high.

Why it failed: Greedy approach fails by missing pairs with XOR at range edges. Fix by counting all pairs and checking XOR in [low, high] inclusively.

✓ Correctly counts pairs including those at XOR range boundaries.

t3_02corner

Input{"nums":[1,1,1,1],"low":0,"high":0}

Expected6

⏱ Performance - must finish in 2000ms

All pairs XOR to 0, which is within [0,0]. Number of pairs = C(4,2) = 6.

💡 Check if your solution handles duplicates correctly.

💡 XOR of identical numbers is zero, count all pairs.

💡 Use combination formula n*(n-1)/2 for identical elements.

Why it failed: Fails by counting pairs incorrectly when duplicates exist. Fix by counting all pairs of identical elements properly.

✓ Correctly counts pairs with duplicates and XOR zero.

t3_03corner

Input{"nums":[0,1023,1024,2047],"low":512,"high":1536}

Expected4

⏱ Performance - must finish in 2000ms

Pairs with XOR in [512,1536]: (0,1023)=1023, (0,1024)=1024, (1023,2047)=1024, (1024,2047)=1023.

💡 Test if your solution handles large values near 10^6 correctly.

💡 Check bit trie implementation for correct bit length handling.

💡 Ensure XOR comparisons use correct bitwise operations.

Why it failed: Fails due to incorrect bit length or trie traversal. Fix by using consistent bit length (e.g. 20 bits) for all numbers.

✓ Correctly counts pairs with large values using bit trie.

t4_01performance

Input{"nums":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80,81,82,83,84,85,86,87,88,89,90,91,92,93,94,95,96,97,98,99],"low":0,"high":1000000}

Expectednull

⏱ Performance - must finish in 2000ms

n=100, O(n*log M) expected to run within 2s; brute force O(n^2) will TLE.

💡 Brute force O(n^2) is too slow for n=100.

💡 Use bit trie to achieve O(n*log M) time complexity.

💡 Implement prefix counting with bitwise trie for efficiency.

Why it failed: TLE due to O(n^2) brute force approach. Fix by implementing bit trie based counting with O(n*log M) complexity.

✓ Efficient bit trie solution runs within time limit.

Practice

1. Given the following Trie implementation, after inserting the words ['cat', 'car', 'dog'], what is the value of node.isEnd after processing the last character of 'car' in the insert method?

easy

A. false, because 'car' shares prefix with 'cat' and 'dog'

B. true, but only if 'car' was inserted before 'cat'

C. false, because isEnd is only set after inserting all words

D. true, because 'car' is a complete inserted word

Solution

1. Step 1: Trace insertion of 'car'

   For each character 'c', 'a', 'r', nodes are created or reused. After last character 'r', node.isEnd is set to true.

2. Step 2: Confirm isEnd flag meaning

   isEnd marks the end of a complete word. Since 'car' was inserted, node.isEnd after 'r' is true.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   isEnd is true after inserting 'car' [OK]

Hint: isEnd is set true after inserting each complete word [OK]

Common Mistakes:

• Assuming isEnd is false if prefix overlaps with other words

2. Consider the following Python code implementing a bit trie for maximum XOR queries. Given nums = [1, 2] inserted into the trie, what is the value of max_xor after calling max_xor(3)?

easy

A. 2

B. 1

C. 3

D. 0

Solution

1. Step 1: Trace insertion of nums 1 and 2

   Bits of 1 and 2 inserted from MSB to LSB into trie.

2. Step 2: Trace max_xor(3)

   For each bit of 3, try toggled bit in trie to maximize XOR. At highest differing bit, toggled bit exists, setting max_xor bits accordingly.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   3 XOR 0 = 3, 3 XOR 1 = 2, 3 XOR 2 = 1; max is 3 [OK]

Hint: XOR with 3 and 0,1,2 yields max 3 [OK]

Common Mistakes:

• Confusing bit order in trie traversal
• Off-by-one in bit indexing
• Returning wrong max_xor value

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. You are given a 2D grid of letters and a list of words. You need to find all words from the list that can be formed by sequentially adjacent letters on the board, where adjacent letters are horizontally or vertically neighboring. Each letter cell may only be used once per word. Which algorithmic approach is best suited to efficiently solve this problem when the word list is large?

easy

A. Build a Trie from the word list and perform a single DFS on the board with prefix pruning using the Trie.

B. Use a brute force DFS for each word independently, checking all paths on the board.

C. Use dynamic programming to store subproblems of partial word matches on the board.

D. Sort the words and use binary search combined with DFS to find matches on the board.

Solution

1. Step 1: Understand problem constraints and prefix sharing

   Multiple words need to be found on the board, and many words may share prefixes, making prefix pruning beneficial.

2. Step 2: Use Trie for efficient prefix pruning during DFS

   Building a Trie allows pruning DFS paths early if no word starts with the current prefix, avoiding redundant searches and improving efficiency.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Trie + DFS reduces search space drastically compared to checking each word separately [OK]

Hint: Trie + DFS prunes invalid prefixes early [OK]

Common Mistakes:

• Thinking brute force DFS per word is efficient
• Assuming DP applies here
• Using sorting and binary search without prefix pruning

5. Examine the following buggy code snippet for building the trie and collecting suggestions. Which line contains the subtle bug that can cause incorrect suggestion ordering or runtime errors?

medium

A. Line with 'products.sort()' missing before building trie

B. Line with 'for product in products:' loop

C. Line with 'if len(node.suggestions) < 3:' condition

D. Line with 'if node and ch in node.children:' during search

Solution

1. Step 1: Identify missing sorting

   The code does not sort products before building the trie, so suggestions are appended in input order, not lex order.

2. Step 2: Consequences of missing sort

   Without sorting, the suggestions list at each node is not guaranteed to be lexicographically smallest three, violating problem requirements.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Sorting before trie build is essential for correct lex order [OK]

Hint: Always sort products before building trie for lex order [OK]

Common Mistakes:

• Forgetting to sort products
• Appending suggestions without limit
• Not handling missing prefix nodes