Practice
1. Given the following Python code snippet implementing the optimized trie approach, what is the output of
maximizeXor(nums, queries) for nums = [0,1,2] and queries = [[3,1],[1,2]]?
class TrieNode:
def __init__(self):
self.children = [null, null]
class Trie:
def __init__(self):
self.root = TrieNode()
def insert(self, num):
node = self.root
for i in range(31, -1, -1):
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):
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()
queries = sorted([(x, m, i) for i, (x, m) in enumerate(queries)], key=lambda x: x[1])
res = [-1] * len(queries)
trie = Trie()
idx = 0
n = len(nums)
for x, m, i in queries:
while idx < n and nums[idx] <= m:
trie.insert(nums[idx])
idx += 1
res[i] = trie.maxXor(x)
return res
print(maximizeXor(nums, queries))easy
Solution
Step 1: Process first query (3,1)
Only nums ≤ 1 are [0,1]. Insert 0 and 1 into trie. maxXor(3) with {0,1} is max(3^0=3,3^1=2)=3.Step 2: Process second query (1,2)
Now nums ≤ 2 are [0,1,2]. Insert 2. maxXor(1) with {0,1,2} is max(1^0=1,1^1=0,1^2=3)=3.Final Answer:
Option B -> Option BQuick Check:
Results match expected XOR values [OK]
Hint: Insert nums ≤ m incrementally, then maxXor query [OK]
Common Mistakes:
- Forgetting to sort queries and nums
- Not inserting all eligible nums before query
- Returning -1 when trie is empty incorrectly
2. Given the following code snippet implementing the optimal approach to find the median of two sorted arrays, what is the returned median for input nums1 = [1, 3], nums2 = [2]?
easy
Solution
Step 1: Trace initial variables
nums1=[1,3], nums2=[2], m=2, n=1, swap occurs so nums1=[2], nums2=[1,3], m=1, n=2, half_len=2.Step 2: First iteration of binary search
low=0, high=1, i=0, j=2-0=2; max_left_nums1=-inf, min_right_nums1=2; max_left_nums2=nums2[1]=3, min_right_nums2=inf. Condition fails, low updated to i+1=1.Step 3: Second iteration
i=1, j=1; max_left_nums1=2, min_right_nums1=inf; max_left_nums2=nums2[0]=1, min_right_nums2=nums2[1]=3. Condition satisfied.Step 4: Calculate median
Total length is odd (3), median is max(max_left_nums1, max_left_nums2) = max(2,1) = 2.0.Final Answer:
Option C -> Option CQuick Check:
Median of [1,2,3] is 2.0 [OK]
Hint: Swap arrays to binary search smaller one; median is 2.0 [OK]
Common Mistakes:
- Not swapping arrays leads to wrong partition
- Off-by-one in i or j calculation
- Confusing odd/even median calculation
3. Given a list of unique words, you need to find all pairs of distinct indices (i, j) such that the concatenation of words[i] + words[j] forms a palindrome. Which approach guarantees an optimal solution in terms of time complexity for large inputs?
easy
Solution
Step 1: Understand brute force limitations
Brute force checks all pairs, resulting in O(n^2 * k) time, which is inefficient for large n.Step 2: Recognize Trie-based optimization
Building a Trie of reversed words allows prefix matching and palindrome suffix tracking, reducing redundant checks and achieving O(n * k^2) time.Final Answer:
Option B -> Option BQuick Check:
Trie-based approach is known optimal for palindrome pairs [OK]
Hint: Trie with palindrome suffixes optimizes palindrome pair search [OK]
Common Mistakes:
- Assuming brute force is acceptable for large inputs
- Confusing palindrome substring DP with pair finding
- Believing greedy prefix-suffix matching suffices
4. Given the following code snippet implementing the two-pointer approach for trapping rain water, what is the value of the variable
water after the first iteration of the while loop when the input is [2, 0, 2]?easy
Solution
Step 1: Initialize variables
left=0, right=2, left_max=0, right_max=0, water=0Step 2: First iteration of while loop
height[left]=2, height[right]=2, since height[left] is not less than height[right], else branch runs. height[right]=2 ≥ right_max=0, so right_max=2, right decrements to 1, water remains 0.Final Answer:
Option A -> Option AQuick Check:
Water not added on first iteration because max updated [OK]
Hint: Water added only when current height < max height on that side [OK]
Common Mistakes:
- Adding water before max height is updated
- Confusing left and right pointer conditions
- Off-by-one errors in pointer movement
5. What is the time complexity of the bitmask backtracking solution for the N-Queens problem, and why is it not simply O(n^n)?
medium
Solution
Step 1: Understand pruning effect on search space
Each queen placed reduces available columns, so permutations of columns are tried, not all combinations.Step 2: Recognize factorial complexity
Backtracking tries permutations of n columns for n rows, leading to O(n!) complexity, not exponential or polynomial.Final Answer:
Option C -> Option CQuick Check:
Pruning reduces search to permutations, not all subsets [OK]
Hint: N-Queens backtracking explores permutations -> O(n!) [OK]
Common Mistakes:
- Confusing subsets (2^n) with permutations (n!)
- Assuming polynomial due to board size
- Ignoring pruning effect