How do you solve Find Building Where Alice and Bob Can Meet on LeetCode?

Find Building Where Alice and Bob Can Meet (LeetCode #2940) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n + q) time complexity and O(n) space complexity. Key insight: Understanding the movement constraints is crucial for determining valid meeting points.

What is the time complexity of Find Building Where Alice and Bob Can Meet?

The optimal solution for Find Building Where Alice and Bob Can Meet runs in O(n + q) time and uses O(n) space. The preprocessing step to find the next taller building takes O(n), and each query can be resolved in O(1) time on average, leading to O(n + q) overall.

What is the brute force solution for Find Building Where Alice and Bob Can Meet?

The brute force approach for Find Building Where Alice and Bob Can Meet is Brute Force, with O(n²) time complexity and O(1) space complexity. We will check each query by examining all buildings that Alice and Bob can potentially move to. This means for each query, we will look for the first building that both can reach based on their heights. The optimal Optimal Solution approach improves this to O(n + q).

What algorithm or data structure is used to solve Find Building Where Alice and Bob Can Meet?

Find Building Where Alice and Bob Can Meet uses the following concepts: Array, Binary Search, Stack, Binary Indexed Tree, Segment Tree, Heap (Priority Queue), Monotonic Stack. The recommended patterns to study are: Stack for finding the next greater element., Array manipulation for efficient query handling..

What are the interview tips for Find Building Where Alice and Bob Can Meet?

Always clarify the movement rules before jumping into coding. Consider edge cases, such as when Alice and Bob are in the same building. Discuss your thought process and approach before coding to ensure clarity.

What are common mistakes when solving Find Building Where Alice and Bob Can Meet?

Not considering the case when a > b and failing to swap indices. Overlooking the need to check if both Alice and Bob can reach the same building.

#2940

Find Building Where Alice and Bob Can Meet

Hard

ArrayBinary SearchStackBinary Indexed TreeSegment TreeHeap (Priority Queue)Monotonic StackStack for finding the next greater element.Array manipulation for efficient query handling.

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n + q)
Space	O(1)	O(n)

💡

Intuition

Time O(n + q)Space O(n)

We can preprocess the heights to find the next taller building for each building, allowing us to quickly determine where Alice and Bob can meet without checking each building for every query.

⚙️

Algorithm

5 steps

1Step 1: Create an array nextTaller to store the index of the next taller building for each building.
2Step 2: Use a stack to efficiently find the next taller building for each building in a single pass.
3Step 3: For each query, use the nextTaller array to find the leftmost building they can both reach.
4Step 4: If heights[a] < heights[next] and heights[b] < heights[next], return next.
5Step 5: If no valid building is found, return -1.

solution.py21 lines

1def findMeetingBuildings(heights, queries):
2    n = len(heights)
3    nextTaller = [-1] * n
4    stack = []
5    for i in range(n):
6        while stack and heights[stack[-1]] < heights[i]:
7            nextTaller[stack.pop()] = i
8        stack.append(i)
9    ans = []
10    for a, b in queries:
11        if a > b:
12            a, b = b, a
13        leftmost = -1
14        next = nextTaller[a]
15        while next != -1:
16            if heights[b] < heights[next]:
17                leftmost = next
18                break
19            next = nextTaller[next]
20        ans.append(leftmost)
21    return ans

ℹ

Complexity note: The preprocessing step to find the next taller building takes O(n), and each query can be resolved in O(1) time on average, leading to O(n + q) overall.

1Understanding the movement constraints is crucial for determining valid meeting points.
2Preprocessing data can significantly reduce the time complexity for multiple queries.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.