Leetcode #2940: Find Building Where Alice and Bob Can Meet
In this guide, we solve Leetcode #2940 Find Building Where Alice and Bob Can Meet in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
You are given a 0-indexed array heights of positive integers, where heights[i] represents the height of the ith building. If a person is in building i, they can move to any other building j if and only if i < j and heights[i] < heights[j].
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Stack, Binary Indexed Tree, Segment Tree, Array, Binary Search, Monotonic Stack, Heap (Priority Queue)
Intuition
The problem structure suggests a monotonic decision, which makes binary search a natural fit.
By halving the search space each step, we reach the answer efficiently.
Approach
Search either directly on a sorted array or on the answer space using a check function.
Each check is fast, and the logarithmic search keeps the overall runtime low.
Steps:
- Define the search bounds.
- Check the mid point condition.
- Narrow the bounds until convergence.
Example
Input: heights = [6,4,8,5,2,7], queries = [[0,1],[0,3],[2,4],[3,4],[2,2]]
Output: [2,5,-1,5,2]
Explanation: In the first query, Alice and Bob can move to building 2 since heights[0] < heights[2] and heights[1] < heights[2].
In the second query, Alice and Bob can move to building 5 since heights[0] < heights[5] and heights[3] < heights[5].
In the third query, Alice cannot meet Bob since Alice cannot move to any other building.
In the fourth query, Alice and Bob can move to building 5 since heights[3] < heights[5] and heights[4] < heights[5].
In the fifth query, Alice and Bob are already in the same building.
For ans[i] != -1, It can be shown that ans[i] is the leftmost building where Alice and Bob can meet.
For ans[i] == -1, It can be shown that there is no building where Alice and Bob can meet.
Python Solution
class BinaryIndexedTree:
__slots__ = ["n", "c"]
def __init__(self, n: int):
self.n = n
self.c = [inf] * (n + 1)
def update(self, x: int, v: int):
while x <= self.n:
self.c[x] = min(self.c[x], v)
x += x & -x
def query(self, x: int) -> int:
mi = inf
while x:
mi = min(mi, self.c[x])
x -= x & -x
return -1 if mi == inf else mi
class Solution:
def leftmostBuildingQueries(
self, heights: List[int], queries: List[List[int]]
) -> List[int]:
n, m = len(heights), len(queries)
for i in range(m):
queries[i] = [min(queries[i]), max(queries[i])]
j = n - 1
s = sorted(set(heights))
ans = [-1] * m
tree = BinaryIndexedTree(n)
for i in sorted(range(m), key=lambda i: -queries[i][1]):
l, r = queries[i]
while j > r:
k = n - bisect_left(s, heights[j]) + 1
tree.update(k, j)
j -= 1
if l == r or heights[l] < heights[r]:
ans[i] = r
else:
k = n - bisect_left(s, heights[l])
ans[i] = tree.query(k)
return ans
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.