Solving Leetcode Interviews in Seconds with AI: Find Building Where Alice and Bob Can Meet
Introduction
In this blog post, we will explore how to solve the LeetCode problem "2940" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.
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]. You are also given another array queries where queries[i] = [ai, bi]. On the ith query, Alice is in building ai while Bob is in building bi. Return an array ans where ans[i] is the index of the leftmost building where Alice and Bob can meet on the ith query. If Alice and Bob cannot move to a common building on query i, set ans[i] to -1. Example 1: 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. Example 2: Input: heights = [5,3,8,2,6,1,4,6], queries = [[0,7],[3,5],[5,2],[3,0],[1,6]] Output: [7,6,-1,4,6] Explanation: In the first query, Alice can directly move to Bob's building since heights[0] < heights[7]. In the second query, Alice and Bob can move to building 6 since heights[3] < heights[6] and heights[5] < heights[6]. In the third query, Alice cannot meet Bob since Bob cannot move to any other building. In the fourth query, Alice and Bob can move to building 4 since heights[3] < heights[4] and heights[0] < heights[4]. In the fifth query, Alice can directly move to Bob's building since heights[1] < heights[6]. 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. Constraints: 1 <= heights.length <= 5 104 1 <= heights[i] <= 109 1 <= queries.length <= 5 104 queries[i] = [ai, bi] 0 <= ai, bi <= heights.length - 1
Explanation
Here's a breakdown of the solution:
- Iterate through queries: Process each query independently to find the leftmost meeting point.
- Linear scan for each query: For each query (a, b), iterate through potential meeting points 'j' (starting from max(a,b)). Check if both Alice (from 'a') and Bob (from 'b') can reach 'j'. If so, return j immediately as it's the leftmost.
Early exit optimization: if a == b, return a immediately since they are already in the same building.
Runtime Complexity: O(Q * N), where Q is the number of queries and N is the length of the
heightsarray.- Storage Complexity: O(Q), where Q is the number of queries (to store the
ansarray).
def solve():
def leftmost_meeting_building(heights, queries):
n = len(heights)
ans = []
def can_reach(start, end, heights):
return start < end and heights[start] < heights[end]
for a, b in queries:
if a == b:
ans.append(a)
continue
meeting_point = -1
for j in range(max(a, b), n):
alice_can_reach = can_reach(a, j, heights)
bob_can_reach = can_reach(b, j, heights)
if alice_can_reach and bob_can_reach:
meeting_point = j
break
elif a == j and can_reach(b, j, heights):
meeting_point = j
break
elif b == j and can_reach(a, j, heights):
meeting_point = j
break
ans.append(meeting_point)
return ans
heights = [6,4,8,5,2,7]
queries = [[0,1],[0,3],[2,4],[3,4],[2,2]]
print(leftmost_meeting_building(heights, queries))
heights = [5,3,8,2,6,1,4,6]
queries = [[0,7],[3,5],[5,2],[3,0],[1,6]]
print(leftmost_meeting_building(heights, queries))
def leftmost_building_where_alice_and_bob_can_meet(heights, queries):
n = len(heights)
ans = []
def can_reach(start, end, heights):
return start < end and heights[start] < heights[end]
for a, b in queries:
if a == b:
ans.append(a)
continue
meeting_point = -1
for j in range(max(a, b), n):
alice_can_reach = can_reach(a, j, heights)
bob_can_reach = can_reach(b, j, heights)
if alice_can_reach and bob_can_reach:
meeting_point = j
break
elif a == j and can_reach(b, j, heights):
meeting_point = j
break
elif b == j and can_reach(a, j, heights):
meeting_point = j
break
ans.append(meeting_point)
return ans
# Code
```python
def leftmost_building_where_alie_and_bob_an_meet(heights, queries): n = len(heights)
ans = []
def can_reach(start, end, heights):
return start < end and heights[start] < heights[end]
for a, b in queries:
if a == b:
ans.append(a)
continue
meeting_point = -1
for j in range(max(a, b), n):
alice_can_reach = can_reach(a, j, heights)
bob_can_reach = can_reach(b, j, heights)
if alice_can_reach and bob_can_reach:
meeting_point = j
break
elif a == j and can_reach(b, j, heights):
meeting_point = j
break
elif b == j and can_reach(a, j, heights):
meeting_point = j
break
ans.append(meeting_point)
return ans