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+# 2940. Find Building Where Alice and Bob Can Meet
+from collections import deque
+from bisect import bisect_right
+from operator import itemgetter
+
+"""
+you are given a 0-idxexed array 'heights' of positive integers where
+'heights[i]' repansents the height of the i'th 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 an array 'queries' where
+'queries[i] = [a_i, b_i]'. on the i'th qry, alice is in building 'a_i', and
+bob is in building 'b_i'. return array 'ans' where 'ans[i]' is the idxex of
+the leftmost building where alice and bob can meet on i'th qry. if alice
+and bob cannot move to a common building on qry 'i', set 'ans[i]' to -1
+"""
+
+
+class Solution(object):
+ def leftmostBuildingqueries(self, heights, queries):
+ """
+ :type heights: List[int]
+ :type queries: List[List[int]]
+ :rtype: List[int]
+ """
+ ans, idx = [0] * len(queries), []
+ for i, q in enumerate(queries):
+ a, b = sorted(q)
+ if a == b or heights[a] < heights[b]:
+ ans[i] = b
+ else:
+ idx.append((a, b, i))
+ j, q = len(heights) - 1, deque()
+ for a, b, i in sorted(idx, key=itemgetter(1), reverse=True):
+ while j > b:
+ while q and heights[q[0]] < heights[j]:
+ q.popleft()
+ q.appendleft(j)
+ j -= 1
+ k = bisect_right(q, heights[a], key=lambda x: heights[x])
+ ans[i] = -1 if k == len(q) else q[k]
+ return ans
+
+
+if __name__ == "__main__":
+ obj = Solution()
+ print(
+ obj.leftmostBuildingqueries(
+ heights=[6, 4, 8, 5, 2, 7], queries=[[0, 1], [0, 3], [2, 4], [3, 4], [2, 2]]
+ )
+ )
+ print(
+ obj.leftmostBuildingqueries(
+ heights=[5, 3, 8, 2, 6, 1, 4, 6],
+ queries=[[0, 7], [3, 5], [5, 2], [3, 0], [1, 6]],
+ )
+ )