【面试题】找到 Alice 和 Bob 可以相遇的建筑

def find_building(heights, queries):
    n = len(heights)

    # 动态规划计算每个建筑的最远到达建筑
    farthest = list(range(n))

    for i in range(n):
        j = i + 1
        while j < n:
            if heights[j] > heights[i]:
                farthest[i] = j
            j += 1

    results = []

    # 处理每个查询
    for ai, bi in queries:
        if ai >= n or bi >= n:
            results.append(-1)
            continue

        # 找到 Alice 能到达的范围
        max_reach_alice = farthest[ai]
        for in_n in farthest[ai:]:

            if n < in_n or bi < in_n:
                break
            max_reach_alice = in_n
        print(ai, bi, max_reach_alice)
        # 找到 Bob 能到达的范围
        max_reach_bob = farthest[bi]
        for in_n in farthest[bi:]:

            if n < in_n or ai < in_n:
                break
            max_reach_bob = in_n
        print(ai, bi, max_reach_bob)

        # 判断是否可以相遇
        if max_reach_alice >= bi and max_reach_bob >= ai:
            results.append(min(max_reach_alice, max_reach_bob))
        else:
            results.append(-1)

    return results