Find Building Where Alice and Bob Can Meet

class Solution {
public:
    vector<int> processBuildingQueries(vector<int>& elevations,
                                       vector<vector<int>>& queries) {
        vector<vector<vector<int>>> indexedQueries(elevations.size());
        priority_queue<vector<int>, vector<vector<int>>, greater<vector<int>>>
            minHeap;
        vector<int> results(queries.size(), -1);

        // Processing each query and storing them in indexedQueries.
        for (int i = 0; i < queries.size(); i++) {
            int start = queries[i][0], end = queries[i][1];
            if (start < end && elevations[start] < elevations[end]) {
                results[i] = end;
            } else if (start > end && elevations[start] > elevations[end]) {
                results[i] = start;
            } else if (start == end) {
                results[i] = start;
            } else {
                indexedQueries[max(start, end)].push_back(
                    {max(elevations[start], elevations[end]), i});
            }
        }

        // Processing each height in elevations against the queries.
        for (int j = 0; j < elevations.size(); j++) {
            while (!minHeap.empty() && minHeap.top()[0] < elevations[j]) {
                results[minHeap.top()[1]] = j;
                minHeap.pop();
            }
            for (auto& query : indexedQueries[j]) {
                minHeap.push(query);
            }
        }
        return results;
    }
};

The provided solution in C++ addresses a problem where you need to find an optimal building for Alice and Bob to meet, given building heights and a range of building indices they might consider meeting at. The solution leverages several data structures and a priority queue to efficiently tackle this problem.

• A two-dimensional vector indexedQueries is used to relate height queries to building indices.
• A priority queue minHeap is employed to process the incoming queries in a way that always considers the smallest elevation.

Follow these steps to understand how the solution handles the input queries:

1. Iterate over each query to determine quick outcomes based on simple conditions, such as when Alice’s and Bob’s starting positions or the relative elevation differences point directly to an optimal solution.

2. For queries not resolved by the initial pass (those where neither end is immediately decisive), store potential buildings in indexedQueries for later resolution.

3. Traverse each building's height and use minHeap to compare current elevations with outstanding queries. If the current building height exceeds the height needed by a query, resolve that query with the current building index.

4. Continue to process queries from the indexed list for each building, ensuring that all considerations for minimum building height constraints are handled as the loop progresses.

The main result is a vector where each position corresponds to the original query array's indices and contains the index of the building where Alice and Bob could potentially meet. The -1 value indicates no suitable building was found for that query. By organizing and evaluating the queries efficiently via the priority queue mechanism and direct checks, this solution aims to minimize the operational complexity, achieving dynamic handling of conditions with varying requirements for both immediate and deferred query resolution.

class Solution {

    public int[] processQueries(int[] heights, int[][] queries) {
        List<List<List<Integer>>> queriesStore = new ArrayList<>(heights.length);
        for (int i = 0; i < heights.length; i++) queriesStore.add(new ArrayList<>());
        PriorityQueue<List<Integer>> maxHeap = new PriorityQueue<>((x, y) -> x.get(0) - y.get(0));
        int[] answer = new int[queries.length];
        Arrays.fill(answer, -1);

        for (int qIdx = 0; qIdx < queries.length; qIdx++) {
            int pos1 = queries[qIdx][0], pos2 = queries[qIdx][1];
            if (pos1 < pos2 && heights[pos1] < heights[pos2]) {
                answer[qIdx] = pos2;
            } else if (pos1 > pos2 && heights[pos1] > heights[pos2]) {
                answer[qIdx] = pos1;
            } else if (pos1 == pos2) {
                answer[qIdx] = pos1;
            } else {
                queriesStore.get(Math.max(pos1, pos2)).add(Arrays.asList(Math.max(heights[pos1], heights[pos2]), qIdx));
            }
        }

        for (int idx = 0; idx < heights.length; idx++) {
            while (!maxHeap.isEmpty() && maxHeap.peek().get(0) < heights[idx]) {
                answer[maxHeap.peek().get(1)] = idx;
                maxHeap.poll();
            }
            for (List<Integer> ele : queriesStore.get(idx)) {
                maxHeap.offer(ele);
            }
        }
        return answer;
    }
}

This Java solution focuses on addressing the "Find Building Where Alice and Bob Can Meet" problem by analyzing the heights and positions to determine the best meeting point. The processQueries method receives two parameters: heights, an array representing the heights of buildings, and queries, a 2D array where each query includes two positions representing buildings Alice and Bob start from.

Here's an outline of the solution approach:

• Initialize a dynamic storage (queriesStore) to hold potential queries and a maxHeap to manage the active greatest height queries for efficiency.
• Iterate over each query to assess direct comparisons:
  • If Alice's position is lower than Bob's, and her building is shorter, then Bob's position is stored as the meeting point.
  • If Alice's position is higher and her building is taller, then her position is stored as the meeting point.
  • If their positions are identical, that position is immediately considered the meeting point.
• If none of the above conditions is met, the query's information is stored in queriesStore.

The next component of the method:

• Iterates over each building:
  • Checks maxHeap to find if the top building height is less than the current building's height and updates the answer for the queries accordingly.
  • New potential highest buildings are added to maxHeap from queriesStore.

By leveraging PriorityQueue and dynamic list storage, this approach ensures the method efficiently handles scenarios where multiple potential meeting buildings need assessing over a range of queries. The answer for each query is stored in the answer array where direct answers are populated instantly, and others determined through comparative logic leveraging heap properties for maximum height determination. This method returns the final answer array, indicating the optimal meeting buildings for Alice and Bob for each query.

class BuildingInspector:
    def findLeftmostBuilding(self, building_heights, search_queries):
        priority_queue = []
        query_results = [-1] * len(search_queries)
        query_associations = [[] for _ in building_heights]

        for index, query in enumerate(search_queries):
            start, end = query
            if start < end and building_heights[start] < building_heights[end]:
                query_results[index] = end
            elif start > end and building_heights[start] > building_heights[end]:
                query_results[index] = start
            elif start == end:
                query_results[index] = start
            else:
                query_associations[max(start, end)].append(
                    (max(building_heights[start], building_heights[end]), index)
                )

        for idx, height in enumerate(building_heights):
            while priority_queue and priority_queue[0][0] < height:
                _, query_idx = heapq.heappop(priority_queue)
                query_results[query_idx] = idx
            for item in query_associations[idx]:
                heapq.heappush(priority_queue, item)

        return query_results

The provided Python solution focuses on finding the leftmost building that satisfies certain conditions for given pairs of building indexes, called search queries. This is done using the BuildingInspector class with the method findLeftmostBuilding.

Here's how the method operates:

• Initializes a priority queue to handle the logic of determining the leftmost building based on conditions defined in the search queries.
• Maintains an array query_results to store the results of each query, initialized with -1 indicating unresolved queries.
• Creates a list of list query_associations which stores associations between building indices and search queries.

The method does the following:

1. Iterates over each search query (a pair of start and end indices) along with their respective indices:

   • Directly determines and assigns the appropriate building depending on specific conditions like relative positions and heights of the start and end buildings.
   • In cases where starting indices have the same position or do not immediately satisfy the condition, the method makes associations for further processing.

2. Iterates over each building by its index and height:

   • Resolves queries using the priority queue when a building's height surpasses conditions stored in the priority queue.
   • Pushes unresolved queries into the priority queue based on their associated conditions from query_associations.

The method finally returns the query_results, where each index corresponds to the result of the respective search query in the input. The solution efficiently handles multiple queries using priority queues to keep track of and resolve conditions based on building heights and indices. This approach ensures that even with a complex set of conditions, the building indices that satisfy search queries are determined optimally.