What is the time complexity of Find Building Where Alice and Bob Can Meet?

The optimal solution has a time complexity of O(n log n + q log n).

What is the space complexity of Find Building Where Alice and Bob Can Meet?

The space complexity is O(n + q).

What data structures are used to solve Find Building Where Alice and Bob Can Meet?

This problem uses Array, Binary Search, Stack, Binary Indexed Tree, Segment Tree, Heap (Priority Queue), Monotonic Stack. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Find Building Where Alice and Bob Can Meet?

Insight 1: Sort all possible heights and the threshold heights to allow binary search, improving efficiency over iterating through all heights. Insight 2: Use a segment tree to efficiently find the leftmost building index that satisfies the height condition. This avoids brute-force checking of each building.

Which companies ask LeetCode 2940. Find Building Where Alice and Bob Can Meet in interviews?

This problem is frequently asked at Infosys.

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Find Building Where Alice and Bob Can Meet - LeetCode 2940 Solution

Q: How to solve Find Building Where Alice and Bob Can Meet?

The solution employs a segment tree and binary search to efficiently find the meeting point for each query. The key idea is to first handle trivial cases, then group the queries based on the maximum index of the building pair. We sort all the distinct heights and threshold heights for queries. We then iterate through the buildings in reverse order, update the segment tree with each building's index based on its height rank, and process the queries associated with each building to find the leftmost meeting point.

Find Building Where Alice and Bob Can Meet - Complete Solution Guide

Find Building Where Alice and Bob Can Meet is LeetCode problem 2940, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given a 0-indexed array heights of positive integers, where heights[i] represents 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 another array queries where queries[i] = [a i , b i ] . On the i th query, Alice is in building a i while Bob is in building b i . Return an array ans where ans[i] is the index of the leftmost building where Alice and Bob can meet on the

Detailed Explanation

The problem asks us to find, for each query (a, b), the leftmost building index 'k' such that heights[a] < heights[k] and heights[b] < heights[k]. In other words, we need to find a meeting point 'k' to the right of both building 'a' and building 'b', where both Alice (starting at building 'a') and Bob (starting at building 'b') can move. If no such building exists, we return -1. If a and b are the same we return a, if a < b and heights[a] < heights[b], then the meeting point is b. We are given an array 'heights' representing the heights of the buildings and an array 'queries', where each query consists of a pair of building indices (a, b).

Solution Approach

The solution employs a segment tree and binary search to efficiently find the meeting point for each query. The key idea is to first handle trivial cases, then group the queries based on the maximum index of the building pair. We sort all the distinct heights and threshold heights for queries. We then iterate through the buildings in reverse order, update the segment tree with each building's index based on its height rank, and process the queries associated with each building to find the leftmost meeting point.

Step-by-Step Algorithm

Step 1: Initialize the answer array with -1. Also, create lists to group queries according to the larger of the two indices in each query. This optimization helps processing queries efficiently.
Step 2: Handle trivial cases: If a == b, the answer is a. If heights[min(a,b)] < heights[max(a,b)] the answer is max(a,b).
Step 3: Store unique heights from the `heights` array and any threshold heights derived from the queries into a set. Sort this set to enable binary search later.
Step 4: Build a segment tree. The segment tree will store the minimum index for a range of sorted height indices.
Step 5: Iterate backwards through the buildings. This allows us to build the segment tree from right to left.
Step 6: For each building, process any queries associated with that building. This involves binary searching to find the correct index into sorted_h for the threshold height. Then, query the segment tree to find the leftmost building index greater than this threshold height.
Step 7: Update the segment tree with the current building's height and index. This ensures that we maintain the correct information for future queries.
Step 8: Return the answer array.

Key Insights

Insight 1: Sort all possible heights and the threshold heights to allow binary search, improving efficiency over iterating through all heights.
Insight 2: Use a segment tree to efficiently find the leftmost building index that satisfies the height condition. This avoids brute-force checking of each building.
Insight 3: Preprocess the queries by grouping them according to the taller building index. This allows us to traverse buildings in reverse order and apply the segment tree updates in the correct sequence.

Complexity Analysis

Time Complexity: O(n log n + q log n)

Space Complexity: O(n + q)

Topics

This problem involves: Array, Binary Search, Stack, Binary Indexed Tree, Segment Tree, Heap (Priority Queue), Monotonic Stack.

Companies

Asked at: Infosys.

Find Building Where Alice and Bob Can Meet | LeetCode 2940

Hard

Array Binary Search Stack Binary Indexed Tree Segment Tree Heap (Priority Queue)Monotonic Stack

Time: O(n log n + q log n)Space: O(n + q)

Solution Code

Available in:

import java.util.*;

class Solution {
    public int[] leftmostBuildingQueries(int[] heights, int[][] queries) {
        int n = heights.length;
        int q = queries.length;
        int[] ans = new int[q];
        Arrays.fill(ans, -1);

        List<List<Pair<Integer, Integer>>> q_groups = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            q_groups.add(new ArrayList<>());
        }

        Set<Integer> all_heights_set = new HashSet<>();

        for (int i = 0; i < q; i++) {
            int a = queries[i][0];
            int b = queries[i][1];

            if (a == b) {
                ans[i] = a;
                continue;
            }

            int max_idx = (a > b) ? a : b;
            int min_idx = (a > b) ? b : a;

            if (heights[min_idx] < heights[max_idx]) {
                ans[i] = max_idx;
                continue;
            }

            int h_thresh = Math.max(heights[a], heights[b]);
            q_groups.get(max_idx).add(new Pair<>(h_thresh, i));
            all_heights_set.add(h_thresh);
        }

        List<Integer> sorted_h = new ArrayList<>(all_heights_set);
        Collections.sort(sorted_h);
        Map<Integer, Integer> h_map = new HashMap<>();
        for (int i = 0; i < sorted_h.size(); i++) {
            h_map.put(sorted_h.get(i), i);
        }
        int m = sorted_h.size();

        if (m == 0) {
            return ans;
        }

        int[] seg_tree = new int[2 * m];
        Arrays.fill(seg_tree, n);

        class SegTree {
            public void update(int p, int val) {
                p += m;
                seg_tree[p] = Math.min(seg_tree[p], val);
                while (p > 1) {
                    p /= 2;
                    seg_tree[p] = Math.min(seg_tree[p * 2], seg_tree[p * 2 + 1]);
                }
            }

            public int query(int l, int r) { // range [l, r)
                int res = n;
                l += m;
                r += m;
                while (l < r) {
                    if ((l & 1) == 1) {
                        res = Math.min(res, seg_tree[l]);
                        l += 1;
                    }
                    if ((r & 1) == 1) {
                        r -= 1;
                        res = Math.min(res, seg_tree[r]);
                    }
                    l /= 2;
                    r /= 2;
                }
                return res;
            }
        }

        SegTree st = new SegTree();

        for (int i = n - 1; i >= 0; i--) {
            for (Pair<Integer, Integer> pair : q_groups.get(i)) {
                int h_thresh = pair.getKey();
                int q_idx = pair.getValue();
                
                int p = binarySearchRight(sorted_h, h_thresh);
                
                if (p < m) {
                    int min_k = st.query(p, m);
                    if (min_k < n) {
                        ans[q_idx] = min_k;
                    }
                }
            }

            int h_rank = h_map.get(heights[i]);
            st.update(h_rank, i);
        }

        return ans;
    }

    private int binarySearchRight(List<Integer> sorted_h, int h_thresh) {
        int low = 0;
        int high = sorted_h.size();

        while (low < high) {
            int mid = low + (high - low) / 2;
            if (sorted_h.get(mid) <= h_thresh) {
                low = mid + 1;
            } else {
                high = mid;
            }
        }

        return low;
    }

    static class Pair<K, V> {
        private final K key;
        private final V value;

        public Pair(K key, V value) {
            this.key = key;
            this.value = value;
        }

        public K getKey() {
            return key;
        }

        public V getValue() {
            return value;
        }
    }
}

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Problem Info

Description

Problem 2940: Find Building Where Alice and Bob Can Meet - Detailed explanation and solution approaches.

Topics

Array Binary Search Stack Binary Indexed Tree Segment Tree Heap (Priority Queue)Monotonic Stack

Companies

Infosys

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:2940

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

8 steps

1
Step 1: Initialize the answer array with -1. Also, create lists to group queries according to the larger of the two indices in each query. This optimization helps processing queries efficiently.
2
Step 2: Handle trivial cases: If a == b, the answer is a. If heights[min(a,b)] < heights[max(a,b)] the answer is max(a,b).
3
Step 3: Store unique heights from the `heights` array and any threshold heights derived from the queries into a set. Sort this set to enable binary search later.
4
Step 4: Build a segment tree. The segment tree will store the minimum index for a range of sorted height indices.
5
Step 5: Iterate backwards through the buildings. This allows us to build the segment tree from right to left.
6
Step 6: For each building, process any queries associated with that building. This involves binary searching to find the correct index into sorted_h for the threshold height. Then, query the segment tree to find the leftmost building index greater than this threshold height.
7
Step 7: Update the segment tree with the current building's height and index. This ensures that we maintain the correct information for future queries.
8
Step 8: Return the answer array.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n log n + q log n)

✨ Log-linear - Great for sorting algorithms!

💾

Space Complexity

O(n + q)

📝

Why?

Sorting heights contributes O(n log n). Building and querying the segment tree each take O(log n) time. The overall time complexity is dominated by the sorting step and the query processing.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Brute-Force Search

For each query (a, b), iterate through all buildings to the right of both 'a' and 'b', checking if heights[a] < heights[k] and heights[b] < heights[k].

Time: O(n*q), where n is the number of buildings and q is the number of queries.Space: O(q), for storing the answer array.

Trade-offs: This approach is simple to implement but highly inefficient for large datasets, due to its quadratic time complexity. It should only be used for very small input sizes.

✨

Key Insights

1Insight 1: Sort all possible heights and the threshold heights to allow binary search, improving efficiency over iterating through all heights.
2Insight 2: Use a segment tree to efficiently find the leftmost building index that satisfies the height condition. This avoids brute-force checking of each building.
3Insight 3: Preprocess the queries by grouping them according to the taller building index. This allows us to traverse buildings in reverse order and apply the segment tree updates in the correct sequence.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Stack

LIFO structure - Last In, First Out. Like a stack of plates!

Edge Cases

Edge case 1: a == b. The solution directly handles this case by returning 'a' as the meeting point.
Edge case 2: heights[min(a, b)] < heights[max(a, b)]. The solution returns the index of the higher building.
Edge case 3: No possible meeting point. The solution correctly assigns -1 to the corresponding element in the result array.
Edge case 4: Empty list of heights. While not explicitly covered, the code handles the case where m (number of distinct heights) is zero by returning the ans, that is filled with -1 by default

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the edge cases (a == b) and heights[min(a, b)] < heights[max(a, b)] directly, which leads to unnecessary computations.

✗Common mistake 2: Incorrect segment tree implementation, resulting in incorrect query results.

✗Common mistake 3: Using linear search to find meeting points instead of binary search and segment tree, leading to a time limit exceeded error.

✗Common mistake 4: Not sorting the heights, or heights combined with threshold values which can lead to problems querying the segment tree.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Resource allocation: Given a set of resources with varying capacities, determine the best resource to allocate to two tasks based on their minimum capacity requirements.

🏦Real-world scenario 2: Meeting scheduling: Determine a suitable meeting room that can accommodate two groups of people, given constraints on the minimum room size required by each group.

🎮Real-world scenario 3: Data Analysis: Find the closest data point that is higher than a certain threshold for two different datasets. Useful in anomaly detection or trend identification.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

👨‍💻 From Experience

I encountered a variation of this problem in a system design interview. The key was recognizing the underlying pattern.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: July 31, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 14

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 7

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I avoid off-by-one errors in binary search?

Use the pattern: left = 0, right = n-1, while (left <= right), mid = left + (right - left) / 2. Decide if you need to find exact match, first occurrence, or last occurrence, and adjust your conditions accordingly. Test with arrays of size 0, 1, and 2.

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Find Building Where Alice and Bob Can Meet"?

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether binary search techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?