LeetCode #2940 — HARD

Find Building Where Alice and Bob Can Meet

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

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 i^th 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] = [a_i, b_i]
0 <= a_i, b_i <= heights.length - 1

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Binary Search · Stack · Segment Tree

Example 1

[6,4,8,5,2,7]
[[0,1],[0,3],[2,4],[3,4],[2,2]]

Example 2

[5,3,8,2,6,1,4,6]
[[0,7],[3,5],[5,2],[3,0],[1,6]]

Core Insight

What unlocks the optimal approach

For each query <code>[x, y]</code>, if <code>x > y</code>, swap <code>x</code> and <code>y</code>. Now, we can assume that <code>x <= y</code>.
For each query <code>[x, y]</code>, if <code>x == y</code> or <code>heights[x] < heights[y]</code>, then the answer is <code>y</code> since <code>x ≤ y</code>.
Otherwise, we need to find the smallest index <code>t</code> such that <code>y < t</code> and <code>heights[x] < heights[t]</code>. Note that <code>heights[y] <= heights[x]</code>, so <code>heights[x] < heights[t]</code> is a sufficient condition.
To find index <code>t</code> for each query, sort the queries in descending order of <code>y</code>. Iterate over the queries while maintaining a monotonic stack which we can binary search over to find index <code>t</code>.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2940: Find Building Where Alice and Bob Can Meet
class BinaryIndexedTree {
    private final int inf = 1 << 30;
    private int n;
    private int[] c;

    public BinaryIndexedTree(int n) {
        this.n = n;
        c = new int[n + 1];
        Arrays.fill(c, inf);
    }

    public void update(int x, int v) {
        while (x <= n) {
            c[x] = Math.min(c[x], v);
            x += x & -x;
        }
    }

    public int query(int x) {
        int mi = inf;
        while (x > 0) {
            mi = Math.min(mi, c[x]);
            x -= x & -x;
        }
        return mi == inf ? -1 : mi;
    }
}

class Solution {
    public int[] leftmostBuildingQueries(int[] heights, int[][] queries) {
        int n = heights.length;
        int m = queries.length;
        for (int i = 0; i < m; ++i) {
            if (queries[i][0] > queries[i][1]) {
                queries[i] = new int[] {queries[i][1], queries[i][0]};
            }
        }
        Integer[] idx = new Integer[m];
        for (int i = 0; i < m; ++i) {
            idx[i] = i;
        }
        Arrays.sort(idx, (i, j) -> queries[j][1] - queries[i][1]);
        int[] s = heights.clone();
        Arrays.sort(s);
        int[] ans = new int[m];
        int j = n - 1;
        BinaryIndexedTree tree = new BinaryIndexedTree(n);
        for (int i : idx) {
            int l = queries[i][0], r = queries[i][1];
            while (j > r) {
                int k = n - Arrays.binarySearch(s, heights[j]) + 1;
                tree.update(k, j);
                --j;
            }
            if (l == r || heights[l] < heights[r]) {
                ans[i] = r;
            } else {
                int k = n - Arrays.binarySearch(s, heights[l]);
                ans[i] = tree.query(k);
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2940: Find Building Where Alice and Bob Can Meet
const inf int = 1 << 30

type BinaryIndexedTree struct {
	n int
	c []int
}

func NewBinaryIndexedTree(n int) BinaryIndexedTree {
	c := make([]int, n+1)
	for i := range c {
		c[i] = inf
	}
	return BinaryIndexedTree{n: n, c: c}
}

func (bit *BinaryIndexedTree) update(x, v int) {
	for x <= bit.n {
		bit.c[x] = min(bit.c[x], v)
		x += x & -x
	}
}

func (bit *BinaryIndexedTree) query(x int) int {
	mi := inf
	for x > 0 {
		mi = min(mi, bit.c[x])
		x -= x & -x
	}
	if mi == inf {
		return -1
	}
	return mi
}

func leftmostBuildingQueries(heights []int, queries [][]int) []int {
	n, m := len(heights), len(queries)
	for _, q := range queries {
		if q[0] > q[1] {
			q[0], q[1] = q[1], q[0]
		}
	}
	idx := make([]int, m)
	for i := range idx {
		idx[i] = i
	}
	sort.Slice(idx, func(i, j int) bool { return queries[idx[j]][1] < queries[idx[i]][1] })
	s := make([]int, n)
	copy(s, heights)
	sort.Ints(s)
	ans := make([]int, m)
	tree := NewBinaryIndexedTree(n)
	j := n - 1
	for _, i := range idx {
		l, r := queries[i][0], queries[i][1]
		for ; j > r; j-- {
			k := n - sort.SearchInts(s, heights[j]) + 1
			tree.update(k, j)
		}
		if l == r || heights[l] < heights[r] {
			ans[i] = r
		} else {
			k := n - sort.SearchInts(s, heights[l])
			ans[i] = tree.query(k)
		}
	}
	return ans
}

# Accepted solution for LeetCode #2940: Find Building Where Alice and Bob Can Meet
class BinaryIndexedTree:
    __slots__ = ["n", "c"]

    def __init__(self, n: int):
        self.n = n
        self.c = [inf] * (n + 1)

    def update(self, x: int, v: int):
        while x <= self.n:
            self.c[x] = min(self.c[x], v)
            x += x & -x

    def query(self, x: int) -> int:
        mi = inf
        while x:
            mi = min(mi, self.c[x])
            x -= x & -x
        return -1 if mi == inf else mi


class Solution:
    def leftmostBuildingQueries(
        self, heights: List[int], queries: List[List[int]]
    ) -> List[int]:
        n, m = len(heights), len(queries)
        for i in range(m):
            queries[i] = [min(queries[i]), max(queries[i])]
        j = n - 1
        s = sorted(set(heights))
        ans = [-1] * m
        tree = BinaryIndexedTree(n)
        for i in sorted(range(m), key=lambda i: -queries[i][1]):
            l, r = queries[i]
            while j > r:
                k = n - bisect_left(s, heights[j]) + 1
                tree.update(k, j)
                j -= 1
            if l == r or heights[l] < heights[r]:
                ans[i] = r
            else:
                k = n - bisect_left(s, heights[l])
                ans[i] = tree.query(k)
        return ans

// Accepted solution for LeetCode #2940: Find Building Where Alice and Bob Can Meet
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2940: Find Building Where Alice and Bob Can Meet
// class BinaryIndexedTree {
//     private final int inf = 1 << 30;
//     private int n;
//     private int[] c;
// 
//     public BinaryIndexedTree(int n) {
//         this.n = n;
//         c = new int[n + 1];
//         Arrays.fill(c, inf);
//     }
// 
//     public void update(int x, int v) {
//         while (x <= n) {
//             c[x] = Math.min(c[x], v);
//             x += x & -x;
//         }
//     }
// 
//     public int query(int x) {
//         int mi = inf;
//         while (x > 0) {
//             mi = Math.min(mi, c[x]);
//             x -= x & -x;
//         }
//         return mi == inf ? -1 : mi;
//     }
// }
// 
// class Solution {
//     public int[] leftmostBuildingQueries(int[] heights, int[][] queries) {
//         int n = heights.length;
//         int m = queries.length;
//         for (int i = 0; i < m; ++i) {
//             if (queries[i][0] > queries[i][1]) {
//                 queries[i] = new int[] {queries[i][1], queries[i][0]};
//             }
//         }
//         Integer[] idx = new Integer[m];
//         for (int i = 0; i < m; ++i) {
//             idx[i] = i;
//         }
//         Arrays.sort(idx, (i, j) -> queries[j][1] - queries[i][1]);
//         int[] s = heights.clone();
//         Arrays.sort(s);
//         int[] ans = new int[m];
//         int j = n - 1;
//         BinaryIndexedTree tree = new BinaryIndexedTree(n);
//         for (int i : idx) {
//             int l = queries[i][0], r = queries[i][1];
//             while (j > r) {
//                 int k = n - Arrays.binarySearch(s, heights[j]) + 1;
//                 tree.update(k, j);
//                 --j;
//             }
//             if (l == r || heights[l] < heights[r]) {
//                 ans[i] = r;
//             } else {
//                 int k = n - Arrays.binarySearch(s, heights[l]);
//                 ans[i] = tree.query(k);
//             }
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #2940: Find Building Where Alice and Bob Can Meet
class BinaryIndexedTree {
    private n: number;
    private c: number[];
    private inf: number = 1 << 30;

    constructor(n: number) {
        this.n = n;
        this.c = Array(n + 1).fill(this.inf);
    }

    update(x: number, v: number): void {
        while (x <= this.n) {
            this.c[x] = Math.min(this.c[x], v);
            x += x & -x;
        }
    }

    query(x: number): number {
        let mi = this.inf;
        while (x > 0) {
            mi = Math.min(mi, this.c[x]);
            x -= x & -x;
        }
        return mi === this.inf ? -1 : mi;
    }
}

function leftmostBuildingQueries(heights: number[], queries: number[][]): number[] {
    const n = heights.length;
    const m = queries.length;
    for (const q of queries) {
        if (q[0] > q[1]) {
            [q[0], q[1]] = [q[1], q[0]];
        }
    }
    const idx: number[] = Array(m)
        .fill(0)
        .map((_, i) => i);
    idx.sort((i, j) => queries[j][1] - queries[i][1]);
    const tree = new BinaryIndexedTree(n);
    const ans: number[] = Array(m).fill(-1);
    const s = [...heights];
    s.sort((a, b) => a - b);
    const search = (x: number) => {
        let [l, r] = [0, n];
        while (l < r) {
            const mid = (l + r) >> 1;
            if (s[mid] >= x) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l;
    };
    let j = n - 1;
    for (const i of idx) {
        const [l, r] = queries[i];
        while (j > r) {
            const k = n - search(heights[j]) + 1;
            tree.update(k, j);
            --j;
        }
        if (l === r || heights[l] < heights[r]) {
            ans[i] = r;
        } else {
            const k = n - search(heights[l]);
            ans[i] = tree.query(k);
        }
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O((n + m)

Space

O(n + m)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.

Breaking monotonic invariant

Wrong move: Pushing without popping stale elements invalidates next-greater/next-smaller logic.

Usually fails on: Indices point to blocked elements and outputs shift.

Fix: Pop while invariant is violated before pushing current element.