LeetCode #1642 — MEDIUM

Furthest Building You Can Reach

Move from brute-force thinking to an efficient approach using array strategy.

Solve on LeetCode

The Problem

Problem Statement

You are given an integer array heights representing the heights of buildings, some bricks, and some ladders.

You start your journey from building 0 and move to the next building by possibly using bricks or ladders.

While moving from building i to building i+1 (0-indexed),

If the current building's height is greater than or equal to the next building's height, you do not need a ladder or bricks.
If the current building's height is less than the next building's height, you can either use one ladder or (h[i+1] - h[i]) bricks.

Return the furthest building index (0-indexed) you can reach if you use the given ladders and bricks optimally.

Example 1:

Input: heights = [4,2,7,6,9,14,12], bricks = 5, ladders = 1
Output: 4
Explanation: Starting at building 0, you can follow these steps:
- Go to building 1 without using ladders nor bricks since 4 >= 2.
- Go to building 2 using 5 bricks. You must use either bricks or ladders because 2 < 7.
- Go to building 3 without using ladders nor bricks since 7 >= 6.
- Go to building 4 using your only ladder. You must use either bricks or ladders because 6 < 9.
It is impossible to go beyond building 4 because you do not have any more bricks or ladders.

Example 2:

Input: heights = [4,12,2,7,3,18,20,3,19], bricks = 10, ladders = 2
Output: 7

Example 3:

Input: heights = [14,3,19,3], bricks = 17, ladders = 0
Output: 3

Constraints:

1 <= heights.length <= 10⁵
1 <= heights[i] <= 10⁶
0 <= bricks <= 10⁹
0 <= ladders <= heights.length

Step 01

Brute Force Baseline

Problem summary: You are given an integer array heights representing the heights of buildings, some bricks, and some ladders. You start your journey from building 0 and move to the next building by possibly using bricks or ladders. While moving from building i to building i+1 (0-indexed), If the current building's height is greater than or equal to the next building's height, you do not need a ladder or bricks. If the current building's height is less than the next building's height, you can either use one ladder or (h[i+1] - h[i]) bricks. Return the furthest building index (0-indexed) you can reach if you use the given ladders and bricks optimally.

Baseline thinking

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

Pattern signal: Array · Greedy

Example 1

[4,2,7,6,9,14,12]
5
1

Example 2

[4,12,2,7,3,18,20,3,19]
10
2

Example 3

[14,3,19,3]
17
0

Core Insight

What unlocks the optimal approach

Assume the problem is to check whether you can reach the last building or not.
You'll have to do a set of jumps, and choose for each one whether to do it using a ladder or bricks. It's always optimal to use ladders in the largest jumps.
Iterate on the buildings, maintaining the largest r jumps and the sum of the remaining ones so far, and stop whenever this sum exceeds b.

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

Upper-end input sizes

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 #1642: Furthest Building You Can Reach
class Solution {
    public int furthestBuilding(int[] heights, int bricks, int ladders) {
        PriorityQueue<Integer> q = new PriorityQueue<>();
        int n = heights.length;
        for (int i = 0; i < n - 1; ++i) {
            int a = heights[i], b = heights[i + 1];
            int d = b - a;
            if (d > 0) {
                q.offer(d);
                if (q.size() > ladders) {
                    bricks -= q.poll();
                    if (bricks < 0) {
                        return i;
                    }
                }
            }
        }
        return n - 1;
    }
}

// Accepted solution for LeetCode #1642: Furthest Building You Can Reach
func furthestBuilding(heights []int, bricks int, ladders int) int {
	q := hp{}
	n := len(heights)
	for i, a := range heights[:n-1] {
		b := heights[i+1]
		d := b - a
		if d > 0 {
			heap.Push(&q, d)
			if q.Len() > ladders {
				bricks -= heap.Pop(&q).(int)
				if bricks < 0 {
					return i
				}
			}
		}
	}
	return n - 1
}

type hp struct{ sort.IntSlice }

func (h *hp) Push(v any) { h.IntSlice = append(h.IntSlice, v.(int)) }
func (h *hp) Pop() any {
	a := h.IntSlice
	v := a[len(a)-1]
	h.IntSlice = a[:len(a)-1]
	return v
}

# Accepted solution for LeetCode #1642: Furthest Building You Can Reach
class Solution:
    def furthestBuilding(self, heights: List[int], bricks: int, ladders: int) -> int:
        h = []
        for i, a in enumerate(heights[:-1]):
            b = heights[i + 1]
            d = b - a
            if d > 0:
                heappush(h, d)
                if len(h) > ladders:
                    bricks -= heappop(h)
                    if bricks < 0:
                        return i
        return len(heights) - 1

// Accepted solution for LeetCode #1642: Furthest Building You Can Reach
struct Solution;

use std::cmp::Reverse;
use std::collections::BinaryHeap;

impl Solution {
    fn furthest_building(heights: Vec<i32>, mut bricks: i32, mut ladders: i32) -> i32 {
        let n = heights.len();
        let mut replacement: BinaryHeap<Reverse<i32>> = BinaryHeap::new();
        for i in 1..n {
            if heights[i - 1] < heights[i] {
                replacement.push(Reverse(heights[i] - heights[i - 1]));
                if ladders > 0 {
                    ladders -= 1;
                } else {
                    let min = replacement.pop().unwrap().0;
                    if min <= bricks {
                        bricks -= min;
                    } else {
                        return (i - 1) as i32;
                    }
                }
            }
        }
        (n - 1) as i32
    }
}

#[test]
fn test() {
    let heights = vec![4, 2, 7, 6, 9, 14, 12];
    let bricks = 5;
    let ladders = 1;
    let res = 4;
    assert_eq!(Solution::furthest_building(heights, bricks, ladders), res);
    let heights = vec![4, 12, 2, 7, 3, 18, 20, 3, 19];
    let bricks = 10;
    let ladders = 2;
    let res = 7;
    assert_eq!(Solution::furthest_building(heights, bricks, ladders), res);
    let heights = vec![14, 3, 19, 3];
    let bricks = 17;
    let ladders = 0;
    let res = 3;
    assert_eq!(Solution::furthest_building(heights, bricks, ladders), res);
}

// Accepted solution for LeetCode #1642: Furthest Building You Can Reach
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #1642: Furthest Building You Can Reach
// class Solution {
//     public int furthestBuilding(int[] heights, int bricks, int ladders) {
//         PriorityQueue<Integer> q = new PriorityQueue<>();
//         int n = heights.length;
//         for (int i = 0; i < n - 1; ++i) {
//             int a = heights[i], b = heights[i + 1];
//             int d = b - a;
//             if (d > 0) {
//                 q.offer(d);
//                 if (q.size() > ladders) {
//                     bricks -= q.poll();
//                     if (bricks < 0) {
//                         return i;
//                     }
//                 }
//             }
//         }
//         return n - 1;
//     }
// }

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 log n)

Space

O(1)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

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.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.