1642. Furthest Building You Can Reach

Description

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

 

Solutions

Solution: Greedy + Binary Search

• Time complexity: O(nlogn)
• Space complexity: O(n)

 

JavaScript

/**
 * @param {number[]} heights
 * @param {number} bricks
 * @param {number} ladders
 * @return {number}
 */
const furthestBuilding = function (heights, bricks, ladders) {
  const diffs = heights.map((height, index) => {
    return Math.max((heights[index + 1] ?? 0) - height, 0);
  });

  let left = ladders;
  let right = heights.length - 1;

  while (left <= right) {
    const mid = Math.floor((left + right) / 2);
    const current = diffs.slice(0, mid);
    current.sort((a, b) => b - a);
    const sumDiff = current.slice(ladders).reduce((sum, diff) => sum + diff, 0);

    sumDiff > bricks ? (right = mid - 1) : (left = mid + 1);
  }
  return right;
};