1642

1642. Furthest Building You Can Reach (Medium)

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

Related Topics:
Binary Search, Heap

Solution 1. Min heap

Firstly we only consider positive diffs, and ignore non-positive ones.

Secondly, we use brick for small diffs, and ladder for large diffs.

We can turn the original array into the diffs.

// original array
[4,2,7,6,9,14,12] 
// diff array
[0,5,0,3,5,0,x]

Now the problem becomes "find the maximum index i that sum( A[0]...A[i] ) - sum( top L elements ) <= B.

We can use a min heap to keep track of the top L items.

// OJ: https://leetcode.com/problems/furthest-building-you-can-reach/
// Author: github.com/lzl124631x
// Time: O(NlogL)
// Space: O(L)
class Solution {
public:
    int furthestBuilding(vector<int>& A, int B, int L) {
        long N = A.size(), sum = 0, sumTop = 0;
        for (int i = 0; i < N - 1; ++i) { // convert to diff array
            A[i] = max(0, A[i + 1] - A[i]);
        }
        priority_queue<int, vector<int>, greater<>> pq; // min-heap storing the top L diffs
        for (int i = 0; i < N; ++i) {
            sum += A[i];
            sumTop += A[i];
            pq.push(A[i]);
            if (pq.size() > L) {
                sumTop -= pq.top();
                pq.pop();
            }
            if (sum - sumTop > B) return i;
        }
        return N - 1;
    }
};

Or

// OJ: https://leetcode.com/problems/furthest-building-you-can-reach/
// Author: github.com/lzl124631x
// Time: O(NlogL)
// Space: O(L)
class Solution {
public:
    int furthestBuilding(vector<int>& H, int B, int L) {
        int N = H.size(), sum = 0; // sum is the sum of the diffs that we want to use bricks to handle
        priority_queue<int, vector<int>, greater<>> pq; // min-heap, keep track of the L largest diffs
        for (int i = 1; i < N; ++i) {
            int diff = H[i] - H[i - 1]; // compute diff on the fly
            if (diff <= 0) continue; // ignore negative diffs
            pq.push(diff);
            if (pq.size() <= L) continue; // doesn't need any brick, continue
            sum += pq.top();
            pq.pop();
            if (sum > B) return i - 1;
        }
        return N - 1;
    }
};