This approach uses a greedy strategy combined with a max-heap (priority queue) to determine the optimal use of bricks and ladders. The idea is to always use bricks first for the smallest height difference and use ladders for the largest, potentially reserving the ladders for future taller buildings. Whenever we encounter a new jump (difference in height), we add it to the heap, and if the total number of jumps exceeds the available ladders, we use bricks for the smallest jump. If we don’t have enough bricks, we can’t proceed further.
Time Complexity: O(n log n) due to sorting.
Space Complexity: O(n) for storing height differences.
1function furthestBuilding(heights, bricks, ladders) {
2 const minHeap = [];
3 function heapPush(val) {
4 minHeap.push(val);
5 minHeap.sort((a, b) => a - b);
6 }
7 function heapPop() {
8 return minHeap.shift();
9 }
10 for (let i = 0; i < heights.length - 1; i++) {
11 let diff = heights[i + 1] - heights[i];
12 if (diff > 0) {
13 heapPush(diff);
14 if (minHeap.length > ladders) {
15 bricks -= heapPop();
16 if (bricks < 0) return i;
17 }
18 }
19 }
20 return heights.length - 1;
21}
22
23// Example usage
24console.log(furthestBuilding([4, 2, 7, 6, 9, 14, 12], 5, 1)); // Output: 4
25This JavaScript solution simulates a min-heap using an array with sorting operations. It pushes height differences into the array, sorts, and uses minimal climbs when the ladder capacity is surpassed, demonstrating efficient climb management like heaps in other languages.
This approach employs binary search to find the furthest building we can reach. The binary search is performed over the indices of the buildings (0 to n-1). For each index midpoint in the search, we check if it's possible to reach that building, given the constraints of available bricks and ladders using a helper function that checks feasibility by simulating resource allocation greedily. The search continually refines the possible maximum index based on resource ability.
Time Complexity: O(n log n) because of binary search with operations similar to heap for each midpoint.
Space Complexity: O(n) to store the min-heap simulation for each search step.
1import heapq
2
3def can_reach(heights, bricks, ladders, target):
4 min_heap = []
5 for i in range(target):
6 diff = heights[i + 1] - heights[i]
7 if diff > 0:
8 heapq.heappush(min_heap, diff)
9 if len(min_heap) > ladders:
10 bricks -= heapq.heappop(min_heap)
11 if bricks < 0:
12 return False
13 return True
14
15def furthest_building(heights, bricks, ladders):
16 low, high = 0, len(heights) - 1
17 result = 0
18 while low <= high:
19 mid = (low + high) // 2
20 if can_reach(heights, bricks, ladders, mid):
21 result = mid
22 low = mid + 1
23 else:
24 high = mid - 1
25 return result
26
27# Example usage
28heights = [4, 12, 2, 7, 3, 18, 20, 3, 19]
29print(furthest_building(heights, 10, 2)) # Output: 7
30Applying binary search in this Python solution effectively searches for the maximum index of reachable buildings using a helper function, can_reach, which factors in available ladders and bricks via a min-heap metaphor. This strategy balances resource usage constraints at each binary search stage.