Minimum Cost to Connect Sticks - LeetCode
You have some number of sticks with positive integer lengths. These lengths are given as an array sticks, where sticks[i] is the length of the ith stick. You can connect any two sticks of lengths x and y into one stick by paying a cost of x + y. You must connect all the sticks until there is only one stick remaining. Return the minimum cost of connecting all the given sticks into one stick in this way.
Example 1: Input: sticks = [2,4,3] Output: 14 Explanation: You start with sticks = [2,4,3]. 1. Combine sticks 2 and 3 for a cost of 2 + 3 = 5. Now you have sticks = [5,4]. 2. Combine sticks 5 and 4 for a cost of 5 + 4 = 9. Now you have sticks = [9]. There is only one stick left, so you are done. The total cost is 5 + 9 = 14. Example 2: Input: sticks = [1,8,3,5] Output: 30 Explanation: You start with sticks = [1,8,3,5]. 1. Combine sticks 1 and 3 for a cost of 1 + 3 = 4. Now you have sticks = [4,8,5]. 2. Combine sticks 4 and 5 for a cost of 4 + 5 = 9. Now you have sticks = [9,8]. 3. Combine sticks 9 and 8 for a cost of 9 + 8 = 17. Now you have sticks = [17]. There is only one stick left, so you are done. The total cost is 4 + 9 + 17 = 30. Example 3: Input: sticks = [5] Output: 0 Explanation: There is only one stick, so you don’t need to do anything. The total cost is 0.
Constraints:
1 <= sticks.length <= 104
1 <= sticks[i] <= 104
- code
class Solution:
def connectSticks(self, sticks: List[int]) -> int:
if len(sticks) == 1: return 0
heapq.heapify(sticks)
res = 0
while len(sticks) > 1:
last, cur = heapq.heappop(sticks), heapq.heappop(sticks)
res += cur + last
heapq.heappush(sticks, last + cur)
return res