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Formatted question description: https://leetcode.ca/all/1167.html
1167. Minimum Cost to Connect Sticks
Level
Medium
Description
You have some sticks
with positive integer lengths.
You can connect any two sticks of lengths X
and Y
into one stick by paying a cost of X + Y
. You perform this action until there is 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
Example 2:
Input: sticks = [1,8,3,5]
Output: 30
Constraints:
1 <= sticks.length <= 10^4
1 <= sticks[i] <= 10^4
Solution
Use the greedy approach. Use a priority queue to store all elements in sticks
, where the minimum element is polled from the priority queue each time. While the priority queue contains at least 2 elements, poll 2 elements, calculate the sum, add the sum to the cost, and offer the sum to the priority queue. Repeat the process until the priority queue only contains 1 element. Then return the cost.
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class Solution { public int connectSticks(int[] sticks) { PriorityQueue<Integer> priorityQueue = new PriorityQueue<Integer>(); for (int stick : sticks) priorityQueue.offer(stick); int cost = 0; while (priorityQueue.size() > 1) { int stick1 = priorityQueue.poll(); int stick2 = priorityQueue.poll(); int sum = stick1 + stick2; cost += sum; priorityQueue.offer(sum); } return cost; } }
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// OJ: https://leetcode.com/problems/minimum-cost-to-connect-sticks/ // Time: O(NlogN) // Space: O(N) class Solution { public: int connectSticks(vector<int>& A) { priority_queue<int, vector<int>, greater<>> pq; for (int n : A) pq.push(n); int ans = 0; while (pq.size() > 1) { int a = pq.top(); pq.pop(); int b = pq.top(); pq.pop(); pq.push(a + b); ans += a + b; } return ans; } };
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class Solution: def connectSticks(self, sticks: List[int]) -> int: h = [] for s in sticks: heappush(h, s) res = 0 while len(h) > 1: val = heappop(h) + heappop(h) res += val heappush(h, val) return res
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func connectSticks(sticks []int) int { h := IntHeap(sticks) heap.Init(&h) res := 0 for h.Len() > 1 { val := heap.Pop(&h).(int) val += heap.Pop(&h).(int) res += val heap.Push(&h, val) } return res } type IntHeap []int func (h IntHeap) Len() int { return len(h) } func (h IntHeap) Less(i, j int) bool { return h[i] < h[j] } func (h IntHeap) Swap(i, j int) { h[i], h[j] = h[j], h[i] } func (h *IntHeap) Push(x interface{}) { *h = append(*h, x.(int)) } func (h *IntHeap) Pop() interface{} { old := *h n := len(old) x := old[n-1] *h = old[0 : n-1] return x }