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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.

  • 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;
    }
}

  • // 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;
    }
};

  • 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



  • 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
}

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