# 1167. Minimum Cost to Connect Sticks

https://leetcode.com/problems/minimum-cost-to-connect-sticks

## Description

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`

## ac: Greedy

```java
class Solution {
    public int connectSticks(int[] sticks) {
        // Edge cases
        if (sticks.length <= 1) {
            return 0;
        }
        
        PriorityQueue<Integer> q = new PriorityQueue<>();
        for (int s : sticks) {
            q.offer(s);
        }
        
        int res = 0;
        while (q.size() > 1) {
            int s1 = q.poll();
            int s2 = q.poll();
            res = res + s1 + s2;
            q.offer(s1+s2);
        }
        
        return res;
    }
}
// O(NlogN) time, O(N) space.
```
