## 1167. Minimum Cost to Connect Sticks

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

## Rust Solution

struct Solution;
use std::cmp::Reverse;
use std::collections::BinaryHeap;

impl Solution {
fn connect_sticks(sticks: Vec<i32>) -> i32 {
let mut queue: BinaryHeap<Reverse<i32>> = BinaryHeap::new();
let mut res = 0;
for stick in sticks {
queue.push(Reverse(stick));
}
while queue.len() > 1 {
let x = queue.pop().unwrap().0;
let y = queue.pop().unwrap().0;
let z = x + y;
res += z;
queue.push(Reverse(z));
}
res
}
}

#[test]
fn test() {
let sticks = vec![2, 4, 3];
let res = 14;
assert_eq!(Solution::connect_sticks(sticks), res);
let sticks = vec![1, 8, 3, 5];
let res = 30;
assert_eq!(Solution::connect_sticks(sticks), res);
}

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