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Formatted question description: https://leetcode.ca/all/1167.html

# 1167. Minimum Cost to Connect Sticks

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
}

• function connectSticks(sticks: number[]): number {
const pq = new Heap(sticks);
let ans = 0;
while (pq.size() > 1) {
const x = pq.pop();
const y = pq.pop();
ans += x + y;
pq.push(x + y);
}
return ans;
}

type Compare<T> = (lhs: T, rhs: T) => number;

class Heap<T = number> {
data: Array<T | null>;
lt: (i: number, j: number) => boolean;
constructor();
constructor(data: T[]);
constructor(compare: Compare<T>);
constructor(data: T[], compare: Compare<T>);
constructor(data: T[] | Compare<T>, compare?: (lhs: T, rhs: T) => number);
constructor(
data: T[] | Compare<T> = [],
compare: Compare<T> = (lhs: T, rhs: T) =>
lhs < rhs ? -1 : lhs > rhs ? 1 : 0,
) {
if (typeof data === 'function') {
compare = data;
data = [];
}
this.data = [null, ...data];
this.lt = (i, j) => compare(this.data[i]!, this.data[j]!) < 0;
for (let i = this.size(); i > 0; i--) this.heapify(i);
}

size(): number {
return this.data.length - 1;
}

push(v: T): void {
this.data.push(v);
let i = this.size();
while (i >> 1 !== 0 && this.lt(i, i >> 1)) this.swap(i, (i >>= 1));
}

pop(): T {
this.swap(1, this.size());
const top = this.data.pop();
this.heapify(1);
}

top(): T {
return this.data[1]!;
}
heapify(i: number): void {
while (true) {
let min = i;
const [l, r, n] = [i * 2, i * 2 + 1, this.data.length];
if (l < n && this.lt(l, min)) min = l;
if (r < n && this.lt(r, min)) min = r;
if (min !== i) {
this.swap(i, min);
i = min;
} else break;
}
}

clear(): void {
this.data = [null];
}

private swap(i: number, j: number): void {
const d = this.data;
[d[i], d[j]] = [d[j], d[i]];
}
}