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 <= 1041 <= sticks[i] <= 104
Companies:
Amazon
Related Topics:
Greedy
Similar Questions:
// OJ: https://leetcode.com/problems/minimum-cost-to-connect-sticks/
// Author: github.com/lzl124631x
// 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;
}
};