LC: 1167. Minimum Cost to Connect Sticks
https://leetcode.com/problems/minimum-cost-to-connect-sticks/
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:
Constraints:
1 <= sticks.length <= 1041 <= sticks[i] <= 104
1167. Minimum Cost to Connect Sticks:
The Essence:
Wenn man zwei von einer Menge Stöcke auswählt, muss man hier immer die zwei Kürzesten auswählen.
Beweis: Es seien z>y=x die Länge dreier Stöcke. Wenn man die zwei Auswahlwege vergleicht, erkennt man die Ungleichung:
(x+y)+((x+y)+z)=2(x+y)+z < (x+z)+((x+z)+y)=2(x+z)+y
Details:
Der Problemlöser kann eine Prioritätsschlange verwenden, um die kürzesten zwei Stöcke auszuwählen und dann sie wieder zu der Menge addieren. Ein solches Verfahren ist sehr ähnlich wie die Huffman-Kodierung.
Solution(s):
Default Code:
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