leetcode-1167-minimum-cost-to-connect-sticks.swift

/**

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


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

*/

/**
Big O Annotation
Time complexity O(n) where n is the amount of elements in nums.
Space complexity O(n) where n is the amount of elements in nums.
*/
func connectSticks(_ sticks: [Int]) -> Int {
        
    // Init a min heap from the sticks
    var heap = Heap(sticks, areSorted: <)

    // The total costs
    var cost: Int = 0
    
    // While the heap is not empty
    while !heap.isEmpty {
        
        // Get the root (aka the most min element)
        let stick1 = heap.getRoot()!

        // See if there is a second elements
        guard let stick2 = heap.getRoot() else {

            // ... if not, stop
            break
        }
        
        // ... if so, smash them
        let sum = stick1 + stick2

        // And append the costs
        cost += sum

        // And add the smashed stick to the heap
        heap.insert(sum)
    }
    
    return cost
}

// Basic Heap implementation
struct Heap {
    
    private var storage: [Int] = []
    private var areSorted: (Int, Int) -> Bool
    
    init(_ storage: [Int], areSorted: @escaping (Int, Int) -> Bool) {
        self.storage = storage
        self.areSorted = areSorted
        
        guard !storage.isEmpty else {
            return 
        }
        
        for index in stride(from: storage.count / 2 - 1, through: 0 , by: -1) {
            siftDown(from: index)
        }
    }
    
    public var isEmpty: Bool {
        return storage.isEmpty
    } 
    
    public var count: Int {
        return storage.count
    }
    
    public func getChildIndices(forParentAt index: Int) -> (left: Int, right: Int) {
        let left = (index * 2) + 1
        return (left, left + 1)
    }
    
    public func getParentIndex(forChildAt index: Int) -> Int {
        return (index - 1) / 2
    }
    
    public mutating func insert(_ element: Int) {
        storage.append(element)
        siftUp(from: count - 1)
    }
    
    public mutating func getRoot() -> Int? {
        
        guard !isEmpty else {
            return nil
        }
        
        storage.swapAt(0, count - 1)
        let root = storage.removeLast()
        siftDown(from: 0)
        return root
    }
    
    public mutating func siftUp(from index: Int) {
        
        var childIndex = index
        var parentIndex = getParentIndex(forChildAt: childIndex)
        
        while childIndex > 0 && areSorted(storage[childIndex], storage[parentIndex]) {
            
            storage.swapAt(childIndex, parentIndex)
            childIndex = parentIndex
            parentIndex = getParentIndex(forChildAt: childIndex)
        }
    }
    
    public mutating func siftDown(from index: Int) {
        
        var parentIndex = index
        
        while true {
            
            let (left, right) = getChildIndices(forParentAt: parentIndex)
            var swapIndex: Int?
            
            if left < count && areSorted(storage[left], storage[parentIndex]) {
                swapIndex = left
            }
            
            if right < count && areSorted(storage[right], storage[swapIndex ?? parentIndex]) {
                swapIndex = right
            }
            
            guard swapIndex != nil else {
                return 
            }
            
            storage.swapAt(parentIndex, swapIndex!)
            parentIndex = swapIndex!
        }
    }
}