maximum-units-on-a-truck.py

"""

1710. Maximum Units on a Truck
Easy

You are assigned to put some amount of boxes onto one truck. You are given a 2D array boxTypes, where boxTypes[i] = [numberOfBoxesi, numberOfUnitsPerBoxi]:

numberOfBoxesi is the number of boxes of type i.
numberOfUnitsPerBoxi is the number of units in each box of the type i.
You are also given an integer truckSize, which is the maximum number of boxes that can be put on the truck. You can choose any boxes to put on the truck as long as the number of boxes does not exceed truckSize.

Return the maximum total number of units that can be put on the truck.

 

Example 1:

Input: boxTypes = [[1,3],[2,2],[3,1]], truckSize = 4
Output: 8
Explanation: There are:
- 1 box of the first type that contains 3 units.
- 2 boxes of the second type that contain 2 units each.
- 3 boxes of the third type that contain 1 unit each.
You can take all the boxes of the first and second types, and one box of the third type.
The total number of units will be = (1 * 3) + (2 * 2) + (1 * 1) = 8.
Example 2:

Input: boxTypes = [[5,10],[2,5],[4,7],[3,9]], truckSize = 10
Output: 91
 

Constraints:

1 <= boxTypes.length <= 1000
1 <= numberOfBoxesi, numberOfUnitsPerBoxi <= 1000
1 <= truckSize <= 106

"""

# V0
# IDEA : GREEDY + sorting
class Solution(object):
    def maximumUnits(self, boxTypes, truckSize):
        # boxTypes[i] = [numberOfBoxesi, numberOfUnitsPerBoxi]:
        # edge case
        if not boxTypes or not truckSize:
            return 0
        """
        NOTE : we sort on sort(key=lambda x : -x[1])

            -> if unit is bigger, we can get bigger aggregated result (n * unit)
        """
        boxTypes.sort(key=lambda x : -x[1])
        res = 0
        for n, unit in boxTypes:
            # case 1 : truckSize == 0, break for loop and return ans
            if truckSize == 0:
                break
            # case 2 : truckSize < n, we CAN'T add all n to truck, but CAN only add (truckSize * unit) amount to truck
            elif truckSize < n:
                res += (truckSize * unit)
                truckSize = 0
                break
            # case 3 : normal case, it's OK to put all (n * unit) to truck once
            else:      
                res += (n * unit)
                truckSize -= n
        return res

# V1
# IDEA : GREEDY
class Solution(object):
    def maximumUnits(self, boxTypes, truckSize):
        res = 0
        for number, unit in sorted(boxTypes, key=lambda d:d[1], reverse=True):
            if truckSize >= number:
                res += number * unit
                truckSize -= number
            else:
                res += truckSize * unit
                break        
        return res

# V1
# IDEA : GREEDY
# https://leetcode.com/problems/maximum-units-on-a-truck/discuss/1045318/Python-solution
class Solution(object):
    def maximumUnits(self, boxTypes, truckSize):
        boxTypes.sort(key = lambda x: -x[1])
        n = len(boxTypes)
        result = 0
        i = 0
        while truckSize >= boxTypes[i][0]:
            result += boxTypes[i][1] * boxTypes[i][0]
            truckSize -= boxTypes[i][0]
            i += 1
            if i == n:
                return result
        result += truckSize * boxTypes[i][1]
        return result

# V1'
# https://leetcode.com/problems/maximum-units-on-a-truck/discuss/1000278/Python-sort.-Short-and-fast-(100)
class Solution:
    def maximumUnits(self, boxTypes, truckSize):
        boxTypes.sort(reverse=True, key=itemgetter(1))
        
        result = 0
        for boxes, units in boxTypes:
            if boxes < truckSize:
                result += boxes * units
                truckSize -= boxes
            else:
                result += truckSize * units
                break        
        return result

# V2