Contents
Fill The Truck Maximum Units on a Truck Solution
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.
- numberOfUnitsPerBoxiis 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

SOLUTION
Simple Sort Solution
For this problem, we simply need to prioritize the more valuable boxes first. To do this, we should sort the boxtypes array (B) in descending order by the number of units per box (B[i][1]).
Then we can iterate through B and at each step, we should add as many of the boxes as we can, until we reach the truck size (T). We should add the number of boxes added multiplied by the units per box to our answer (ans), and decrease T by the same number of boxes.
Once the truck is full (T == 0), or once the iteration is done, we should return ans.
- Time Complexity: O(N log N) where N is the length of B, for the sort
- Space Complexity: O(1)
Program: Fill The Truck Maximum Units on a Truck Solution in C++
class Solution {
public:
int maximumUnits(vector<vector<int>>& B, int T) {
sort(B.begin(), B.end(), [](auto& a, auto& b) { return b[1] < a[1];});
int ans = 0;
for (auto& b : B) {
int count = min(b[0], T);
ans += count * b[1], T -= count;
if (!T) return ans;
}
return ans;
}
};
Program: Fill The Truck Maximum Units on a Truck Solution in Python
class Solution:
def maximumUnits(self, B: List[List[int]], T: int) -> int:
B.sort(key=lambda x: x[1], reverse=True)
ans = 0
for b,n in B:
boxes = min(b, T)
ans += boxes * n
T -= boxes
if T == 0: return ans
return ans
Program: Fill The Truck Maximum Units on a Truck Solution in Java
class Solution {
public int maximumUnits(int[][] B, int T) {
Arrays.sort(B, (a,b) -> b[1] - a[1]);
int ans = 0;
for (int[] b : B) {
int count = Math.min(b[0], T);
ans += count * b[1];
T -= count;
if (T == 0) return ans;
}
return ans;
}
}
Program: Fill The Truck Maximum Units on a Truck Solution in JavaScript
var maximumUnits = function(B, T) {
B.sort((a,b) => b[1] - a[1])
let ans = 0
for (let i = 0; T && i < B.length; i++) {
let count = Math.min(B[i][0], T)
ans += count * B[i][1], T -= count
}
return ans
};
Greedily Select Max Units/Box Ratio
The given constraints for numberOfUnitsPerBox are small enough that we can use an approach similar to counting sort to reduce the time complexity to O(N).
Here, we can declare an array freq of size=1000 (which is maximum number of units per box) where freq[i] will denote the number of boxes that can hold i number of units. We can iterate through the given boxTypes array and populate the freq array. Then we can iterate over the freq array and greedily choose starting from i=1000 till we run out of truckSize or pick all available boxes.
Time Complexity : O(N)
Space Complexity : O(1)
Program: Maximum Units on a Truck Solution in C++
int maximumUnits(vector<vector<int>>& boxTypes, int truckSize) {
int freq[1001]{0}, maxUnits = 0; // freq[i] = number of boxes that can hold i units
for(auto& box : boxTypes) freq[box[1]] += box[0];
// greedily choose starting from max units till either truckSize runs out or you choose all boxes
for(int units = 1000; truckSize > 0 && ~units; --units) {
maxUnits += min(truckSize, freq[units]) * units;
truckSize -= freq[units];
}
return maxUnits;
}
Program: Maximum Units on a Truck Solution in Python
def maximumUnits(self, boxTypes: List[List[int]], truckSize: int) -> int:
freq, max_units = [0]*1001, 0
for box in boxTypes:
freq[box[1]] += box[0]
for units in range(1000,0,-1):
if truckSize < 0: break
max_units += min(truckSize, freq[units]) * units
truckSize -= freq[units]
return max_units
Program Java: Maximum Units on a Truck Solution in Java
class Solution {
public int maximumUnits(int[][] boxTypes, int truckSize) {
Arrays.sort(boxTypes,(a,b)->(b[1]-a[1]));
int i=0;
int ans=0;
while(i<boxTypes.length && truckSize>0){
if(truckSize-boxTypes[i][0]>=0){
truckSize-=boxTypes[i][0];
ans+=boxTypes[i][0]*boxTypes[i][1];}else{
ans+=boxTypes[i][1]*truckSize;
truckSize=0;
}
System.out.println(ans);
i++;
}
return ans;
}
}
Bucket Sort Solution
This time we will store info into sizeBucket, an array of 1001 elements (to cover all the provided range) all set to 0 and that we will populate while also storing the extremes of said range in minBucket and maxBucket, increasing each bucket of size boxType[1] by boxType[0] (how many we have) as we go. Be sure to add, not to assign here, since we do not know if we would be given multiple entries of the same size.
Once done, we can loop through the boxes we bucket-sorted going from maxBucket to minBucket (included) and following a logic specular to the previous one.
This bucket sorting version, which despite using buckets, turns out to be even more efficient in terms of space too:
Program: Fill The Truck Maximum Units on a Truck Solution in C++
class Solution {
public:
int maximumUnits(vector<vector<int>>& boxes, int truckSize) {
// support variables
int res = 0, sizeBucket[1001] = {}, maxBucket = INT_MIN, minBucket = INT_MAX;
// bucket sorting tthe boxes and recording the bucket range
for (auto &boxType: boxes) {
maxBucket = max(maxBucket, boxType[1]);
minBucket = min(minBucket, boxType[1]);
sizeBucket[boxType[1]] += boxType[0];
}
// carrying as many larger sized boxes as we can first
for (int i = maxBucket, size, currBatch; i >= minBucket; i--) {
size = sizeBucket[i];
if (!size) continue;
currBatch = min(size, truckSize);
truckSize -= currBatch;
res += currBatch * i;
if (!truckSize) break;
}
return res;
}
};
Amazon OA 2023 Questions with Solution
- Shopping Patterns Solution Amazon OA 2023
- Reorder Data in Log Files Solution Amazon OA 2023
- Top K Frequent Words Solution Amazon OA 2023
- Trees Height Solution Amazon OA SDE 2023
- Counting Binary Substrings Amazon OA 2023
- Grid Connections Amazon OA 2023
- Shipment Imbalance Amazon OA 2023
- Max Profit Amazon OA 2023
- Find Lowest Price Amazon OA 2023
- Decode String Frequency Amazon OA 2023
- Simple Cipher Amazon OA 2023
- Valid Discount Coupons Amazon OA 2023 Solution
- Count Maximum Teams Amazon OA 2023
- Minimum Coin Flips Amazon OA 2023
- Max Average Stock Price Amazon OA 2023 Solution
- Robot Bounded In Circle Amazon OA 2023
- Shopping Options Amazon OA 2023 Solution
- Maximize Score After N Operations Number Game Solution Amazon OA 2023
- Slowest Key Amazon OA 2023 Solution
- Five Star Seller Maximum Average Pass Ratio Amazon OA 2023
- Split String Into Unique Primes Amazon OA 2023 Solution
- Storage Optimization Amazon OA 2023 Solution
- Minimum Difficulty of a Job Schedule Amazon OA 2023 Solution
- Autoscale Policy Utilization Check Amazon OA 2023
- Optimal Utilization Solution Amazon OA 2023
- Merge Two Sorted Lists Solution Amazon OA 2023
- Two Sum Unique Pairs Solution Amazon OA 2023
- Amazon Music Pairs Amazon OA 2023 Solution
- Class Grouping Amazon OA 2023 Solution
- Find Max products Amazon OA 2023 Solution
- Get encrypted number Amazon OA 2023 Solution
- Find Total Imbalance Amazon OA 2023 Solution
- Find Total Power Amazon OA 2023 Solution
Same Question came in my test on campus in IP.
Hello, Neat post. There is a problem with your site in web explorer, would test
this? IE still is the market chief and a big component of folks will leave out your
magnificent writing because of this problem.