
Hi! My name is Nhut Nguyen. I am a software engineer and a writer in Copenhagen, Denmark.
Learn more about me at nhutnguyen.com
Search for a command to run...

Hi! My name is Nhut Nguyen. I am a software engineer and a writer in Copenhagen, Denmark.
Learn more about me at nhutnguyen.com
No comments yet. Be the first to comment.
A simple example of using C++ switch

Two dynamic programming techniques to solve Leetcode 120. Triangle. One has space complexity O(n^2). The other is O(n).

An example of using a sliding window approach and an unordered map to track character positions

A simple C++ solution to Leetcode 1695. Maximum Erasure Value using a sliding window approach and prefix sums.

Strategies to avoid them will help you excel in your following technical interview

You are assigned to put some amount of boxes onto one truck. You are given a 2D array boxTypes, where boxTypes[i] = [numberOfBoxes_i, numberOfUnitsPerBox_i]:
numberOfBoxes_i is the number of boxes of type i.
numberOfUnitsPerBox_i 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.
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.
Input: boxTypes = [[5,10],[2,5],[4,7],[3,9]], truckSize = 10
Output: 91
Explanation: (5 * 10) + (3 * 9) + (2 * 7) = 91.
1 <= boxTypes.length <= 1000.
1 <= numberOfBoxes_i, numberOfUnitsPerBox_i <= 1000.
1 <= truckSize <= 10^6.
Put the boxes having more units first.
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int maximumUnits(vector<vector<int>>& boxTypes, int truckSize) {
// sort for the boxes based on their number of units
sort(boxTypes.begin(), boxTypes.end(), [](const vector<int>& a, const vector<int>& b) {
return a[1] > b[1];
});
int maxUnits = 0;
int i = 0;
while (truckSize > 0 && i < boxTypes.size()) {
if (boxTypes[i][0] <= truckSize) {
maxUnits += boxTypes[i][0] * boxTypes[i][1];
truckSize -= boxTypes[i][0];
} else {
maxUnits += truckSize * boxTypes[i][1];
break;
}
i++;
}
return maxUnits;
}
int main() {
vector<vector<int>> boxTypes{{1,3},{2,2},{3,1}};
cout << maximumUnits(boxTypes, 4) << endl;
boxTypes = {{5,10},{2,5},{4,7},{3,9}};
cout << maximumUnits(boxTypes, 10) << endl;
}
Output:
8
91
Runtime: O(NlogN), where N = boxTypes.length.
Extra space: O(1).
Note that two vectors can be compared. That is why you can sort them.
But in this case you want to sort them based on the number of units. That is why you need to define the comparison function like the code above. Otherwise, the sort algorithm will use the dictionary order to sort them by default.
What is your approach? The problem was picked from leetcode.com. You can submit your solution in any programming language and check the performance.
Thanks for reading. Feel free to share your thought about my content and check out my FREE book 10 Classic Coding Challenges.