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1710. Maximum Units on a Truck

Description

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

Solutions

Solution 1: Greedy + Sorting

Thinking

Capacity is measured in boxes, but the objective is units. Boxes of one type share the same unit count, so denser types should be taken first.

Sort by units per box descending and load \(\min(\textit{truckSize},\textit{count})\) of each type, decreasing remaining capacity until it is gone.

According to the problem, we should choose as many units as possible. Therefore, we first sort boxTypes in descending order of the number of units.

Then we traverse boxTypes from front to back, choose up to truckSize boxes, and accumulate the number of units.

The time complexity is \(O(n \times \log n)\), where \(n\) is the length of the two-dimensional array boxTypes.

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class Solution:
    def maximumUnits(self, boxTypes: List[List[int]], truckSize: int) -> int:
        ans = 0
        for a, b in sorted(boxTypes, key=lambda x: -x[1]):
            ans += b * min(truckSize, a)
            truckSize -= a
            if truckSize <= 0:
                break
        return ans

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class Solution {
    public int maximumUnits(int[][] boxTypes, int truckSize) {
        Arrays.sort(boxTypes, (a, b) -> b[1] - a[1]);
        int ans = 0;
        for (var e : boxTypes) {
            int a = e[0], b = e[1];
            ans += b * Math.min(truckSize, a);
            truckSize -= a;
            if (truckSize <= 0) {
                break;
            }
        }
        return ans;
    }
}

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class Solution {
public:
    int maximumUnits(vector<vector<int>>& boxTypes, int truckSize) {
        sort(boxTypes.begin(), boxTypes.end(), [](auto& a, auto& b) { return a[1] > b[1]; });
        int ans = 0;
        for (auto& e : boxTypes) {
            int a = e[0], b = e[1];
            ans += b * min(truckSize, a);
            truckSize -= a;
            if (truckSize <= 0) break;
        }
        return ans;
    }
};

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func maximumUnits(boxTypes [][]int, truckSize int) (ans int) {
    sort.Slice(boxTypes, func(i, j int) bool { return boxTypes[i][1] > boxTypes[j][1] })
    for _, e := range boxTypes {
        a, b := e[0], e[1]
        ans += b * min(truckSize, a)
        truckSize -= a
        if truckSize <= 0 {
            break
        }
    }
    return
}

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export function maximumUnits(boxTypes: number[][], truckSize: number): number {
    boxTypes.sort(([_, a], [__, b]) => b - a);
    let ans = 0;
    for (const [count, size] of boxTypes) {
        ans += Math.min(truckSize, count) * size;
        truckSize -= count;
        if (truckSize < 0) break;
    }
    return ans;
}

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impl Solution {
    pub fn maximum_units(mut box_types: Vec<Vec<i32>>, truck_size: i32) -> i32 {
        box_types.sort_by(|a, b| b[1].cmp(&a[1]));
        let mut sum = 0;
        let mut ans = 0;
        for box_type in box_types.iter() {
            if sum + box_type[0] < truck_size {
                sum += box_type[0];
                ans += box_type[0] * box_type[1];
            } else {
                ans += (truck_size - sum) * box_type[1];
                break;
            }
        }
        ans
    }
}

Solution 2: Counting Sort

Thinking

Solution 1 spends \(O(n\log n)\) on sorting. Units per box are at most \(1000\), so a counting array can store box counts by that value.

Scan from \(1000\) down to \(1\) and load in the same greedy order, now in linear time.

We can also use the idea of counting sort, create an array \(cnt\) of length \(1001\), where \(cnt[b]\) represents the number of boxes with \(b\) units.

Then starting from the box with the maximum number of units, choose up to truckSize boxes, and accumulate the number of units.

The time complexity is \(O(M)\), where \(M\) is the maximum number of units. In this problem, \(M=1000\).

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class Solution:
    def maximumUnits(self, boxTypes: List[List[int]], truckSize: int) -> int:
        cnt = [0] * 1001
        for a, b in boxTypes:
            cnt[b] += a
        ans = 0
        for b in range(1000, 0, -1):
            a = cnt[b]
            if a:
                ans += b * min(truckSize, a)
                truckSize -= a
                if truckSize <= 0:
                    break
        return ans

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class Solution {
    public int maximumUnits(int[][] boxTypes, int truckSize) {
        int[] cnt = new int[1001];
        for (var e : boxTypes) {
            int a = e[0], b = e[1];
            cnt[b] += a;
        }
        int ans = 0;
        for (int b = 1000; b > 0 && truckSize > 0; --b) {
            int a = cnt[b];
            if (a > 0) {
                ans += b * Math.min(truckSize, a);
                truckSize -= a;
            }
        }
        return ans;
    }
}

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class Solution {
public:
    int maximumUnits(vector<vector<int>>& boxTypes, int truckSize) {
        int cnt[1001] = {0};
        for (auto& e : boxTypes) {
            int a = e[0], b = e[1];
            cnt[b] += a;
        }
        int ans = 0;
        for (int b = 1000; b > 0 && truckSize > 0; --b) {
            int a = cnt[b];
            if (a) {
                ans += b * min(truckSize, a);
                truckSize -= a;
            }
        }
        return ans;
    }
};

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func maximumUnits(boxTypes [][]int, truckSize int) (ans int) {
    cnt := [1001]int{}
    for _, e := range boxTypes {
        a, b := e[0], e[1]
        cnt[b] += a
    }
    for b := 1000; b > 0 && truckSize > 0; b-- {
        a := cnt[b]
        if a > 0 {
            ans += b * min(truckSize, a)
            truckSize -= a
        }
    }
    return
}

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function maximumUnits(boxTypes: number[][], truckSize: number): number {
    boxTypes.sort(([_, a], [__, b]) => b - a);
    let ans = 0;
    for (const [count, size] of boxTypes) {
        ans += Math.min(truckSize, count) * size;
        truckSize -= count;
        if (truckSize < 0) {
            break;
        }
    }
    return ans;
}

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