Largest Values From Labels

• Use a recursive or iterative method to generate all 2^n subsets of the n items.
• For each generated subset:
  • Verify its validity:
    • Check if the number of items in the subset is at most numWanted.
    • Use a hash map to count the occurrences of each label in the subset. Check if any label count exceeds useLimit.
  • If the subset is valid, calculate the sum of its values.
  • Update the overall maximum sum if the current subset's sum is greater.
• Return the overall maximum sum after checking all subsets.

The brute-force method systematically generates every single combination of items. For each combination, it performs two checks: first, whether the size of the subset is within the numWanted limit, and second, whether the usage of any single label is within the useLimit. This is done by counting the items and their labels for each subset. If a subset is deemed valid, its total value is computed and compared against the maximum sum found so far, updating it if necessary. While this method is guaranteed to find the optimal solution, its exponential nature makes it impractical for anything but very small inputs.

1// This is a conceptual illustration. A full implementation would be too complex and inefficient.2class Solution {3    int maxSum = 0;4 5    public int largestValsFromLabels(int[] values, int[] labels, int numWanted, int useLimit) {6        // A recursive helper function would generate all subsets.7        // findSubsets(0, new ArrayList<>(), values, labels, numWanted, useLimit);8        // return maxSum;9        // Due to the 2^n complexity, this approach is not feasible and is omitted.10        return -1; // Placeholder for an infeasible solution11    }12 13    // void findSubsets(int index, List<Integer> currentSubsetIndices, ...) {14    //     // Base case: if index reaches the end15    //     if (index == values.length) {16    //         // Check if current subset is valid17    //         // If valid, calculate sum and update maxSum18    //         return;19    //     }20    //     // Recurse without including the current item21    //     findSubsets(index + 1, currentSubsetIndices, ...);22    //     // Recurse including the current item23    //     currentSubsetIndices.add(index);24    //     findSubsets(index + 1, currentSubsetIndices, ...);25    //     currentSubsetIndices.remove(currentSubsetIndices.size() - 1);26    // }27}

• Create a list of custom objects or pairs, where each object holds an item's value and its label.
• Sort this list of items in descending order based on their values.
• Initialize maxSum = 0, itemsChosen = 0, and a hash map labelCounts to track the usage of each label.
• Iterate through the sorted list of items.
• For each item, if adding it does not violate the numWanted and useLimit constraints, add its value to maxSum, increment itemsChosen, and update its label's count in the map.
• Stop early if itemsChosen reaches numWanted.
• Return maxSum.

This approach relies on the greedy principle that to maximize a sum, one should always prioritize adding the largest available numbers.

First, we combine the values and labels arrays into a single structure, like a list of Item objects, to keep them paired. Then, we sort these items based on their value in descending order. This ensures that we process items from most valuable to least valuable.

We then iterate through this sorted list. For each item, we check if we can select it. A selection is valid if we haven't yet reached our quota of numWanted items and if the number of items already chosen with the same label is less than useLimit. We use a hash map to keep track of the counts of each label selected so far. If the item can be selected, we add its value to our total sum and update our counts. We continue this process until we have either considered all items or have selected numWanted items.

1class Item {2    int value;3    int label;4    Item(int value, int label) {5        this.value = value;6        this.label = label;7    }8}9 10class Solution {11    public int largestValsFromLabels(int[] values, int[] labels, int numWanted, int useLimit) {12        int n = values.length;13        List<Item> items = new ArrayList<>();14        for (int i = 0; i < n; i++) {15            items.add(new Item(values[i], labels[i]));16        }17 18        // Sort items by value in descending order19        items.sort((a, b) -> b.value - a.value);20 21        Map<Integer, Integer> labelCounts = new HashMap<>();22        int maxSum = 0;23        int itemsChosen = 0;24 25        for (Item item : items) {26            if (itemsChosen == numWanted) {27                break;28            }29 30            int currentLabelCount = labelCounts.getOrDefault(item.label, 0);31            if (currentLabelCount < useLimit) {32                maxSum += item.value;33                itemsChosen++;34                labelCounts.put(item.label, currentLabelCount + 1);35            }36        }37 38        return maxSum;39    }40}

• Determine the maximum possible value maxVal from the constraints.
• Create an array of lists (buckets) of size maxVal + 1.
• Populate the buckets: for each item (values[i], labels[i]), add labels[i] to the list at buckets[values[i]].
• Initialize maxSum = 0, itemsChosen = 0, and a hash map labelCounts.
• Iterate downwards from v = maxVal - 1 to 0.
• For each value v, iterate through the labels in buckets[v].
• For each label, check if adding the item is valid (respects numWanted and useLimit).
• If valid, add v to maxSum, increment itemsChosen, and update the label count.
• Stop if itemsChosen reaches numWanted.
• Return maxSum.

The core of this optimized greedy approach is to avoid a comparison-based sort. Since item values are bounded, we can use an array as a set of 'buckets' to group items by value. The index of the array represents a value, and the element at that index is a list of labels of all items having that value.

1. We create an array of lists, buckets, where the size is determined by the maximum possible value (20001 according to constraints).
2. We iterate through the input items once, placing each item's label into the corresponding value's bucket. For example, an item with value=50, label=1 results in 1 being added to buckets[50].
3. After populating the buckets, we iterate through the buckets array from the highest index (highest value) down to zero. This naturally processes items in descending order of value.
4. For each value, we look at the labels in its bucket. For each label, we apply the same greedy logic as before: if we have room for more items (itemsChosen < numWanted) and haven't exhausted the limit for that label (labelCounts.get(...) < useLimit), we select the item.
5. This continues until we have selected numWanted items or have checked all buckets.

This method achieves a linear time complexity because it avoids the O(n log n) sorting step.

1class Solution {2    public int largestValsFromLabels(int[] values, int[] labels, int numWanted, int useLimit) {3        // The maximum value is constrained to 200004        int maxVal = 20001;5        List<Integer>[] buckets = new ArrayList[maxVal];6        for (int i = 0; i < maxVal; i++) {7            buckets[i] = new ArrayList<>();8        }9 10        for (int i = 0; i < values.length; i++) {11            buckets[values[i]].add(labels[i]);12        }13 14        Map<Integer, Integer> labelCounts = new HashMap<>();15        int maxSum = 0;16        int itemsChosen = 0;17 18        for (int v = maxVal - 1; v >= 0; v--) {19            if (itemsChosen == numWanted) {20                break;21            }22            if (buckets[v].isEmpty()) {23                continue;24            }25 26            for (int label : buckets[v]) {27                if (itemsChosen == numWanted) {28                    break;29                }30                int currentLabelCount = labelCounts.getOrDefault(label, 0);31                if (currentLabelCount < useLimit) {32                    maxSum += v;33                    itemsChosen++;34                    labelCounts.put(label, currentLabelCount + 1);35                }36            }37        }38        return maxSum;39    }40}