Maximum Number of Integers to Choose From a Range I

class Solution {
public:
    int countValidNumbers(vector<int>& restricts, int limit, int upperSum) {
        unordered_set<int> restrictedNums(restricts.begin(), restricts.end());

        int validCount = 0;

        for (int number = 1; number <= limit; number++) {
            if (restrictedNums.find(number) != restrictedNums.end()) continue;

            if (upperSum - number < 0) return validCount;

            upperSum -= number;
            validCount++;
        }
        return validCount;
    }
};

The C++ solution presented above is designed to calculate how many numbers you can choose from a range starting from 1 to the given limit, such that the sum of these chosen numbers does not exceed a specified upperSum, and the numbers are not in a given list of restricted numbers.

The following captures the process:

• The code first initializes an unordered_set named restrictedNums with the numbers from the restricts vector to efficiently check if a number is restricted.
• It initializes a counter validCount to keep track of how many numbers can be selected.
• The function then iterates through all numbers from 1 to the specified limit.
• For each number, it checks if the number is in the restrictedNums set. If it is, the loop continues to the next number, skipping the current one.
• Next, it checks if the remaining allowance in upperSum after choosing the current number would be negative. If true, it terminates and returns the count (validCount) of numbers already chosen.
• If valid, the current number is subtracted from upperSum, and validCount is incremented.
• Finally, after the loop has processed all numbers up to limit, the total count of valid numbers that can be chosen is returned.

This method is efficient in terms of both time and space, as it only iterates through the numbers once and uses a set for quick membership checking. The use of early termination when the upperSum is exceeded, further optimizes the solution.

class Solution {

    public int maximumUsable(int[] exclude, int limit, int totalSum) {
        Set<Integer> excludedNumbers = new HashSet<>();
        for (int item : exclude) {
            excludedNumbers.add(item);
        }

        int validCount = 0;

        for (int i = 1; i <= limit; i++) {
            if (excludedNumbers.contains(i)) continue;

            if (totalSum - i < 0) return validCount;

            totalSum -= i;
            validCount++;
        }
        return validCount;
    }
}

This Java solution tackles the challenge of finding the maximum number of integers that can be chosen from a given range up to a specified limit, given a certain total sum and an array of numbers to exclude. Here's how the solution works:

• Initialize a HashSet to store the excluded numbers for quick look-up.
• Iterate over each number within the given range (from 1 to limit).
  • If the current number is in the set of excluded numbers, skip it.
  • Check if the total sum minus the current number is a negative value. If it is, the function returns the count of valid numbers found so far.
  • If the total sum is still zero or positive after subtracting the current number, decrease the total sum by the current integer and increment the count of valid numbers.
• If the loop completes without the total sum becoming negative, return the count of valid numbers accumulated so far.

This ensures you dynamically exclude certain numbers and consider only the permissible sum for counting valid integers, making it efficient in checking conditions and managing the total sum dynamically. The use of a HashSet allows for fast exclusions checks, and careful iteration ensures that only viable options reduce the allowed sum.

class Solution:
    def maximumAvailable(self, forbidden: list[int], limit: int, totalMax: int) -> int:
        forbidden_nums = set(forbidden)
        allowed_count = 0

        for value in range(1, limit + 1):
            if value in forbidden_nums:
                continue

            if totalMax - value < 0:
                return allowed_count

            totalMax -= value
            allowed_count += 1

        return allowed_count

The given Python solution defines a method to calculate how many integers can be chosen from a range limited by a given limit and totalMax value, while avoiding a list of forbidden numbers. Here's how the solution works:

• Convert the list of forbidden numbers into a set for efficient lookup.
• Initialize a counter for the allowed integers.
• Iterate over the range from 1 to limit.
  • Skip the iteration if the current number is in the set of forbidden numbers.
  • Check if subtracting the current number from totalMax would result in a negative value. If true, return the count of allowed integers so far.
  • Deduct the value of the current number from totalMax and increase the allowed integers counter.
• Return the count of allowed integers after the loop ends.

This approach ensures that only integers that are not forbidden and do not exceed the totalMax when cumulatively added are counted.