2554. Maximum Number of Integers to Choose From a Range I

You are given an integer array banned and two integers n and maxSum. You are choosing some number of integers following the below rules:

• The chosen integers have to be in the range [1, n].
• Each integer can be chosen at most once.
• The chosen integers should not be in the array banned.
• The sum of the chosen integers should not exceed maxSum.

Return the maximum number of integers you can choose following the mentioned rules.

Example 1:

Input: banned = [1,6,5], n = 5, maxSum = 6
Output: 2
Explanation: You can choose the integers 2 and 4.
2 and 4 are from the range [1, 5], both did not appear in banned, and their sum is 6, which did not exceed maxSum.

Example 2:

Input: banned = [1,2,3,4,5,6,7], n = 8, maxSum = 1
Output: 0
Explanation: You cannot choose any integer while following the mentioned conditions.

Example 3:

Input: banned = [11], n = 7, maxSum = 50
Output: 7
Explanation: You can choose the integers 1, 2, 3, 4, 5, 6, and 7.
They are from the range [1, 7], all did not appear in banned, and their sum is 28, which did not exceed maxSum.

Constraints:

• 1 <= banned.length <= 104
• 1 <= banned[i], n <= 104
• 1 <= maxSum <= 109

Approach 01:

class Solution {
public:
    int maxCount(vector<int>& banned, int n, int maxSum) {
        unordered_set<int> bannedSet(begin(banned), end(banned));
        int validCount = 0;
        int currentSum = 0;

        for (int currentNum = 1; currentNum <= n; currentNum++) {
            if (bannedSet.count(currentNum)) {
                continue;
            }

            if (currentSum + currentNum <= maxSum) {
                validCount++;
                currentSum += currentNum;
            } else {
                break;
            }
        }

        return validCount;
    }
};

class Solution:
    def maxCount(self, banned: list[int], n: int, maxSum: int) -> int:
        bannedSet = set(banned)
        validCount = 0
        currentSum = 0

        for currentNum in range(1, n + 1):
            if currentNum in bannedSet:
                continue

            if currentSum + currentNum <= maxSum:
                validCount += 1
                currentSum += currentNum
            else:
                break

        return validCount

Time Complexity

• Creating the banned set:

  The banned vector is converted into an unordered set, which requires iterating over the elements of banned. If the size of banned is \(M\), this step takes \(O(M)\).

• Iterating from 1 to n:

  The loop iterates at most \(n\) times. Each iteration involves checking membership in the bannedSet, which is an \(O(1)\) operation on average due to hash-based lookups. Thus, the loop takes \(O(n)\).

• Overall Time Complexity:

  The overall time complexity is \(O(M + n)\), where \(M\) is the size of the banned list and \(n\) is the maximum number to consider.

Space Complexity

• Space for the bannedSet:

  The bannedSet stores all elements from the banned vector. If the size of banned is \(M\), this requires \(O(M)\) space.

• Auxiliary variables:

  The function uses a few integer variables (validCount, currentSum, and currentNum), which require \(O(1)\) space.

• Overall Space Complexity:

  The overall space complexity is \(O(M)\) to store the banned set.