Maximum Number of Integers to Choose From a Range II

Problem Description

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.

Examples

Example 1:

Input: banned = [1,4,6], n = 6, maxSum = 4
Output: 1
Explanation: You can choose the integer 3.
3 is in the range [1, 6], and do not appear in banned. The sum of the chosen integers is 3, which does not exceed maxSum.

Example 2:

Input: banned = [4,3,5,6], n = 7, maxSum = 18
Output: 3
Explanation: You can choose the integers 1, 2, and 7.
All these integers are in the range [1, 7], all do not appear in banned, and their sum is 18, which does not exceed maxSum.

Constraints

• 1 <= banned.length <= 105
• 1 <= banned[i] <= n <= 109
• 1 <= maxSum <= 1015

Approach

We can add $0$ and $n + 1$ to the array banned, then deduplicate and sort the array banned.

Next, we enumerate every two adjacent elements $i$ and $j$ in the array banned. The range of selectable integers is $[i + 1, j - 1]$. We use binary search to enumerate the number of elements we can select in this range, find the maximum number of selectable elements, and then add it to $ans$. At the same time, we subtract the sum of these elements from maxSum. If maxSum is less than $0$, we break the loop. Return the answer.

The time complexity is $O(n \times \log n)$, and the space complexity is $O(n)$. Where $n$ is the length of the array banned.

Python3

class Solution:
    def maxCount(self, banned: List[int], n: int, maxSum: int) -> int:
        banned.extend([0, n + 1])
        ban = sorted(set(banned))
        ans = 0
        for i, j in pairwise(ban):
            left, right = 0, j - i - 1
            while left < right:
                mid = (left + right + 1) >> 1
                if (i + 1 + i + mid) * mid // 2 <= maxSum:
                    left = mid
                else:
                    right = mid - 1
            ans += left
            maxSum -= (i + 1 + i + left) * left // 2
            if maxSum <= 0:
                break
        return ans

Java

class Solution {
    public int maxCount(int[] banned, int n, long maxSum) {
        Set<Integer> black = new HashSet<>();
        black.add(0);
        black.add(n + 1);
        for (int x : banned) {
            black.add(x);
        }
        List<Integer> ban = new ArrayList<>(black);
        Collections.sort(ban);
        int ans = 0;
        for (int k = 1; k < ban.size(); ++k) {
            int i = ban.get(k - 1), j = ban.get(k);
            int left = 0, right = j - i - 1;
            while (left < right) {
                int mid = (left + right + 1) >>> 1;
                if ((i + 1 + i + mid) * 1L * mid / 2 <= maxSum) {
                    left = mid;
                } else {
                    right = mid - 1;
                }
            }
            ans += left;
            maxSum -= (i + 1 + i + left) * 1L * left / 2;
            if (maxSum <= 0) {
                break;
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    int maxCount(vector<int>& banned, int n, long long maxSum) {
        banned.push_back(0);
        banned.push_back(n + 1);
        sort(banned.begin(), banned.end());
        banned.erase(unique(banned.begin(), banned.end()), banned.end());
        int ans = 0;
        for (int k = 1; k < banned.size(); ++k) {
            int i = banned[k - 1], j = banned[k];
            int left = 0, right = j - i - 1;
            while (left < right) {
                int mid = left + ((right - left + 1) / 2);
                if ((i + 1 + i + mid) * 1LL * mid / 2 <= maxSum) {
                    left = mid;
                } else {
                    right = mid - 1;
                }
            }
            ans += left;
            maxSum -= (i + 1 + i + left) * 1LL * left / 2;
            if (maxSum <= 0) {
                break;
            }
        }
        return ans;
    }
};