2554

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

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

Companies: PayPal

Related Topics:
Array, Hash Table, Binary Search, Greedy, Sorting

Similar Questions:

• First Missing Positive (Hard)
• Find All Numbers Disappeared in an Array (Easy)
• Append K Integers With Minimal Sum (Medium)
• Replace Elements in an Array (Medium)
• Maximum Number of Integers to Choose From a Range II (Medium)

Hints:

• Keep the banned numbers that are less than n in a set.
• Loop over the numbers from 1 to n and if the number is not banned, use it.
• Keep adding numbers while they are not banned, and their sum is less than k.

Solution 1.

// OJ: https://leetcode.com/problems/maximum-number-of-integers-to-choose-from-a-range-i
// Author: github.com/lzl124631x
// Time: O(BlogB + N)
// Space: O(1)
class Solution {
public:
    int maxCount(vector<int>& banned, int n, int maxSum) {
        sort(begin(banned), end(banned));
        long long ans = 0, sum = 0, i = 0, N = banned.size();
        for (int j = 1; j <= n; ++j) {
            if (i < N && banned[i] == j) {
                while (i < N && banned[i] == j) ++i;
            } else if (sum + j <= maxSum) {
                sum += j;
                ++ans;
            } else break;
        }
        return ans;
    }
};