2640. Maximum Number Of Integers To Choose From A Range I¶

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2640. 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

Solution¶

class Solution {
    public int maxCount(int[] banned, int n, int maxSum) {
        int freq[] = new int[n + 1];
        for (int ele : banned) if (ele <= n) freq[ele]++;
        int current_sum = 0, count = 0;
        for (int i = 1; i <= n; i++) {
            if (current_sum + i <= maxSum && freq[i] == 0) {
                current_sum += i;
                count++;
            }
        }
        return count;
    }
}

Complexity Analysis¶

• Time Complexity: O(?)
• Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here