A product manager has to organize n meetings with different people. Meeting with each person results in an increase or decrease in the effectiveness index of the manager. The manager wants to organize the meetings such that the index remains positive for as many meetings as possible. Find the maximum number of meetings for which the effectiveness index is positive. The index at the beginning is 0.
Note: After the meetings begin, the index must remain above 0 to be positive.
Example
n = 4
effectiveness = [1, -20, 3, -2]
One optimal meeting order is [3, -2, 1, -20]. The index is positive for the first three meetings, after which it is 3 - 2 + 1 = 1. So, the answer is 3. There is no way to keep the index positive for all meetings.
Sample Output
3
Explanation
One optimal rearrangement is [2, 0, 1, -3].
Sample Case 1
Sample Input For Custom Testing
STDIN
3
-3
0, -2
FUNCTION
n = 3
effectiveness = [-3, 0, -2]
Sample Output
0
Explanation
There is no arrangement that produces a positive index.
Given an array of strings, return the number of pair of strings that when concatenated, can be rearranged to palindrome.The arrays of strings will only contain lowercase English characters.
Example
n = 4
effectiveness = [1, -20, 3, -2]
One optimal meeting order is [3, -2, 1, -20]. The index is positive for the first three meetings, after which it is 3 - 2 + 1 = 1. So, the answer is 3. There is no way to have 4 meetings with a positive index.
Function Description
Complete the function maxMeetings in the editor.
maxMeetings has the following parameter:
int effectiveness[n]: the increase or decrease effectiveness for each meeting.
Returns
int: the maximum possible number of meetings while maintaining a positive index.
Constraints
- 1 ≤ n ≤ 10^5
- -10^9 ≤ effectiveness[i] ≤ 10^9
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