Easy
Given an array nums
sorted in non-decreasing order, return the maximum between the number of positive integers and the number of negative integers.
nums
is pos
and the number of negative integers is neg
, then return the maximum of pos
and neg
.Note that 0
is neither positive nor negative.
Example 1:
Input: nums = [-2,-1,-1,1,2,3]
Output: 3
Explanation: There are 3 positive integers and 3 negative integers. The maximum count among them is 3.
Example 2:
Input: nums = [-3,-2,-1,0,0,1,2]
Output: 3
Explanation: There are 2 positive integers and 3 negative integers. The maximum count among them is 3.
Example 3:
Input: nums = [5,20,66,1314]
Output: 4
Explanation: There are 4 positive integers and 0 negative integers. The maximum count among them is 4.
Constraints:
1 <= nums.length <= 2000
-2000 <= nums[i] <= 2000
nums
is sorted in a non-decreasing order.Follow up: Can you solve the problem in O(log(n))
time complexity?
public class Solution {
public int maximumCount(int[] nums) {
int plus = 0;
int minus = 0;
for (int num : nums) {
if (num > 0) {
plus++;
}
if (num < 0) {
minus++;
}
}
return Math.max(plus, minus);
}
}