2529. Maximum Count of Positive Integer and Negative Integer

Given an array nums sorted in non-decreasing order, return the maximum between the number of positive integers and the number of negative integers.

• In other words, if the number of positive integers in 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?

JAVA

• class Solution {
      public int maximumCount(int[] nums) {
          int a = nums.length - search(nums, 1);
          int b = search(nums, 0);
          return Math.max(a, b);
      }
  
      private int search(int[] nums, int x) {
          int left = 0, right = nums.length;
          while (left < right) {
              int mid = (left + right) >> 1;
              if (nums[mid] >= x) {
                  right = mid;
              } else {
                  left = mid + 1;
              }
          }
          return left;
      }
  }

C++

• class Solution {
  public:
      int maximumCount(vector<int>& nums) {
          int a = nums.end() - lower_bound(nums.begin(), nums.end(), 1);
          int b = lower_bound(nums.begin(), nums.end(), 0) - nums.begin();
          return max(a, b);
      }
  };