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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?

Solution:

class Solution {
    public int maximumCount(int[] nums) {
        // >= 0
        int zero = binarySearch(nums, 0);
        if (zero == nums.length){
            return nums.length;
        }

        if (nums[zero] != 0){
            return Math.max(zero, nums.length - zero);
        }

        int neg = zero;// 0

        int lastZero = binarySearch(nums, 1) - 1;

        int pos = nums.length -1 - lastZero;// 2 -1 =1 
        return Math.max(neg, pos);
    }

    public int binarySearch(int[] nums, int target){
        int left = 0;
        int right = nums.length - 1;

        while(left <= right){
            int mid = left + (right - left)/2;
            if (nums[mid] < target){
                left = mid + 1;
            }else{
                right = mid - 1;
            }
        }

        return left;
    }
}

// TC: O(logn)
// SC: O(n)