2529. Maximum Count of Positive Integer and Negative Integer

Problem

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

Iterative - O(n) time - O(1) space

class Solution {
public:
    int maximumCount(vector<int>& nums) {
        int negatives = 0, positives = 0;
        for (int i = 0; i < nums.size(); i++) {
            if (nums[i] < 0) negatives++;
            if (nums[i] > 0) positives++;
        }
        return max(negatives, positives);
    }
};

Binary Search - O(log n) time - O(1) space

class Solution {
public:
    int maximumCount(vector<int>& nums) {
        int left = 0, n = nums.size(), right = n - 1;
 
        while (left <= right) {
            int mid = left + (right - left) / 2;
            if (nums[mid] > 0) right = mid - 1;
            else left = mid + 1;
        }
        int positives = n - left;
 
        left = 0, right = n -1;
        while (left <= right) {
            int mid = left + (right - left) / 2;
            if (nums[mid] < 0) left = mid +1;
            else right = mid - 1;
        }
        int negatives = right + 1;
 
        return max(negatives, positives);
    }
};