2529

2529. Maximum Count of Positive Integer and Negative Integer (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.

• 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?

Related Topics:
Array, Binary Search, Counting

Similar Questions:

• Binary Search (Easy)
• Count Negative Numbers in a Sorted Matrix (Easy)

Solution 1. Binary Search

// OJ: https://leetcode.com/problems/maximum-count-of-positive-integer-and-negative-integer
// Author: github.com/lzl124631x
// Time: O(logN)
// Space: O(1)
class Solution {
public:
    int maximumCount(vector<int>& A) {
        int N = A.size(), L = 0, R = N - 1;
        while (L <= R) {
            int M = (L + R) / 2;
            if (A[M] >= 0) R = M - 1;
            else L = M + 1;
        }
        int neg = L;
        R = N - 1;
        while (L <= R) {
            int M = (L + R) / 2;
            if (A[M] <= 0) L = M + 1;
            else R = M - 1;
        }
        return max(neg, N - L);
    }
};