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Formatted question description: https://leetcode.ca/all/2529.html

# 2529. Maximum Count of Positive Integer and Negative Integer

## Description

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?

## Solutions

• 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;
}
}

• 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);
}
};

• class Solution:
def maximumCount(self, nums: List[int]) -> int:
a = len(nums) - bisect_left(nums, 1)
b = bisect_left(nums, 0)
return max(a, b)


• func maximumCount(nums []int) int {
a := len(nums) - sort.SearchInts(nums, 1)
b := sort.SearchInts(nums, 0)
return max(a, b)
}

func max(a, b int) int {
if a > b {
return a
}
return b
}

• function maximumCount(nums: number[]): number {
const search = (target: number) => {
let left = 0;
let right = n;
while (left < right) {
const mid = (left + right) >>> 1;
if (nums[mid] < target) {
left = mid + 1;
} else {
right = mid;
}
}
return left;
};
const n = nums.length;
const i = search(0);
const j = search(1);
return Math.max(i, n - j);
}


• impl Solution {
fn search(nums: &Vec<i32>, target: i32) -> usize {
let mut left = 0;
let mut right = nums.len();
while left < right {
let mid = (left + right) >> 1;
if nums[mid] < target {
left = mid + 1;
} else {
right = mid;
}
}
left
}

pub fn maximum_count(nums: Vec<i32>) -> i32 {
let n = nums.len();
let i = Self::search(&nums, 0);
let j = Self::search(&nums, 1);
i.max(n - j) as i32
}
}