maximum-count-of-positive-integer-and-negative-integer

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Algorithms

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

<ul>
	<li>In other words, if the number of positive integers in <code>nums</code> is <code>pos</code> and the number of negative integers is <code>neg</code>, then return the maximum of <code>pos</code> and <code>neg</code>.</li>
</ul>

Note that <code>0</code> is neither positive nor negative.

&nbsp;
Example 1:

<pre>
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.
</pre>

Example 2:

<pre>
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.
</pre>

Example 3:

<pre>
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.
</pre>

&nbsp;
Constraints:

<ul>
	<li><code>1 &lt;= nums.length &lt;= 2000</code></li>
	<li><code>-2000 &lt;= nums[i] &lt;= 2000</code></li>
	<li><code>nums</code> is sorted in a non-decreasing order.</li>
</ul>

&nbsp;
Follow up: Can you solve the problem in <code>O(log(n))</code> time complexity?

maximumCount

Maximum Count of Positive Integer and Negative Integer

Binary Search

binary-search

二分查找

Count Negative Numbers in a Sorted Matrix

count-negative-numbers-in-a-sorted-matrix

统计有序矩阵中的负数

Array

Counting

给你一个按 非递减顺序 排列的数组 <code>nums</code> ，返回正整数数目和负整数数目中的最大值。

<ul>
	<li>换句话讲，如果 <code>nums</code> 中正整数的数目是 <code>pos</code> ，而负整数的数目是 <code>neg</code> ，返回 <code>pos</code> 和 <code>neg</code>二者中的最大值。</li>
</ul>

注意：<code>0</code> 既不是正整数也不是负整数。

&nbsp;

示例 1：

<pre>
输入：nums = [-2,-1,-1,1,2,3]
输出：3
解释：共有 3 个正整数和 3 个负整数。计数得到的最大值是 3 。
</pre>

示例 2：

<pre>
输入：nums = [-3,-2,-1,0,0,1,2]
输出：3
解释：共有 2 个正整数和 3 个负整数。计数得到的最大值是 3 。
</pre>

示例 3：

<pre>
输入：nums = [5,20,66,1314]
输出：4
解释：共有 4 个正整数和 0 个负整数。计数得到的最大值是 4 。
</pre>

&nbsp;

提示：

<ul>
	<li><code>1 &lt;= nums.length &lt;= 2000</code></li>
	<li><code>-2000 &lt;= nums[i] &lt;= 2000</code></li>
	<li><code>nums</code> 按 非递减顺序 排列。</li>
</ul>

&nbsp;

进阶：你可以设计并实现时间复杂度为 <code>O(log(n))</code> 的算法解决此问题吗？