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] <= 2000nums is sorted in a non-decreasing order.
Follow up: Can you solve the problem in O(log(n)) time complexity?
Solutions
Solution 1: Traversal
We can directly traverse the array, count the number of positive and negative integers a and b, and return the larger of a and b.
The time complexity is O(n), where n is the length of the array. The space complexity is O(1).
PythonJavaC++GoTypeScriptRustC
class Solution:
def maximumCount(self, nums: List[int]) -> int:
a = sum(x > 0 for x in nums)
b = sum(x < 0 for x in nums)
return max(a, b)(code-box)
class Solution {
public int maximumCount(int[] nums) {
int a = 0, b = 0;
for (int x : nums) {
if (x > 0) {
++a;
} else if (x < 0) {
++b;
}
}
return Math.max(a, b);
}
}(code-box)
class Solution {
public:
int maximumCount(vector<int>& nums) {
int a = 0, b = 0;
for (int x : nums) {
if (x > 0) {
++a;
} else if (x < 0) {
++b;
}
}
return max(a, b);
}
};(code-box)
func maximumCount(nums []int) int {
var a, b int
for _, x := range nums {
if x > 0 {
a++
} else if x < 0 {
b++
}
}
return max(a, b)
}(code-box)
function maximumCount(nums: number[]): number {
let [a, b] = [0, 0];
for (const x of nums) {
if (x > 0) {
++a;
} else if (x < 0) {
++b;
}
}
return Math.max(a, b);
}(code-box)
impl Solution {
pub fn maximum_count(nums: Vec<i32>) -> i32 {
let mut a = 0;
let mut b = 0;
for x in nums {
if x > 0 {
a += 1;
} else if x < 0 {
b += 1;
}
}
std::cmp::max(a, b)
}
}(code-box)
#define max(a, b) (a > b ? a : b)
int maximumCount(int* nums, int numsSize) {
int a = 0, b = 0;
for (int i = 0; i < numsSize; ++i) {
if (nums[i] > 0) {
++a;
} else if (nums[i] < 0) {
++b;
}
}
return max(a, b);
}(code-box)
Solution 2: Binary Search
Since the array is sorted in non-decreasing order, we can use binary search to find the index i of the first element that is greater than or equal to 1, and the index j of the first element that is greater than or equal to 0. The number of positive integers is a = n - i, and the number of negative integers is b = j. We return the larger of a and b.
The time complexity is O(log n), where n is the length of the array. The space complexity is O(1).
PythonJavaC++GoTypeScriptRustC
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)(code-box)
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;
}
}(code-box)
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);
}
};(code-box)
func maximumCount(nums []int) int {
a := len(nums) - sort.SearchInts(nums, 1)
b := sort.SearchInts(nums, 0)
return max(a, b)
}(code-box)
function maximumCount(nums: number[]): number {
const i = _.sortedLastIndex(nums, 0);
const j = _.sortedIndex(nums, 0);
const [a, b] = [nums.length - i, j];
return Math.max(a, b);
}(code-box)
impl Solution {
fn search(nums: &Vec<i32>, x: i32) -> usize {
let mut left = 0;
let mut right = nums.len();
while left < right {
let mid = (left + right) >> 1;
if nums[mid] >= x {
right = mid;
} else {
left = mid + 1;
}
}
left
}
pub fn maximum_count(nums: Vec<i32>) -> i32 {
let n = nums.len();
let i = Self::search(&nums, 1);
let j = Self::search(&nums, 0);
(n - i).max(j) as i32
}
}(code-box)
#define max(a, b) (a > b ? a : b)
int search(int* nums, int numsSize, int x) {
int left = 0;
int right = numsSize;
while (left < right) {
int mid = (left + right) >> 1;
if (nums[mid] >= x) {
right = mid;
} else {
left = mid + 1;
}
}
return left;
}
int maximumCount(int* nums, int numsSize) {
int i = search(nums, numsSize, 1);
int j = search(nums, numsSize, 0);
return max(numsSize - i, j);
}(code-box)
