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#
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| class Solution:
def maximumCount(self, nums: List[int]) -> int:
a = sum(v > 0 for v in nums)
b = sum(v < 0 for v in nums)
return max(a, b)
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| class Solution {
public int maximumCount(int[] nums) {
int a = 0, b = 0;
for (int v : nums) {
if (v > 0) {
++a;
}
if (v < 0) {
++b;
}
}
return Math.max(a, b);
}
}
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| class Solution {
public:
int maximumCount(vector<int>& nums) {
int a = 0, b = 0;
for (int& v : nums) {
if (v > 0) {
++a;
}
if (v < 0) {
++b;
}
}
return max(a, b);
}
};
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| func maximumCount(nums []int) int {
a, b := 0, 0
for _, v := range nums {
if v > 0 {
a++
}
if v < 0 {
b++
}
}
return max(a, b)
}
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| function maximumCount(nums: number[]): number {
const count = [0, 0];
for (const num of nums) {
if (num < 0) {
count[0]++;
} else if (num > 0) {
count[1]++;
}
}
return Math.max(...count);
}
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| impl Solution {
pub fn maximum_count(nums: Vec<i32>) -> i32 {
let mut count = [0, 0];
for &num in nums.iter() {
if num < 0 {
count[0] += 1;
} else if num > 0 {
count[1] += 1;
}
}
*count.iter().max().unwrap()
}
}
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| #define max(a, b) (((a) > (b)) ? (a) : (b))
int maximumCount(int* nums, int numsSize) {
int count[2] = {0};
for (int i = 0; i < numsSize; i++) {
if (nums[i] < 0) {
count[0]++;
} else if (nums[i] > 0) {
count[1]++;
}
}
return max(count[0], count[1]);
}
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Solution 2#
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| 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)
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| 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;
}
}
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| 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);
}
};
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| func maximumCount(nums []int) int {
a := len(nums) - sort.SearchInts(nums, 1)
b := sort.SearchInts(nums, 0)
return max(a, b)
}
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| 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);
}
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| 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
}
}
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| #define max(a, b) (((a) > (b)) ? (a) : (b))
int search(int* nums, int numsSize, int target) {
int left = 0;
int right = numsSize;
while (left < right) {
int mid = (left + right) >> 1;
if (nums[mid] < target) {
left = mid + 1;
} else {
right = mid;
}
}
return left;
}
int maximumCount(int* nums, int numsSize) {
int i = search(nums, numsSize, 0);
int j = search(nums, numsSize, 1);
return max(i, numsSize - j);
}
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Solution 3#
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| impl Solution {
pub fn maximum_count(nums: Vec<i32>) -> i32 {
let mut a = 0;
let mut b = 0;
for n in nums {
if n > 0 {
a += 1;
} else if n < 0 {
b += 1;
}
}
std::cmp::max(a, b)
}
}
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