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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
    }
}

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