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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 = 0, b = 0;
        for (int v : nums) {
            if (v > 0) {
                ++a;
            }
            if (v < 0) {
                ++b;
            }
        }
        return Math.max(a, b);
    }
}

  • 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);
    }
};

  • 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)


  • 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)
}

  • 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);
}


  • 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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