# 12.03.2025 [2529. Maximum Count of Positive Integer and Negative Integer]

Max positive count or negative count #easy #binary_search

12.03.2025

2529. Maximum Count of Positive Integer and Negative Integer easy blog post substack youtube

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Problem TLDR

Max positive count or negative count #easy #binary_search

Intuition

The brute-force is accepted. However, it is interesting to explore built-in solutions in each language.

Approach

• Kotlin: the shortest is a brute force, no built-in for array Binary Search

• Rust: partition_point

• c++: equal_range is the perfect match

Complexity

• Time complexity: $$O(n)$$ or O(log(n))

• Space complexity: $$O(1)$$

Code

    fun maximumCount(nums: IntArray) =
        max(nums.count { it > 0 }, nums.count { it < 0 })

    fun maximumCount(nums: IntArray) = max(
        -nums.asList().binarySearch { if (it < 0) -1 else 1 } - 1,
        nums.size + nums.asList().binarySearch { if (it < 1) -1 else 1 } + 1)

    pub fn maximum_count(nums: Vec<i32>) -> i32 {
        let a = nums.partition_point(|&x| x < 0);
        let b = nums[a..].partition_point(|&x| x < 1); 
        a.max(nums[a..].len() - b) as i32
    }

    int maximumCount(vector<int>& n) {
        auto [a, b] = equal_range(begin(n), end(n), 0);
        return max(distance(begin(n), a), distance(b, end(n)));
    }

    int maximumCount(vector<int>& nums) {
        int p = 0, n = 0;
        for (int x: nums) p += x > 0, n += x < 0;
        return max(p, n);
    }