Bu masalaga istalgan AI yechim bera oladi. Lekin biz Engineer sifatida uni kodlarini optimizatsiya qila olamizmi?
Oddiy yechimi quyidagi kurinishda bo'lishi mumkin.

```cpp
class ProductOfNumbers {
public:
    ProductOfNumbers() {

    }

    void add(int num) {
        data_.push_back(num);
    }

    int getProduct(int k) {
        long long pr = 1;

        for(int i = 0; i < k; i++) {
            pr *= data_[data_.size() - i - 1];
        }

        return pr;
    }

private:
    std::vector<int> data_;
};

class Solution {
public:
    int maximumCount(vector<int>& nums) {
        return 
            std::max(std::ranges::count_if(nums,[](auto a){ return a < 0; }),
                    std::ranges::count_if(nums,[](auto a){ return a > 0; }));
                
    }
};

nums = {2, 4, 6, 8, 10}

nums[1] -= 3;
nums[2] -= 3;
nums[3] -= 3;

nums = {2, 1, 3, 5, 10}

    diff[l] ga -val qo‘shamiz (bu l dan boshlab kamaytirish boshlanishini bildiradi).
    diff[r+1] ga +val qo‘shamiz (bu r dan keyin qayta tiklashni bildiradi).

diff = {0, -3, 0, 0, 3, 0}

nums[0] = 2
nums[1] = 4 + (-3) = 1
nums[2] = 6 + (-3) = 3
nums[3] = 8 + (-3) = 5
nums[4] = 10 + 0 = 10

int main(){

    vector<int> nums = {2, 4, 6, 8, 10};
    vector<int> diff(nums.size() + 1, 0);
    
    //[1,3] oraligdagi barcha elementlarni 3 kamaytiramiz
    diff[1] -= 3;
    diff[4] += 3;//Qolganlari uchun +3 chunki [4...oxirgacha bu amal bajarilmasligi kerak]
    //Yuqoridagi amalni bir nechi marta bajarishimiz mumkin
    //endi natijaviy massivni hosil qilamiz
    int current = 0;
    for (int i = 0; i < nums.size(); i++) {
        current += diff[i];
        nums[i] += current;
    }

    for(auto& num: nums){
        std::cout << num << "\n";
    }
    return 0;
}