# 713-Subarray Product Less Than K
###### tags: `Medium`
## Question
https://leetcode.com/problems/subarray-product-less-than-k/
## Key
1. Two pointer
對陣列中每個數進行迭代h，每次初始product為乘積總和，再使用一個left pointer(t)去跟product乘積，如果product大於k會再除left pointer對應的數，因此可以避免乘上相同的數值，每次迭代計算subarray數就是用count紀錄，count為h-t+1
## Reference
## Solution

class Solution {
public:
    int numSubarrayProductLessThanK(vector<int>& nums, int k) {
        int p = 1;
        int a = 0;
        int h = 0, t = 0;
        
        for (h=0; h<nums.size(); h++) {
            p = p * nums[h];
            
            while (t<=h && p>=k) {
                p = p / nums[t];
                t++;
            }
            
            a += h-t+1;
        }
        
        return a;
    }
};