[Leetcode] 209. Minimum Size Subarray Sum

https://leetcode.com/problems/minimum-size-subarray-sum/description/?envType=study-plan-v2&envId=top-interview-150

initial

very tricky cuz the length of sliding window is not fixed so I didnt know how. I got a hint from gemini and voila but im calculating the sum of sliding window for each for iteration. So there was runtime but it still worked

class Solution:
    def minSubArrayLen(self, target: int, nums: List[int]) -> int:
        ans=int(10e18)
        left,right=0,0
        while left< len(nums) and right<len(nums):
            window=nums[left:right+1]
            if sum(window)>=target:
                ans=min(ans,right-left+1)
                left+=1
            else:
                if left==0 and right==len(nums)-1:
                    return 0
                else:
                    right+=1
        return ans

solution

so i searched online and the way the conditions are thought are so good here. They didnt make it like if cur_Sum>=target else. The conditions were if cur_sum>=target, elif right<len(nums) cuz this elif checks if we can add as much elements in the list to cur_sum if cur_sum is smaller than target. If those 2 conditions are not met, (i.e. not a valid sliding window cuz we either have cur_sum smaller than target or the right pointer is out of index), then we break out of this True loop cuz theres no valid ans.

class Solution:
    def minSubArrayLen(self, target: int, nums: List[int]) -> int:
        ans=int(10e18)
        left,right=0,0
        cur_sum=0
        while True:
            if cur_sum>=target:
                cur_sum-=nums[left]
                ans=min(ans,right-left)
                left+=1
            elif right<len(nums):
                cur_sum+=nums[right]
                right+=1
            else:
                break
        if ans==int(10e18):
            return 0
        else:
            return ans

complexity

o(n) time
o(1) space