"""
[start] 12:34

Problem Solving Thought Process:

1. Brute Force
Using 2 for loops to find out 2 element that sums up to the target
Constantly update the minimum length in a globalMin

globalMin = infi
for i -> n
 for j -> n
 cur_sum += nums[j]
 if cur_sum > target:
 globalMin = min(globalMin, cur_sum)
 break
return globalMin

Time: O(N^2)
Space: O(1)


2. Two Pointer 
We can potentially only iterate "j" from beginning to end once by moving "i" along the way

Time: O(N)
Space: O(1)
"""
#[Coding Start] 12:39
#[Coding Finish] 12:43
#[Eyeballing code] 12:43 - 12:46
# Bug
"""
globalMin = float('inf')
curSum = 0
j = 0

for i in range(len(nums)):
 while j < len(nums) and curSum < target:
 curSum += nums[j]
 j += 1

 if curSum >= target:
 globalMin = min(globalMin, j - i)

 curSum -= nums[i]

return globalMin -> bug1. 如果没有找到合适的结果,应该返回0而不是初始值的infinite

fail testcases:

3
[1,1]
"""
 
class Solution(object):
 def minSubArrayLen(self, target, nums): 
 globalMin = float('inf')
 curSum = 0
 j = 0
 
 for i in range(len(nums)):
 while j < len(nums) and curSum < target:
 curSum += nums[j]
 j += 1
 
 if curSum >= target:
 globalMin = min(globalMin, j - i)
 
 curSum -= nums[i]
 
 return globalMin if globalMin != float('inf') else 0