# Leetcode 209 - Minimum Size Subarray Sum ### Understanding the problem

Given an array of n positive integers and a positive integer s, find the minimal length of a contiguous subarray of which the sum ≥ s. If there isn’t one, return 0 instead.

For example, given the array [2,3,1,2,4,3] and s = 7, the subarray [4,3] has the minimal length of 2.

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One approach to solve this problem is to use a sliding window approach. We can use two pointers, one to represent the start of the window, and the other to represent the end of the window. We will maintain a variable to keep track of the current sum of the subarray and compare it with the given sum s.

We start by initializing the left pointer to 0, and the right pointer to 0. We also initialize the current sum to 0.

We then iterate through the array by incrementing the right pointer. For each iteration, we add the current element to the current sum, and check if the current sum is greater than or equal to s. If it is, we update the minimum length of the subarray. If it is not, we move the left pointer to the right and remove the element from the current sum.

We continue this process until the right pointer reaches the end of the array. We then return the minimum length of the subarray.

Here is an example of a python code that implements this approach:

``````def minSubArrayLen(s: int, nums: List[int]) -> int:
left = 0 # initialize left pointer to 0
curr_sum = 0 # initialize current sum to 0
min_len = float("inf") # initialize minimum length of subarray to positive infinity
for right in range(len(nums)):
curr_sum += nums[right]  # add current element to current sum
while curr_sum >= s: # check if current sum is greater than or equal to s
min_len = min(min_len, right - left + 1) # update minimum length of subarray
curr_sum -= nums[left] # remove element from current sum
left += 1 # move left pointer to the right
return min_len if min_len != float("inf") else 0 # return minimum length of subarray or 0 if not found
``````

In this approach, we are using two pointers, left and right, to represent the current window. We start by initializing the left pointer to 0 and the right pointer to 0. We also initialize the current sum to 0 and the minimum length of the subarray to positive infinity.

We then iterate through the array by incrementing the right pointer. For each iteration, we add the current element to the current sum and check if the current sum is greater than or equal to s. If it is, we update the minimum length of the subarray by taking the minimum of the current minimum length and the difference between the right and left pointer + 1 (length of the current subarray). If the current sum is not greater than or equal to s, we move the left pointer to the right and remove the element from the current sum.

We continue this process until the right pointer reaches the end of the array. We then return the minimum length of the subarray.

Let’s test our solution with the example inputs:

``````s = 7
nums = [2,3,1,2,4,3]
print(minSubArrayLen(s, nums))
``````

Output: 2

As we can see, the output is correct and matches the expected output.

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