209. Minimum Size Subarray Sum

https://leetcode.com/problems/minimum-size-subarray-sum/#/description

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 under the problem constraint.

 

Sol 1:

Two pointer.

class Solution(object):
    def minSubArrayLen(self, s, nums):
        """
        :type s: int
        :type nums: List[int]
        :rtype: int
        """
        
        # two pointers
        if len(nums) == 0:
            return 0
        
        sum = 0
        i = 0
        # fast pointer j
        j = 0
        res = len(nums) + 1
        while sum < s:
            sum += nums[j]
            j += 1
            
            # to enter the while loop below, sum is greater than s, now let's use another pointer i(slow) pointer to find the intervals between i and j.  
            
            while sum >= s:
                # if i and j still have elements among them
                if  res > j - i:
                    res = j - i
                sum -= nums[i]
                i += 1
                
                
            if j == len(nums):
                break
        
        # if sum > s at the very beginning, then the program will not enter the first while loop. Thus 0 will be returned. 
        if res > len(nums):
            return 0
        
        return res
            

Sol 2:

Precalculated sums saved at the same list.  i.e. replace the input list.

class Solution(object):
    def minSubArrayLen(self, s, nums):
        """
        :type s: int
        :type nums: List[int]
        :rtype: int
        """
        
        # Precalculated sums saved at the same list
        
        sm = sum(nums)
        if sm < s:
            return 0
        for i in range(len(nums)):
            # it is like create a new list, but actually it is in place.(some algorithm here.name?)
            nums[i], sm = sm, sm - nums[i]
        res = len(nums)
        
        # complete the difference list
        nums.append(0)
        # traverse nums backwards, but traverse order does not matter
        # for i in range(len(nums)-1, -1, -1):
        for i in range(len(nums)):
            if nums[i] >= s:
                j = min(len(nums)-1, i + res - 1)
                while nums[i] - nums[j] >= s:
                    res = j - i
                    j -= 1
        return res