Minimum size subarray sum

class Solution:
    def minSubArrayLen(self, target: int, nums: List[int]) -> int:
        res = math.inf
        curr = 0
        left = 0
        right = 0
        while right < len(nums):
            curr += nums[right]
            right += 1
            while curr >= target:
                res = min(right - left, res)
                curr -= nums[left]
                left += 1
        return res if res != math.inf else 0

Minimum Size Subarray Sum

Difficulty: Medium

Given an array of positive integers nums and a positive integer target, return the minimal length of a subarray whose sum is greater than or equal to target. If there is no such subarray, return 0 instead.

 

Example 1:

Input: target = 7, nums = [2,3,1,2,4,3]
Output: 2
Explanation: The subarray [4,3] has the minimal length under the problem constraint.

Example 2:

Input: target = 4, nums = [1,4,4]
Output: 1

Example 3:

Input: target = 11, nums = [1,1,1,1,1,1,1,1]
Output: 0

 

Constraints:

  • 1 <= target <= 109
  • 1 <= nums.length <= 105
  • 1 <= nums[i] <= 104

 

Follow up: If you have figured out the O(n) solution, try coding another solution of which the time complexity is O(n log(n)).