MinimumSizeSubarraySum.py

"""
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)).
"""


class Solution:
    def minSubArrayLen(self, target: int, nums: List[int]) -> int:
        min_len = float('inf')
        left, sum_val = 0, 0

        for right in range(len(nums)):
            sum_val += nums[right]

            while sum_val >= target:
                min_len = min(min_len, right - left + 1)
                sum_val -= nums[left]
                left += 1

        return min_len if min_len != float('inf') else 0


# Example usage
if __name__ == "__main__":
    solution = Solution()
    target = 7
    nums = [2, 3, 1, 2, 4, 3]
    result = solution.minSubArrayLen(target, nums)
    print(result)  # Output: 2