Given an array of positive integers
nums and a positive integer
target, return the minimal length of a contiguous subarray
[numsl, numsl+1, ..., numsr-1, numsr] of which the sum is greater than or equal to
target. If there is no such subarray, return
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.
Input: target = 4, nums = [1,4,4] Output: 1
Input: target = 11, nums = [1,1,1,1,1,1,1,1] Output: 0
1 <= target <= 109
1 <= nums.length <= 105
1 <= nums[i] <= 105
Follow up: If you have figured out the
O(n) solution, try coding another solution of which the time complexity is
Have two pointers, one fast, one slow to move to the right. Keep moving the faster pointer to the right as long as the sum is less than the target. Subtract the slower pointer pointing number when every time moving the slower pointer.
class Solution: def minSubArrayLen(self, target: int, nums: List[int]) -> int: j = 0 current_sum = 0 min_length = float(inf) for i in range(len(nums)): while current_sum < target and j < len(nums): current_sum += nums[j] j += 1 if current_sum >= target: min_length = min(min_length, j - i) current_sum -= nums[i] if min_length == float(inf): return 0 return min_length
- Time Complexity: O(N).
- Space Complexity: O(1).
One Thought to “LeetCode 209. Minimum Size Subarray Sum”
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