Leetcode.209 Minimum Size Subarray Sum

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.

Example:

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

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).

Solution

1. 滑動視窗

• 時間複雜度：\(O(n)\)
• 空間複雜度：\(O(1)\)

class Solution:
    def minSubArrayLen(self, s: int, nums: List[int]) -> int:
        n = len(nums)
        # 判斷為空的時候
        if n == 0:
            return 0
        # res：（1）記錄結果值 （2）處理沒有結果時的情況
        res = n+1
        l = 0
        r = 0
        su = nums[l]
        while l < n:
            if su >= s:
                res = min(r - l + 1, res)
                su -= nums[l]
                l += 1
            elif r < n-1:
                r += 1
                su += nums[r]
            else:
                break
        if res == n+1:
            return 0
        return res
 

2. 二分查詢法

主要思路：利用二分法查詢符合條件的框最小尺寸

相當於在\([l,r]\)中查詢size位置，滿足條件，最後得到的\(size+1\)才是真正的尺寸

class Solution:
    def minSubArrayLen(self, s: int, nums: List[int]) -> int:
        l = 0
        r = len(nums) - 1
        res = 0
        while l <= r:
            size = l + (r - l)//2
            if self._find_aim(s, nums, size):
                r = size-1
                res = size + 1
            else:
                l = size+1
        return 0 if res == 0 else res
                
    def _find_aim(self,s,nums,size):
        sum = 0
        for i in range(len(nums)):
            sum += nums[i]
            if i > size:
                sum -= nums[i-size-1]
            if sum >= s:
                return True
        return False