LeetCode Minimum Size Subarray Sum Solution

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)).

We want to use a sliding window to find the minimum subarray (window). Because the size of the window is unknown, we must use a flexible sliding window that searchs through all the valid windows that meet the requirement. We will apply the flexible sliding window template on this question. Our search starts on interval (0,0) and extends rightwards before the total reaches target. When the total succeeds the target we have found a valid subarray. Then, we start shrinking this subarray from the left finding a smaller subarray until the window is no longer valid. Afterwards, we continue this process until we iterate through the entire array to find the minimum size subarray that has sum >= target.

Implementation

1def minSubArrayLen(self, target: int, nums: List[int]) -> int:
2    size = len(nums)+1
3    total, l = 0, 0
4    for r in range(len(nums)):
5        total += nums[r]
6        while total >= target:       # valid
7            size = min(size, r-l+1)
8            total -= nums[l]
9            l += 1      
10    return size if size != len(nums)+1 else 0

The above solution using a flexible sliding window uses O(n) time complexity. As a follow up, is there an algorithm that solves this question in O(n log(n))? Yes! Consider using the n elements in nums as a starting point of a subarray, and then use O(log(n)) time complexity to find the endpoint of that subarray. This is can be done via a for loop and a binary search on a prefix sum array.

1def minSubArrayLen(self, target: int, nums: List[int]) -> int:
2    prefix_sum = [0]
3    for n in nums:
4        prefix_sum.append(prefix_sum[-1] + n)
5    
6    size = len(nums)+1
7    for start in range(len(nums)):
8        total = 0
9        l, r, end = 0, len(nums)-1, -1
10        while l <= r:
11            mid = (l+r)//2
12            if prefix_sum[mid+1] - prefix_sum[start] >= target:
13                end, r = mid, mid - 1
14            else: l = mid + 1
15        if end != -1: size = min(size, end-start+1) 
16    return size if size != len(nums)+1 else 0