Leetcode: Minimum Size Subarray Sum

Minimum Size Subarray Sum



Similar Problems:


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.

For example, given the array [2,3,1,2,4,3] and s = 7,
the subarray [4,3] has the minimal length under the problem constraint.

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

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.


## Blog link: https://code.dennyzhang.com/minimum-size-subarray-sum
## Basic Ideas: Two pointers + Ge accumulated sum
##
##   Corner case: 0 elements or 1 elements
##   Right pointer move one step each time.
##   Left pointer don't need to move left. Why? We only need to find the minimal length
##
## Complexity: Time O(n), Space O(1)
import sys
class Solution:
    def minSubArrayLen(self, s, nums):
        """
        :type s: int
        :type nums: List[int]
        :rtype: int
        """
        length = len(nums)
        if length == 0: return 0

        min_length = sys.maxsize
        left, sum_v = 0, 0

        # always move right by 1 step
        for right in range(0, length):
            sum_v += nums[right]
            while sum_v >=s:
                min_length = min(min_length, right-left+1)
                # left won't pass right. Why? Because s>0
                sum_v -= nums[left]
                left += 1

        return min_length if min_length != sys.maxsize else 0
linkedin
github
slack

Share It, If You Like It.

Leave a Reply

Your email address will not be published.