0209-minimum-size-subarray-sum¶
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Description¶
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 <= 1091 <= nums.length <= 1051 <= 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)).
Solution(Python)¶
class Solution:
def minSubArrayLen(self, target: int, nums: List[int]) -> int:
# target = 7, nums = [2,3,1,2,4,3]
# minimal length of subarray int sum(subarray) >= target
#
# valid window -> 7
# 2 3 1 2 4 3
#
# nums = [3] k = 3 ; 1
# nums = [1,2,3] k =100; 0
#
# left ..right
# sum = 8
# target = 7
# sum >= target left += 1
# sum -= nums[left]
# expand right
# sum +=nums[right]
# while sum >= k
# update answer
# remove nums[left]
# left ++
#
# target = 7
# 2 3 1 2 4 3
# left =0
# sum = 0
# ans= float(inf)
# right =0 sum 2
# right = 1 sum 5
# right = 2 ans = 2 sum 8 - 2 = 6 left = 1
#
# coorect condition cur_sum >= target remove left
l = 0
cur_sum = 0
ans = float('inf')
for r in range(len(nums)):
cur_sum += nums[r]
while cur_sum >= target and l < len(nums):
ans= min(ans, r - l + 1)
cur_sum -= nums[l]
l+=1
return ans if ans != float('inf') else 0