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.
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
If you have figured out the O(n) solution, try coding another solution of which the time complexity is O(n log(n)).
class Solution
{
public:
int minSubArrayLen(int target, vector<int> &nums)
{
int i = 0;
int j = 0;
int sum = 0;
int minimum = INT_MAX;
while (j < nums.size())
{
sum += nums[j];
while (sum >= target)
{
sum -= nums[i];
minimum = min(j - i + 1, minimum);
i++;
}
j++;
}
if (minimum == INT_MAX)
{
return 0;
}
return minimum;
}
};
Sliding window algorithm is a problem solving technique that transforms two nested loops into one loop. It can reduce the time complexity to O(n).

Reference
[1] https://builtin.com/data-science/sliding-window-algorithm