```
public class Solution {
public int minSubArrayLen(int s, int[] nums) {
if (nums == null || nums.length == 0) {
return 0;
}
int sum = 0;
int ans = Integer.MAX_VALUE;
for (int l = 0, r = 0; r < nums.length; r++) {
sum += nums[r];
while (sum >= s) {
ans = Math.min(ans, r - l + 1);
sum -= nums[l++];
}
}
return ans == Integer.MAX_VALUE ? 0 : ans;
}
}
```

This problem is solved by using the two pointers `l`

and `r`

. In the for loop, we incrementally add the element at `r`

to `sum`

. Whenever the `sum`

is greater or equal to `s`

, we calculate the distance between the two pointers (`r - l + 1`

) and if the distance is smaller than `ans`

, we update `ans`

with the distance. Then subtract the element at `l`

from `sum`

and increment `l`

. At the end of the for loop, we should get the length of minimum subarray.

*Time Complexity*: `O(n)`

*Extra Space*: `O(1)`

## Discussion