```
public class Solution {
public int minSubArrayLen(int s, int[] nums) {
return solveNLogN(s, nums);
}
private int solveN(int s, int[] nums) {
int start = 0, end = 0, sum = 0, minLen = Integer.MAX_VALUE;
while (end < nums.length) {
while (end < nums.length && sum < s) sum += nums[end++];
if (sum < s) break;
while (start < end && sum >= s) sum -= nums[start++];
if (end - start + 1 < minLen) minLen = end - start + 1;
}
return minLen == Integer.MAX_VALUE ? 0 : minLen;
}
private int solveNLogN(int s, int[] nums) {
int[] sums = new int[nums.length + 1];
for (int i = 1; i < sums.length; i++) sums[i] = sums[i - 1] + nums[i - 1];
int minLen = Integer.MAX_VALUE;
for (int i = 0; i < sums.length; i++) {
int end = binarySearch(i + 1, sums.length - 1, sums[i] + s, sums);
if (end == sums.length) break;
if (end - i < minLen) minLen = end - i;
}
return minLen == Integer.MAX_VALUE ? 0 : minLen;
}
private int binarySearch(int lo, int hi, int key, int[] sums) {
while (lo <= hi) {
int mid = (lo + hi) / 2;
if (sums[mid] >= key){
hi = mid - 1;
} else {
lo = mid + 1;
}
}
return lo;
}
}
```

Since the given array contains only positive integers, the subarray sum can only increase by including more elements. Therefore, you don't have to include more elements once the current subarray already has a sum large enough. This gives the linear time complexity solution by maintaining a minimum window with a two indices.

As to NLogN solution, logN immediately reminds you of binary search. In this case, you cannot sort as the current order actually matters. How does one get an ordered array then? Since all elements are positive, the cumulative sum must be strictly increasing. Then, a subarray sum can expressed as the difference between two cumulative sum. Hence, given a start index for the cumulative sum array, the other end index can be searched using binary search.