Use prefix sums and a monotonic deque.

For each prefix sum, check whether subtracting the smallest useful previous prefix gives a sum of at least k. When it does, update the answer and remove that previous prefix because a later prefix can give a shorter length.

Maintain the deque in increasing order of prefix sums. If the current prefix is smaller than or equal to the last stored prefix, the last one is no longer useful.

Pseudocode:

function shortestSubarrayAtLeastKLength(nums, size, k):
    prefix[0] = 0
    for i from 0 to size - 1:
        prefix[i + 1] = prefix[i] + nums[i]
    deque = empty deque of indexes
    answer = infinity
    for i from 0 to size:
        while deque is not empty and prefix[i] - prefix[deque.front] >= k:
            answer = min(answer, i - deque.front)
            remove front from deque
        while deque is not empty and prefix[i] <= prefix[deque.back]:
            remove back from deque
        add i to back of deque
    if answer is infinity:
        return -1
    return answer