Brute Force Solution

A naive approach would be to iterate through all possible subarrays and check if their sum is greater than or equal to the target. We keep track of the minimum length found so far.

Algorithm

1. Initialize min_length to infinity.
2. Iterate through all possible start indices i from 0 to n-1.
3. For each start index i, iterate through all possible end indices j from i to n-1.
4. Calculate the sum of the subarray nums[i:j+1].
5. If the sum is greater than or equal to the target, update min_length with the minimum of min_length and j - i + 1.
6. If min_length is still infinity, return 0. Otherwise, return min_length.

Code (Python)

def min_subarray_len_brute_force(target: int, nums: list[int]) -> int:
    n = len(nums)
    min_length = float('inf')
    for i in range(n):
        for j in range(i, n):
            sub_array_sum = sum(nums[i:j+1])
            if sub_array_sum >= target:
                min_length = min(min_length, j - i + 1)
    return min_length if min_length != float('inf') else 0

Time Complexity

The time complexity of this approach is O(n^2) because we have nested loops to iterate through all possible subarrays.

Space Complexity

The space complexity is O(1) because we only use a constant amount of extra space.

Optimal Solution (Sliding Window)

A more efficient approach is to use the sliding window technique. This allows us to find the minimum length subarray in O(n) time.

Algorithm

1. Initialize min_length to infinity.
2. Initialize window_start to 0 and window_sum to 0.
3. Iterate through the array with window_end from 0 to n-1.
4. Add nums[window_end] to window_sum.
5. While window_sum is greater than or equal to the target: a. Update min_length with the minimum of min_length and window_end - window_start + 1. b. Subtract nums[window_start] from window_sum. c. Increment window_start.
6. If min_length is still infinity, return 0. Otherwise, return min_length.

Code (Python)

def min_subarray_len_optimal(target: int, nums: list[int]) -> int:
    n = len(nums)
    min_length = float('inf')
    window_start = 0
    window_sum = 0

    for window_end in range(n):
        window_sum += nums[window_end]

        while window_sum >= target:
            min_length = min(min_length, window_end - window_start + 1)
            window_sum -= nums[window_start]
            window_start += 1

    return min_length if min_length != float('inf') else 0

Time Complexity

The time complexity of this approach is O(n) because we iterate through the array at most twice (once with window_end and once with window_start).

Space Complexity

The space complexity is O(1) because we only use a constant amount of extra space.

Edge Cases

• If the array is empty, return 0.
• If the target is 0, and the array contains a non-zero number, the minimal length is 1. If all numbers are zero, the minimal length is infinity and 0 is returned.
• If no subarray sum is greater than or equal to the target, return 0.
• If the target is a very large number and all elements in the array are small, return 0.