How can you determine the shortest subarray that meets or exceeds a target sum from a given integer array, using a greedy algorithm?

Question Analysis

This question is asking you to find the shortest contiguous subarray within a given integer array whose sum is equal to or exceeds a specified target sum. The challenge is to determine this subarray efficiently using a greedy algorithm approach.

In the context of greedy algorithms, the idea is to make the locally optimal choice at each step with the hope of finding a global optimum. For this problem, a two-pointer technique or sliding window approach is typically used to maintain a running sum of elements as you traverse the array, attempting to minimize the length of the subarray that meets the target sum.

Answer

To determine the shortest subarray that meets or exceeds a target sum using a greedy algorithm, you can use the sliding window technique. Here's a clear, professional, and concise explanation of how to implement this:

1. Initialize Variables: Start by initializing two pointers, start and end, both set to the beginning of the array. Also, initialize min_length to a large number (to represent infinity) and current_sum to zero.

2. Expand the Window: Move the end pointer through the array, adding each element to current_sum.

3. Contract the Window: Once current_sum meets or exceeds the target sum, attempt to shrink the window from the left by moving the start pointer to minimize the subarray length while still maintaining the sum requirement. For each valid subarray found, update min_length if the current subarray is shorter.

4. Repeat: Continue this process until the end pointer reaches the end of the array.

5. Check Result: After the loop, if min_length remains unchanged, it indicates no valid subarray was found. Otherwise, min_length will contain the length of the shortest subarray that meets the target sum.

Here is a concise implementation in Python:

def minSubArrayLen(target, nums):
    start, current_sum = 0, 0
    min_length = float('inf')

    for end in range(len(nums)):
        current_sum += nums[end]

        while current_sum >= target:
            min_length = min(min_length, end - start + 1)
            current_sum -= nums[start]
            start += 1

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

• Time Complexity: The time complexity of this approach is O(n) because each element is processed at most twice (once added and once removed).
• Space Complexity: The space complexity is O(1) as no additional space is used that grows with the input size.

This solution efficiently finds the shortest subarray using a greedy sliding window approach, aligning with the requirements of the problem.