Find shortest subarray meeting target sum

Given an integer array nums (values may be negative) and an integer T, return the length of the shortest non-empty contiguous subarray whose sum is at least T; return -1 if no such subarray exists. Optimize for n up to 200,000. Explain your approach, prove its correctness, analyze time and space complexity, and implement an O(n) solution.

What a Strong Answer Covers

• A clear algorithm with the right data structures and enough pseudocode or code-level detail to implement it.
• A correctness argument that explains why the algorithm covers all required cases.
• Time and space complexity, plus at least one alternative approach when relevant.
• Focused tests for normal cases, edge cases, and failure modes.

Follow-up Questions

• How would the approach change if the input were streaming or too large for memory?
• What invariants would you assert in production code?
• Which tests would catch off-by-one, duplicate, or tie-breaking bugs?

Find shortest subarray meeting target sum

Given an integer array nums (values may be negative) and an integer T, return the length of the shortest non-empty contiguous subarray whose sum is at least T. Return -1 if no such subarray exists.

The array can be large (n up to 200,000), so you must implement an O(n) solution rather than checking every subarray.

Example 1:

Input: nums = [1], T = 1
Output: 1

Example 2:

Input: nums = [1, 2], T = 4
Output: -1

Example 3:

Input: nums = [2, -1, 2], T = 3
Output: 3

Approach. Because the values can be negative, the classic sliding-window-over-the-array technique does not apply (extending a window can decrease the running sum). Build the prefix-sum array P where P[i] is the sum of the first i elements. A subarray nums[j..i-1] has sum P[i] - P[j], and we want the smallest i - j with P[i] - P[j] >= T.

Scan i from 0 to n while keeping a deque of candidate start indices j whose prefix sums are strictly increasing. For each i: (1) while the front j satisfies P[i] - P[j] >= T, record i - j and pop the front (it can never give a shorter valid window for a later i); (2) while the back's prefix sum is >= P[i], pop it (a smaller or equal prefix sum at a larger index dominates it as a future start); (3) push i. The deque stays monotonically increasing in prefix sum, each index is pushed and popped at most once, giving O(n).

Constraints

• 1 <= n <= 200000 (n may also be 0 for an empty array, in which case return -1)
• -10^4 <= nums[i] <= 10^4
• Subarray sums can exceed 32-bit range; use a 64-bit accumulator (long / long long) in typed languages
• T may be any integer (positive, negative, or zero)
• The subarray must be non-empty and contiguous

Hints

1. Brute force over all O(n^2) subarrays is too slow for n = 200,000. Reframe the problem using prefix sums: subarray nums[j..i-1] has sum P[i] - P[j].
2. Because values can be negative, a plain sliding window fails. Maintain a deque of candidate start indices whose prefix sums are kept strictly increasing.
3. For each i: pop from the FRONT while P[i] - P[front] >= T (record the window length — a valid front can't help a later, longer window), then pop from the BACK any index whose prefix sum is >= P[i] (it's dominated).

Solution

Solution Alignment

The prompt asks for an implementation-level answer. The safest way to present it is to define the state, maintain clear invariants, then walk through complexity and tests.

Problem Restatement

Given an integer array nums (values may be negative) and an integer T, return the length of the shortest non-empty contiguous subarray whose sum is at least T; return -1 if no such subarray exists. Optimize for n up to 200,000. Explain your approach, prove its correctness, analyze time and space complexity, and implement an O(n) solution.

Recommended Approach

Model the states explicitly and use BFS for unweighted shortest paths, Dijkstra for weighted non-negative paths, or topological DP for DAGs. Track visited states at the right granularity so cycles do not cause repeated work.

Correctness

The implementation should maintain an invariant after each loop or operation that directly matches the problem statement. At termination, that invariant implies the returned value has considered every valid candidate exactly once, or has preserved the required data-structure state after every API call.

Complexity

BFS is O(V + E) time and O(V) space for a standard graph. Expanded-state problems multiply those bounds by the number of state dimensions.

Edge Cases and Tests

Disconnected graph, source equals target, cycles, duplicate edges, unreachable target, and whether the answer counts nodes, edges, moves, or transfers.