Determine reachability and minimum jumps

Given an array of non-negative integers nums of length n, where nums[i] is the maximum number of steps you can move forward from index i: (1) Determine whether you can reach index n-1 starting from index 0. (2) If reachable, compute the minimum number of jumps required to reach index n-1. Aim for an O(n) time, O(1) extra space solution if possible. Explain the algorithm, prove its correctness, analyze time/space complexity, and discuss edge cases (e.g., zeros that block progress, n=1, very large jump values, and cases where the last index is unreachable).

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?

Jump Game: Reachability

Given an array of non-negative integers nums of length n, where nums[i] is the maximum number of steps you can move forward from index i, determine whether you can reach the last index n-1 starting from index 0. Return true if reachable, false otherwise.

Aim for an O(n) time, O(1) extra space solution.

Approach (greedy): Track the farthest index reachable so far. Scan left to right; at each index i, if i is already beyond the farthest reachable index, you are stuck and can never reach the end, so return false. Otherwise update farthest = max(farthest, i + nums[i]). If farthest ever reaches or passes n-1, return true. By convention an array of length 0 or 1 is trivially reachable (you are already at the last index).

Correctness: Every index up to farthest is reachable (induction: index 0 is reachable; if index i <= farthest is reachable then so is every index in [i, i+nums[i]]). If the scan never gets stranded (i > farthest), the union of these intervals covers [0, farthest], and once farthest >= n-1 the last index lies in a covered interval. Conversely, getting stranded means a hard gap of zeros blocks all paths.

Constraints

• 1 <= n <= 10^4 (problem also defines the empty array as trivially reachable)
• 0 <= nums[i] <= 10^5
• All values are non-negative integers

Examples

Example 1

Input: ([2, 3, 1, 1, 4],)
Output: True

Explanation: From index 0 (value 2) jump to index 1, which can reach index 4. Reachable.

Example 2

Input: ([3, 2, 1, 0, 4],)
Output: False

Explanation: The best you can do lands on index 3 (value 0), which blocks all progress; index 4 is unreachable.

Hints

1. You do not need to track every path. Track a single number: the farthest index reachable so far.
2. Scan left to right. If the current index i ever exceeds the farthest reachable index, you are stranded behind a wall of zeros and can never reach the end.
3. At each index update farthest = max(farthest, i + nums[i]); as soon as farthest >= n-1 you can stop early and return true.
4. Edge cases: n == 1 (already at the last index, return true) and a leading or interior run of zeros that the farthest reach cannot clear.

Jump Game II: Minimum Jumps

Given an array of non-negative integers nums of length n, where nums[i] is the maximum number of steps you can move forward from index i, compute the minimum number of jumps required to reach the last index n-1 starting from index 0. If the last index is unreachable, return -1.

Aim for an O(n) time, O(1) extra space solution.

Approach (greedy / BFS-by-levels): Treat the array as a layered BFS. Maintain cur_end (the farthest index reachable using the jumps taken so far) and farthest (the farthest index reachable using one more jump from anywhere in the current layer). Scan i from 0 to n-2. If i ever exceeds farthest, the end is unreachable, return -1. Update farthest = max(farthest, i + nums[i]). When i == cur_end, the current jump 'layer' is exhausted, so increment the jump count and advance cur_end = farthest; if cur_end >= n-1 you can return the count immediately.

Correctness: Each jump-count increment corresponds to expanding to the next BFS frontier, which by definition is the minimal number of jumps (BFS finds shortest paths in an unweighted reachability graph). We never overcount because we only increment when we are forced to leave the current frontier. By convention an array of length 0 or 1 needs 0 jumps.

Constraints

• 1 <= n <= 10^4 (problem also defines the empty array as needing 0 jumps)
• 0 <= nums[i] <= 10^5
• All values are non-negative integers; return -1 when the last index cannot be reached

Examples

Example 1

Input: ([2, 3, 1, 1, 4],)
Output: 2

Explanation: Jump 0->1 (value 3), then 1->4. Two jumps is minimal.

Example 2

Input: ([2, 3, 0, 1, 4],)
Output: 2

Explanation: Jump 0->1, then 1->4, skipping the zero at index 2. Two jumps.

Hints

1. Think of it as BFS in layers: every index reachable in k jumps forms one layer.
2. Track cur_end (farthest reachable with the jumps taken so far) and farthest (farthest reachable with one more jump from the current layer).
3. Only increment the jump count when i reaches cur_end — that is the moment you are forced to spend another jump to advance the frontier.
4. Guard against unreachable inputs: if i ever exceeds farthest you are stranded, return -1. Also handle n == 1 (0 jumps) up front.

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 array of non-negative integers nums of length n, where nums[i] is the maximum number of steps you can move forward from index i: ( 1) Determine whether you can reach index n-1 starting from index 0. ( 2) If reachable, compute the minimum number of jumps required to reach index n-1. Aim for an O(n) time, O(1) extra space solution if possible. Explain the algorithm, prove its correctness, analyze time/space complexity, and discuss edge cases (e.g., zeros that block progress, n=1, very large jump values, and cases where the last index is unreachable).

Recommended Approach

Start with a brute-force baseline to confirm correctness, then identify the repeated work or ordering property that enables a better data structure such as a hash map, heap, stack, queue, two pointers, prefix sums, BFS/DFS, or dynamic programming. Write the implementation around a small invariant and test that invariant directly.

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

State the baseline complexity and the optimized complexity. For most interview constraints, justify why the optimized approach meets the expected input size.

Edge Cases and Tests

Empty and singleton inputs, duplicates, ties, invalid inputs, boundary values, and tests that exercise the main invariant.