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10. Greedy Furthest Reach: Jump Game (LC 55) • Invariant Proof & Speed-Run

Theory Context•Miscellaneous & Sweeps

MediumLC 55

Jump Game (LeetCode 55)

You will see how one number, the furthest index any jump can reach, replaces a search over every path.

Target Frequency:AmazonGoogleMicrosoft

You get an integer array nums and begin on index 0. The value nums[i] is the longest jump you may make from index i: from there you can move forward by any whole number of steps from 1 up to nums[i], and a value of 0 means you cannot move on from that index.

Return true if some sequence of jumps lands you on the last index, and false if every sequence gets stuck before it.

Worked Examples

Example 1

Input:nums = [2,3,1,1,4]

Output:true

Explanation: Index 1 is one step away, and its value `3` carries you straight to index 4, the last index. Other routes work too; one is enough.

Example 2

Input:nums = [3,2,1,0,4]

Output:false

Explanation: Indices 0, 1 and 2 can each get as far as index 3 and no further (`0 + 3`, `1 + 2`, `2 + 1`), and index 3 holds `0`, so no jump ever passes it. Index 4 stays out of reach.

⚖️Formal Constraints & Bounds

1 <= nums.length <= 10⁴
0 <= nums[i] <= 10⁵

Deep-Dive & Conceptual Insights

Why It Works & Core Invariant

You never need the path, only how far any path can get: the reachable indices always form one block 0..reach, so one number and one pass decide the answer.

Real-World Scenario & Production Applications

Any chain of forward hops with a limited range per hop: relay stations along a pipeline, charging stops for a drone, stepping stones on a path. Whether the end can be reached never needs the full route, only the furthest point any route could reach so far, so one running number answers it in a single pass.

Step-by-Step Execution Trace Table

Example 2, nums = [3,2,1,0,4] (the trap case):

`i`	`nums[i]`	Check `i > reach`	`reach = max(reach, i + nums[i])`	What it means
0	3	`0 > 0`: no	`max(0, 0 + 3) = 3`	Indices 0 to 3 can be landed on
1	2	`1 > 3`: no	`max(3, 1 + 2) = 3`	No further than before
2	1	`2 > 3`: no	`max(3, 2 + 1) = 3`	No further than before
3	0	`3 > 3`: no	`max(3, 3 + 0) = 3`	Index 3 can't move on
4	4	`4 > 3`: yes	(not reached)	No jump lands on index 4: `return False`

Scroll horizontally to see all columns, or expand to full screen

Core Invariant Specification & Code Shape

Archetype Code Shape

Python

1	You never need the path, only how far any path can get: keep `reach`, the furthest index a run of jumps can land on so far.
2	Every index from `0` to `reach` can be landed on (a jump may be shorter than `nums[i]`), so extend it with `reach = max(reach, i + nums[i])`: the larger reach keeps every option.
3	Start `reach = 0`, loop `for i in range(len(nums))` once, and `return True` after the loop.
4	The trap: check `if i > reach: return False` before reading `nums[i]`, or an index nobody can land on extends `reach`.

Target: Jump Game (LeetCode 55). Before any jump only index 0 is reachable, so the reachable block is 0..0.

Boundary Model: Sorted Coordinate Sweep-Line & Monotonic Stack

Intervals sorted by start coordinate; monotonic stack preserves strictly increasing or decreasing element values.

Loop Invariant Termination

Sweep: if curr.start <= prev.end: merge; Stack: while stack and nums[i] >= stack[-1]: stack.pop().

Conceptual Narrative

🧭 Conceptual Foundation & Pattern Intuition

A path search looks necessary here: from index 0 you could jump 1, 2, ... up to nums[0] steps, and every landing spot branches again. Greedy skips the search. A jump may be shorter than its maximum, so if some jump can land on index k, every index before k can be landed on too. The reachable indices are therefore always one unbroken block 0..reach, and a single number, reach, says everything about the past. Walk the array once, and at each index you can stand on, push reach out to i + nums[i] if that is further.

🪜 The Analogy: Stepping Stones Across a River

You cross a river on a row of stepping stones. Each stone has a number painted on it: how many stones ahead you can leap from there, at most. You don't plan a route. You walk along the stones you can reach and keep one note in your head: "the furthest stone I could get to so far." Each new stone may move that note further. If you ever look at a stone beyond the note, nobody can stand on it, and the far bank is out of reach.

🪄 The Mathematical Harmony / Magic Trick

Code / Blueprint

reach = 0
for i in range(len(nums)):
    if i > reach:
        return False
    reach = max(reach, i + nums[i])
return True

This is the greedy argument in one line: max keeps the larger block, and a larger block contains every index a smaller one does, so choosing it can never cost an option. A locally best choice that never has to be undone is exactly when greedy is correct. The order of the two lines matters: check i > reach first, so an index nobody can land on never extends reach.

💡 Summary

Keep the furthest reach, not a path. Check i > reach before using nums[i], extend with max, and return True once the loop has stood on every index. One pass: $O(N)$ time, $O(1)$ space.

Extending reach from an index you can't stand on: check i > reach before reach = max(reach, i + nums[i]). On [3,2,1,0,4] index 4 is past reach = 3; letting its jump count returns True instead of False.
Assigning instead of taking the max: reach = i + nums[i] lets a short jump after a long one shrink the reach, so [4,1,0,0,0] gets stuck at index 3 and returns False instead of True.
Searching paths: trying every jump length from every index is correct but costs up to O(N · max(nums[i])), about 10^9 steps at these limits. One number, reach, replaces the whole search.
Answering a different question: reach only says yes or no. The fewest jumps, as in Jump Game II (LC 45), needs a second boundary for where the current jump runs out.

Senior SWE Reasoning Architecture

4-Phase Thought Process Model

You will see how a senior engineer spots a greedy reachability question and defends the one-number answer out loud.

Pattern Recognition Signals

The 10-second spot

"The longest jump from each index" and "can you reach the last index": a yes/no reachability question where a longer jump is never worse than a shorter one. When the locally best option (the furthest reach) keeps every choice any other option keeps, that is the signal for Greedy.

Formulating the Predicate & Invariants

Turning intuition into a boolean rule

Before index i is checked, reach is the furthest index any run of jumps from the indices before i can land on, and every index up to reach can be landed on. If i > reach, return False; otherwise reach = max(reach, i + nums[i]).

Silent Failure Traps & Edge Cases

Where confident candidates still lose points

if i > reach: return False must come before reach = max(reach, i + nums[i]): an index nobody can land on must not extend reach (on [3,2,1,0,4] that would return True).
max, not reach = i + nums[i]: a short jump after a long one would shrink the reach ([4,1,0,0,0] would stop at index 3).
Don't try every jump length from every index: that is correct but O(N · max(nums[i])), up to 10^9 steps at these limits; a single reach makes it O(N).

The 60-Second Interview Pitch

Say this out loud before you type a single line

I'd solve this with Greedy. I never need the actual path, only the furthest index any sequence of jumps could land on so far; I call it reach. A jump can be shorter than its maximum, so every index up to reach can be landed on, and the reachable indices always form one block from zero to reach. I walk the array once. At each index I first check whether i is past reach: if it is, nobody can stand there, so I return false. Otherwise I extend reach to the max of reach and i plus nums of i. Taking the max never loses an option, because a longer block contains every shorter one. The trap is the order: checking first stops an unreachable index from pushing reach forward. If the loop finishes, I return true. One pass, so O(N) time and O(1) space.

So: keep reach, check i > reach first, extend with max, and say why the larger reach never loses an option.

Big-O Invariant Derivation

Complexity & Mathematical Proof

TIME COMPLEXITY

O(N)

Look at the code: reach = 0 is one assignment, and for i in range(len(nums)) visits each index at most once. Its body does one comparison (i > reach) and one max, a constant amount of work, and the loop may stop early at return False. Total: O(N).

SPACE COMPLEXITY

O(1)

The code keeps only reach and i, whatever the input size, and returns one boolean: O(1) extra space.

Formal Recurrence Relation

T(N) = O(1) + N · O(1) = O(N)

Derivation Progression

Start

O(1)

reach = 0 is one assignment.

Index loop

at most N iterations

for i in range(len(nums)) visits each index once, and stops early if return False runs.

Per-index work

O(1) per iteration

One comparison i > reach and one max(reach, i + nums[i]), whatever N is.

Total

O(N)

Constant work per index over at most N indices.

Variable Definitions

$N$

Number of indices, len(nums)

$reach$

Furthest index some run of jumps can land on so far

Memory Architecture & Bounds

🟣 Call Stack

O(1) Iterative, no recursion

🔵 Auxiliary Heap

O(1): reach, i

🟢 Output Space

O(1): one boolean

Boundary Best / Worst Cases

Best Case

$O(1)$ : nums[0] = 0 with more than one index, so the loop returns False at i = 1

Average Case

$O(N)$

Worst Case

$O(N)$ : the last index is reachable, so every index is checked

Recurrence Tree Topology

Synthesizing vector architecture diagram...

Staff+ Engineering Perspective

Senior SWE Deconstruction & Hardware Caveats

ARCHITECTURAL SIGNALS & INTERVIEW TRIGGERS

Triggers: "the longest jump allowed from each index", "can you reach the last index". A yes/no reachability question where a longer jump is never worse than a shorter one: Greedy with one running number, reach.

CONSTRAINTS & BOUNDS

$N \le 10^4$ indices and jumps up to $10^5$ . Trying every jump length from every index costs up to $N \cdot \max(nums[i]) \approx 10^9$ steps; the greedy reach is $O(N)$ time and $O(1)$ space, and it reads nums once, left to right, so it also works on a stream.

FAANG PRODUCTION TRAPS & EDGE CASES

Checking i > reach after the update lets an unreachable index extend the reach. On a stream the same rule means a record that arrives beyond the covered prefix must wait: it can't move the prefix forward. i + nums[i] is at most $10^4 + 10^5$ , so it fits a 32-bit integer with room to spare.

Core Algorithmic State Invariants

1. One-Block Reach Invariant

Before index `i` is checked, every index from `0` to `reach` can be landed on, because a jump may be shorter than `nums[i]`. The reachable indices are always one unbroken block.

2. Check, Then Extend

`if i > reach: return False` comes first: an index past the block can't be stood on and must never extend it. Otherwise `reach = max(reach, i + nums[i])` keeps the larger block, which contains every option of the smaller one.

3. One Linear Pass

Each index is checked once with O(1) work, so the answer comes in O(N) time and O(1) space; `return True` after the loop means the last index was stood on.

Theory Context•Miscellaneous & Sweeps

MediumLC 55

Jump Game (LeetCode 55)

You will see how one number, the furthest index any jump can reach, replaces a search over every path.

Target Frequency:AmazonGoogleMicrosoft

Return true if some sequence of jumps lands you on the last index, and false if every sequence gets stuck before it.

Worked Examples

Example 1

Input:nums = [2,3,1,1,4]

Output:true

Explanation: Index 1 is one step away, and its value `3` carries you straight to index 4, the last index. Other routes work too; one is enough.

Example 2

Input:nums = [3,2,1,0,4]

Output:false

Explanation: Indices 0, 1 and 2 can each get as far as index 3 and no further (`0 + 3`, `1 + 2`, `2 + 1`), and index 3 holds `0`, so no jump ever passes it. Index 4 stays out of reach.

⚖️Formal Constraints & Bounds

1 <= nums.length <= 10⁴
0 <= nums[i] <= 10⁵

Deep-Dive & Conceptual Insights

Why It Works & Core Invariant

You never need the path, only how far any path can get: the reachable indices always form one block 0..reach, so one number and one pass decide the answer.

Real-World Scenario & Production Applications

Step-by-Step Execution Trace Table

Example 2, nums = [3,2,1,0,4] (the trap case):

`i`	`nums[i]`	Check `i > reach`	`reach = max(reach, i + nums[i])`	What it means
0	3	`0 > 0`: no	`max(0, 0 + 3) = 3`	Indices 0 to 3 can be landed on
1	2	`1 > 3`: no	`max(3, 1 + 2) = 3`	No further than before
2	1	`2 > 3`: no	`max(3, 2 + 1) = 3`	No further than before
3	0	`3 > 3`: no	`max(3, 3 + 0) = 3`	Index 3 can't move on
4	4	`4 > 3`: yes	(not reached)	No jump lands on index 4: `return False`

Scroll horizontally to see all columns, or expand to full screen

Core Invariant Specification & Code Shape

Archetype Code Shape

Python

1	You never need the path, only how far any path can get: keep `reach`, the furthest index a run of jumps can land on so far.
2	Every index from `0` to `reach` can be landed on (a jump may be shorter than `nums[i]`), so extend it with `reach = max(reach, i + nums[i])`: the larger reach keeps every option.
3	Start `reach = 0`, loop `for i in range(len(nums))` once, and `return True` after the loop.
4	The trap: check `if i > reach: return False` before reading `nums[i]`, or an index nobody can land on extends `reach`.

Target: Jump Game (LeetCode 55). Before any jump only index 0 is reachable, so the reachable block is 0..0.

Boundary Model: Sorted Coordinate Sweep-Line & Monotonic Stack

Intervals sorted by start coordinate; monotonic stack preserves strictly increasing or decreasing element values.

Loop Invariant Termination

Sweep: if curr.start <= prev.end: merge; Stack: while stack and nums[i] >= stack[-1]: stack.pop().

Conceptual Narrative

🧭 Conceptual Foundation & Pattern Intuition

🪜 The Analogy: Stepping Stones Across a River

🪄 The Mathematical Harmony / Magic Trick

Code / Blueprint

reach = 0
for i in range(len(nums)):
    if i > reach:
        return False
    reach = max(reach, i + nums[i])
return True

💡 Summary

Keep the furthest reach, not a path. Check i > reach before using nums[i], extend with max, and return True once the loop has stood on every index. One pass: $O(N)$ time, $O(1)$ space.

Extending reach from an index you can't stand on: check i > reach before reach = max(reach, i + nums[i]). On [3,2,1,0,4] index 4 is past reach = 3; letting its jump count returns True instead of False.
Assigning instead of taking the max: reach = i + nums[i] lets a short jump after a long one shrink the reach, so [4,1,0,0,0] gets stuck at index 3 and returns False instead of True.
Searching paths: trying every jump length from every index is correct but costs up to O(N · max(nums[i])), about 10^9 steps at these limits. One number, reach, replaces the whole search.
Answering a different question: reach only says yes or no. The fewest jumps, as in Jump Game II (LC 45), needs a second boundary for where the current jump runs out.

Senior SWE Reasoning Architecture

4-Phase Thought Process Model

You will see how a senior engineer spots a greedy reachability question and defends the one-number answer out loud.

Pattern Recognition Signals

The 10-second spot

Formulating the Predicate & Invariants

Turning intuition into a boolean rule

Silent Failure Traps & Edge Cases

Where confident candidates still lose points

if i > reach: return False must come before reach = max(reach, i + nums[i]): an index nobody can land on must not extend reach (on [3,2,1,0,4] that would return True).
max, not reach = i + nums[i]: a short jump after a long one would shrink the reach ([4,1,0,0,0] would stop at index 3).
Don't try every jump length from every index: that is correct but O(N · max(nums[i])), up to 10^9 steps at these limits; a single reach makes it O(N).

The 60-Second Interview Pitch

Say this out loud before you type a single line

I'd solve this with Greedy. I never need the actual path, only the furthest index any sequence of jumps could land on so far; I call it reach. A jump can be shorter than its maximum, so every index up to reach can be landed on, and the reachable indices always form one block from zero to reach. I walk the array once. At each index I first check whether i is past reach: if it is, nobody can stand there, so I return false. Otherwise I extend reach to the max of reach and i plus nums of i. Taking the max never loses an option, because a longer block contains every shorter one. The trap is the order: checking first stops an unreachable index from pushing reach forward. If the loop finishes, I return true. One pass, so O(N) time and O(1) space.

So: keep reach, check i > reach first, extend with max, and say why the larger reach never loses an option.

Big-O Invariant Derivation

Complexity & Mathematical Proof

TIME COMPLEXITY

O(N)

SPACE COMPLEXITY

O(1)

The code keeps only reach and i, whatever the input size, and returns one boolean: O(1) extra space.

Formal Recurrence Relation

T(N) = O(1) + N · O(1) = O(N)

Derivation Progression

Start

O(1)

reach = 0 is one assignment.

Index loop

at most N iterations

for i in range(len(nums)) visits each index once, and stops early if return False runs.

Per-index work

O(1) per iteration

One comparison i > reach and one max(reach, i + nums[i]), whatever N is.

Total

O(N)

Constant work per index over at most N indices.

Variable Definitions

$N$

Number of indices, len(nums)

$reach$

Furthest index some run of jumps can land on so far

Memory Architecture & Bounds

🟣 Call Stack

O(1) Iterative, no recursion

🔵 Auxiliary Heap

O(1): reach, i

🟢 Output Space

O(1): one boolean

Boundary Best / Worst Cases

Best Case

$O(1)$ : nums[0] = 0 with more than one index, so the loop returns False at i = 1

Average Case

$O(N)$

Worst Case

$O(N)$ : the last index is reachable, so every index is checked

Recurrence Tree Topology

Synthesizing vector architecture diagram...

Staff+ Engineering Perspective

Senior SWE Deconstruction & Hardware Caveats

ARCHITECTURAL SIGNALS & INTERVIEW TRIGGERS

CONSTRAINTS & BOUNDS

FAANG PRODUCTION TRAPS & EDGE CASES

Core Algorithmic State Invariants

1. One-Block Reach Invariant

Before index `i` is checked, every index from `0` to `reach` can be landed on, because a jump may be shorter than `nums[i]`. The reachable indices are always one unbroken block.

2. Check, Then Extend

3. One Linear Pass

Each index is checked once with O(1) work, so the answer comes in O(N) time and O(1) space; `return True` after the loop means the last index was stood on.

Rosetta Dual-Monaco Comparison

Python 3

CANONICAL INVARIANT TEMPLATE

CONCRETE: JUMP GAME (LEETCODE 55)

T = O(N)S = O(1)

Pattern Implementation Mapping TableCanonical Invariant ⟷ Concrete Code ⟷ Engineering Rationale

Canonical Invariant	Concrete Code	Engineering Rationale
Summarize the past in one number: the furthest reachable index	`reach = 0`	Before any jump only index 0 is reachable, so the reachable block is 0..0.
Visit every index once, left to right	`for i in range(len(nums)):`	Each index is either inside the reachable block (it can be stood on) or past it.
Stop when the block ends before i: check before using index i	`if i > reach: return False`	No jump lands on index i, so no later index can be reached either. Checking first keeps an unreachable index from extending reach.
Extend with the locally best choice	`reach = max(reach, i + nums[i])`	A longer reach contains every index a shorter one does, so max never loses an option.
Every index was reachable	`return True`	The loop stood on the last index, so the answer is yes.

i

nums[i]

Check i > reach

reach = max(reach, i + nums[i])

What it means

0 > 0: no

max(0, 0 + 3) = 3

Indices 0 to 3 can be landed on

1 > 3: no

max(3, 1 + 2) = 3

No further than before

2 > 3: no

max(3, 2 + 1) = 3

No further than before

3 > 3: no

max(3, 3 + 0) = 3

Index 3 can't move on

4 > 3: yes

(not reached)

No jump lands on index 4: return False

i

nums[i]

Check i > reach

reach = max(reach, i + nums[i])

What it means

0 > 0: no

max(0, 0 + 3) = 3

Indices 0 to 3 can be landed on

1 > 3: no

max(3, 1 + 2) = 3

No further than before

2 > 3: no

max(3, 2 + 1) = 3

No further than before

3 > 3: no

max(3, 3 + 0) = 3

Index 3 can't move on

4 > 3: yes

(not reached)

No jump lands on index 4: return False

🧭 Conceptual Foundation & Pattern Intuition

🪜 The Analogy: Stepping Stones Across a River

🪄 The Mathematical Harmony / Magic Trick

Code / Blueprint

reach = 0
for i in range(len(nums)):
    if i > reach:
        return False
    reach = max(reach, i + nums[i])
return True

💡 Summary

Keep the furthest reach, not a path. Check i > reach before using nums[i], extend with max, and return True once the loop has stood on every index. One pass: $O(N)$ time, $O(1)$ space.

Canonical Invariant

Concrete Code

Engineering Rationale

Summarize the past in one number: the furthest reachable index

reach = 0

Before any jump only index 0 is reachable, so the reachable block is 0..0.

Visit every index once, left to right

for i in range(len(nums)):

Each index is either inside the reachable block (it can be stood on) or past it.

Stop when the block ends before i: check before using index i

if i > reach:
    return False

No jump lands on index i, so no later index can be reached either. Checking first keeps an unreachable index from extending reach.

Extend with the locally best choice

reach = max(reach, i + nums[i])

A longer reach contains every index a shorter one does, so max never loses an option.

Every index was reachable

return True

The loop stood on the last index, so the answer is yes.