Jump Game II Explained: Min Jumps on [2, 3, 1, 1, 4]

Imagine you are standing at the start of an array, tasked with reaching the final index using the fewest jumps possible. Consider our first running example, the array nums = [2, 3, 1, 1, 4], where a value of 2 at index 0 means you can leap to index 1 or index 2. If you simply pick the largest single step available on every turn, you might bypass a launcher that unlocks a much wider horizon later. Conversely, look at our second array, nums = [3, 2, 1, 0, 4], where a zero blocks all forward momentum, leaving you stranded before the finish line. Without a systematic way to measure how far your current pool of choices can carry you, you end up blindly guessing or writing sluggish $O(n^2)$ nested loops.

To solve Jump game II (min jumps) explained, we must stop thinking about individual jump paths and start tracking the outer boundaries of what is reachable level by level. Every index within your current reach represents a landing pad for your current jump count. The moment you cross the boundary of that current reach, you are forced to spend another jump, extending your frontier to the maximum distance scouted so far.

To solve our locked example [2, 3, 1, 1, 4] efficiently, we need a model that avoids exploring every single path. Instead of picking the single best next index blindly, we think in terms of reachability intervals or "waves" of expansion. Imagine standing at index 0 with a max jump of 2. You can land anywhere in the window from index 1 to index 2. That entire window forms your current "jump boundary."

As you scan through indices 1 and 2, you do not commit to a final destination immediately. Instead, you look ahead at everyone's potential. You calculate the maximum reachable index (far) from any position within this current window. Only when your loop iterator hits the exact edge of your current jump boundary (end) do you formally spend a jump, officially expanding your boundary to whatever the farthest reach was during that wave.

Let us map this mental model to our two contrasting arrays:

- For Jump II ([2, 3, 1, 1, 4]), our intervals expand progressively until we touch or exceed the final index 4.
- For Jump I ([3, 2, 1, 0, 4]), our far reach stalls out at index 3 (where nums[3] = 0), creating an impassable barrier where the current index exceeds our maximum possible reach.

# Conceptual boundary tracking during the scan
far = 0
end = 0
jumps = 0

for i in range(len(nums) - 1):
    far = max(far, i + nums[i])
    if i == end:
        jumps += 1
        end = far

def trace_intervals(nums):
    far = 0
    end = 0
    jumps = 0
    # Walk through the loop and track when i hits end
    pass

Let us apply the expanding reach model to our primary example: nums = [2, 3, 1, 1, 4]. We want to find the minimum number of jumps required to reach the final index 4. Recall our tracking variables: jumps = 0, currentEnd = 0, and farthest = 0. We iterate up to index n - 2 (index 3), because we only need to trigger a jump if our current window boundary forces us out.

Given

nums = [2, 3, 1, 1, 4] with length $n=5$.

- Initialization: jumps = 0, currentEnd = 0, farthest = 0.
- Iteration i = 0 (nums[0] = 2):
- Update farthest = max(0, 0 + 2) = 2.
- Since i == currentEnd (0 == 0), we must make a jump! Increment jumps to 1 and set currentEnd = farthest (2).
- State: jumps = 1, currentEnd = 2, farthest = 2.

- Iteration i = 1 (nums[1] = 3):
- Update farthest = max(2, 1 + 3) = 4.
- Is i == currentEnd (1 == 2)? No. Do not increment jumps yet.
- State: jumps = 1, currentEnd = 2, farthest = 4.

- Iteration i = 2 (nums[2] = 1):
- Update farthest = max(4, 2 + 1) = 4.
- Since i == currentEnd (2 == 2), we have exhausted our first-jump window. Increment jumps to 2 and set currentEnd = farthest (4).
- State: jumps = 2, currentEnd = 4, farthest = 4.

- Iteration i = 3 (nums[3] = 1):
- Loop terminates because we stop at $n-2$.

Result

The loop finishes, returning jumps = 2. The path corresponds to index 0 -> 1 -> 4.

Try this: Trace the variables for nums = [1, 2, 1, 1, 1] through each index until i = 3. What are the final values of jumps, currentEnd, and farthest?

Now that you have traced the greedy window on [2, 3, 1, 1, 4] through its exact index transitions, let us test your mental model on a modified input. Consider the array nums = [1, 3, 2, 0, 4].

In this variation, the first element is 1, meaning your initial jump choice is forced to land on index 1 (3). From there, the window expands differently than our baseline working example. Try to manually compute the minimum jumps required to reach the last index by keeping strict track of jumps, currentEnd, and farthest at each step before looking at the solution.

Question

For the array nums = [1, 3, 2, 0, 4], what is the exact value of jumps, currentEnd, and farthest immediately after the loop processes index 1 (nums[1] = 3), and how many total minimum jumps are required to reach the final index 4?

nums = [1, 3, 2, 0, 4]
# Trace: jumps=?, currentEnd=?, farthest=?

Now that you have traced [2, 3, 1, 1, 4] and conquered modified single-array traversals, it is time to witness how this exact range-expansion philosophy transfers to neighboring algorithmic challenges. The core mental model you built—distinguishing between the current boundary of exploration (end) and the farthest possible horizon (far)—is not locked to linear arrays. It reappears whenever you need to minimize steps across expanding reachable frontiers.

Consider a variation where you are given a matrix of jump capacities or a string manipulation problem where each character dictates how many indices you may skip forward. When faced with these expanded domains, do not panic and restart your design from scratch. Instead, isolate the exact same three variables: your current frontier boundary, your maximum reachable index within that frontier, and your step counter. The transition from one window boundary to the next remains invariant, whether your indices are laid out in a single row or mapped across a grid.

By replacing blind backtracking with structured window expansion, you turn exponential combinatorial explosions into predictable linear sweeps. Keep this window-tracking blueprint ready whenever an optimization problem asks for the minimum steps to cross a threshold.