High Greedy Question 189 of 224

How do you find the minimum number of jumps to reach the last index?

DSA interview set · Speak this in 60–90 seconds · Faridabad & Delhi NCR

PICTURE THIS: ARRAY IN MEMORY

01234

Index starts at 0. Scan once for max — O(n).

Simple meaning

Greedy BFS-on-array: while in the current jump range, track the farthest you can reach, then increment jumps when the range ends.

1

WHY — Greedy instead of guessing?

Why interviewers care about Greedy:

This is a process

question about Greedy.

Panels listen for order,

trade-offs, and what you would actually do on a DSA project - not buzzwords.

Stay structured

Name the idea, why it exists, then one short example.

Close cleanly

End with when you use it and one common pitfall.

2

STEPS — What happens step by step?

Before you speak the answer, walk the interviewer through these steps:

  1. 1
    Greedy BFS-on-array: while in

    the current jump range, track the farthest you can reach, then increment jumps when the range ends.

  2. 2
    Time is O(n) and

    space is O(1).

  3. 3
    How it works

    DP min-jumps is O(n^2).

  4. 4
    If farthest ever stalls

    before the end, it is unreachable.

  5. 5
    Common mistake

    What juniors usually get wrong.

  6. 6
    Close

    When you pick this over the alternative.

3

EXAMPLE — See it in action

Here's a short line you can speak, broken into clear beats:

Say this line
“DP min-jumps is O(n^2).”
Break into beats
DPminjumpsisOn
Speaking order
2987408337471632900

Note: Adapt this scaffold to your own project — keep it under 60–90 seconds.

Key takeaway

Greedy BFS-on-array: while in the current jump range, track the farthest you can reach, then increment jumps when the range ends. Time is O(n) and space is O(1).

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