0088MediumArrayDynamic ProgrammingGreedy

Fewest Drone Flights

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Trains the technique from

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A courier drone sits on the first of a row of rooftop landing pads, indexed from 0 to len(reach) - 1.

You are given an integer array reach. When the drone lifts off from pad i, its remaining battery lets it settle on any pad from i + 1 up to and including i + reach[i], whichever of those pads exist. A single takeoff-and-landing counts as one flight. A pad whose value is 0 has no charging coil, so a drone that lands there can never lift off again.

Return the fewest flights that carry the drone from pad 0 to the last pad. The route planner only ever hands you a row in which the last pad is attainable, and a drone that already starts on the last pad needs no flights at all.

Examples

Example 1

Input: reach = [3, 1, 1, 4, 2]
Output: 2

One flight of three pads lands on pad 3, then a flight of a single pad finishes the trip. No single flight spans four pads from the start.

Example 2

Input: reach = [1, 4, 0, 0, 2, 1]
Output: 2

Hop to pad 1, whose charge of 4 covers the two dead pads and lands directly on pad 5.

Example 3

Input: reach = [0]
Output: 0

The only pad in the row is also the destination.

Constraints

1 <= reach.length <= 10^4
0 <= reach[i] <= 1000
The last pad is always attainable from pad 0

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def fewest_flights(reach: list[int]) -> int:

Java

public int fewestFlights(int[] reach)

The intended approach

Time O(n)Space O(1)

Sweep left to right while tracking two frontiers: the last pad the current number of flights can still cover, and the farthest pad any takeoff seen so far could land on. Reaching the first frontier means the current flight has to be spent, so bump the count and push the frontier out to the farthest value. That treats every flight as one breadth-first layer without ever storing the layers.

Pythonentry point: fewest_flights

def fewest_flights(reach):
    n = len(reach)
    if n <= 1:
        return 0
    flights = 0
    boundary = 0
    farthest = 0
    for i in range(n - 1):
        if i + reach[i] > farthest:
            farthest = i + reach[i]
        if i == boundary:
            flights += 1
            boundary = farthest
            if boundary >= n - 1:
                break
    return flights

The obvious version

Relax every landing pad from every takeoff pad, O(n * maxReach) table fill.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_fewest_flights(reach):
    n = len(reach)
    INF = float("inf")
    best = [INF] * n
    best[0] = 0
    for i in range(n):
        if best[i] == INF:
            continue
        for step in range(1, reach[i] + 1):
            j = i + step
            if j < n and best[i] + 1 < best[j]:
                best[j] = best[i] + 1
    return int(best[n - 1])

How this problem was checked

Reference solution executed against 10 test cases — 10 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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