Minimum multiplications to reach end

Hard

BFS over the 100000 residues

Problem

From start, repeatedly multiply by array values mod 100000 to reach end; return the fewest steps.

In simple words

BFS over numbers mod 100000, where each multiplier is an edge; the fewest multiplications is shortest.

The idea

Each reachable value modulo 100000 is a vertex, and multiplying by an array element is an edge of cost one. BFS therefore gives the fewest multiplications, over a state space small enough to enumerate.

The trick

The state space is bounded at 100000 by the modulus.
All edges cost one, so BFS suffices.
Return -1 when the target is unreachable.

←→stepspaceplay

minute0

fresh6

1/6

Optimal

timeO(rows·cols)spaceO(rows·cols)

All sources start at minute 0.

1// enqueue all rotten cells; BFS by layers,2// converting fresh neighbors, counting minutes.3while (q.length && fresh > 0) {4  minutes++;5  for (let n = q.length; n > 0; n--) spread(q.shift());6}7return fresh === 0 ? minutes : -1;

Input

grid: 3 × 3

Memory

minute: 0
fresh: 6

Output

minute: 0
fresh: 6
answer: —

Check yourself

3 quick questions about this walkthrough. A wrong answer costs nothing.

Examples

Example 1

Input:: start=3, end=30, arr=[2,5,7]
Output:: 2
Explanation:: 3×2=6, 6×5=30 → 2 steps.

Example 2

Input:: start=7, end=66175, arr=[3,4,65]
Output:: 4
Explanation:: Reached in 4 multiplications.

Example 3

Input:: start=5, end=5, arr=[2,3]
Output:: 0
Explanation:: Already there → 0.

Practice this problem:GeeksforGeeks(opens in a new tab)

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