Understanding the Problem
We are given a start
number and an end
number. Also, we are provided with an array arr[]
of multipliers. In each move, we can multiply the current number with any number from arr[]
, and then take the result modulo 100000
.
The goal is to find the minimum number of such multiplications required to reach the end
number starting from start
. If it's not possible to reach end
, we should return -1
.
This is a classic shortest path problem in disguise, where each number from 0
to 99999
can be thought of as a node in a graph, and we can move from one node to another by performing a multiplication and taking modulo 100000
.
Step-by-Step Solution with Example
Step 1: Visualize as a Graph
We treat every number from 0
to 99999
as a node. From a node u
, we can go to another node v
if v = (u * x) % 100000
where x
is any element from arr[]
.
This gives us a huge graph with up to 100000
nodes, but we only need to explore reachable nodes starting from start
.
Step 2: Use BFS to Find Minimum Steps
Since all transitions are uniform (1 step each), we can apply Breadth-First Search (BFS), which always finds the shortest path in an unweighted graph.
Step 3: Track Visited Nodes
We keep an array dist[100000]
to track the number of steps it takes to reach each node. Initialize all entries to -1
except dist[start]
which is 0
.
Step 4: Process Queue and Apply Multiplications
While processing the BFS queue, for each number curr
, we try all multipliers x
from arr
. We calculate next = (curr * x) % 100000
. If dist[next]
is -1
, it means we haven't reached it yet, so we set dist[next] = dist[curr] + 1
and add next
to the queue.
Step 5: Stop When Target is Reached
If at any point curr
becomes equal to end
, we return dist[curr]
as the answer. If the queue becomes empty and we haven’t reached end
, we return -1
.
Example:
Suppose start = 3
, end = 30
, and arr = [2, 5]
.
- From
3
, multiply by 2
→ 6
- From
3
, multiply by 5
→ 15
- From
15
, multiply by 2
→ 30
We reached 30
in 2 steps: 3 → 15 → 30
. So, the answer is 2
.
Edge Cases
- Start is equal to End: If
start == end
, we return 0
immediately because no steps are needed.
- Multiplier array has 0: Since multiplying by
0
always leads to 0
, it can trap the process. It should be handled carefully.
- Unreachable Target: If the
end
number is not reachable using the given multipliers and modulo arithmetic, BFS will finish without reaching end
, and we return -1
.
- Duplicates in arr[]: No problem — duplicates don’t affect the correctness but may cause redundant attempts, so we can use a set if needed.
Finally
This problem beautifully shows how arithmetic operations can be transformed into a graph traversal problem. By understanding that each multiplication step creates a connection between nodes, and using BFS to ensure minimum steps, we solve it efficiently even with a large search space. Always begin by modeling the problem intuitively and thinking about how to simulate the allowed operations as graph edges.
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