Minimum Multiplications to Reach End

This chapter explores how to find the minimum number of multiplications required to transform a starting number (start) into a target number (end) using an array of multipliers under a modulo constraint. By framing state transformations as unweighted graph transitions, we can efficiently compute the shortest path using a Breadth-First Search (BFS) / Dijkstra's paradigm.

1. Why Do We Need This? (Problem Statement & Status Quo)

Problem Overview

• Inputs:
  • start: An initial integer value.
  • end: The target integer value we want to reach.
  • arr: An array of $N$ numbers available for multiplication at each step.
• Rules:
  • At any step, you can pick any number from arr and multiply it with your current value.
  • After multiplication, you must immediately apply a modulo operation: (current_value * multiplier) % 10^5.
  • Your goal is to find the minimum number of multiplication steps needed to reach end from start.
  • If it is impossible to reach end, return -1.

Status Quo & The Catch

A naive recursive or backtracking approach would explode exponentially because we can reuse array elements infinitely, and numbers can loop through various combinations. We need a systematic way to guarantee that we find the shortest sequence of multiplications without exploring redundant paths.

2. Core Mental Model & Analogy

Real-Life Analogy: The Combination Lock / Teleporter Network

Think of every number from 0 to 99,999 (enforced by the modulo 10^5) as a room in a massive labyrinth. Your start number is Room 3, and your end is Room 30. Each number in the array represents a specialized teleporter key. When you use a key (multiply), it instantly teleports you to a new room number (Current Room * Key) % 10^5.

Because you want the minimum steps, you are looking for the shortest sequence of teleporter activations to walk from the start room to the destination room.

3. Naive vs. Better Approach

Why Standard Permutations Fail

• Naive Exhaustive Search: Trying all combinations of multipliers leads to infinite loops or deep redundant calculations. Because multipliers can be reused infinitely, recursion without state-visitation tracking will result in Time Limit Exceeded (TLE).
• The Better Approach (BFS / Shortest Path): Since every multiplication step carries an equal weight (cost of 1 step), finding the minimum steps is equivalent to finding the shortest path in an unweighted graph.

4. Step-by-Step Evolution of the Algorithm

Step 4.1: Identifying the Graph Nodes

• Because of the modulo operation (% 10^5), any number generated will always stay within the range 0 to 99,999 (10^5 - 1).
• Therefore, the total possible unique states (nodes in our graph) are capped at $10^5$ ($0$ to $99999$).

Step 4.2: Distance Array Configuration

• We maintain a dist array of size $10^5$, initialized to a very large number (infinity, e.g., 1e9).
• dist[node] stores the minimum number of multiplications required to reach that specific node from start.
• dist[start] = 0 (0 steps to reach the start itself).

Step 4.3: Why a Standard Queue Suffices Over a Priority Queue

• In Dijkstra's algorithm, edge weights can vary. However, in this problem, every single multiplication step increases the step count by exactly 1.
• Because step costs increase uniformly (steps + 1), a standard FIFO (First-In, First-Out) Queue naturally processes nodes in increasing order of steps. A Priority Queue is unnecessary overhead.

5. Walkthrough of the Example

Let's trace start = 3, end = 30, and arr = [2, 5, 7].

Queue Initial State: [ (3, steps: 0) ]
Distance Array: dist[3] = 0, all other nodes = Infinity

1. Pop (3, 0) from Queue:

   • Multiply 3 with 2 $\rightarrow$ 6. Steps = $0 + 1 = 1$. Update dist[6] = 1. Push (6, 1).
   • Multiply 3 with 5 $\rightarrow$ 15. Steps = $0 + 1 = 1$. Update dist[15] = 1. Push (15, 1).
   • Multiply 3 with 7 $\rightarrow$ 21. Steps = $0 + 1 = 1$. Update dist[21] = 1. Push (21, 1).

2. Pop (6, 1) from Queue:

   • Multiply 6 with 2 $\rightarrow$ 12 (steps 2). Push (12, 2).
   • Multiply 6 with 5 $\rightarrow$ 30 (steps 2). Since 3 $\times$ 10 or similar isn't here, 30 is reached with steps 2. Push (30, 2).
   • ...and so on.

Eventually, when the algorithm generates 30 via a shorter path or evaluates remaining states, it hits the end node (30) and returns the accumulated minimum steps.

6. Complete Code Implementation

C++ Solution

#include <bits/stdc++.h>
using class_name {
public:
    int minimumMultiplications(vector<int>& arr, int start, int end) {
        // Edge case: if start is already equal to end
        if (start == end) return 0;
        
        // Queue stores pairs of (current_node, steps)
        queue<pair<int, int>> q;
        q.push({start, 0});
        
        int mod = 100000;
        
        // Distance array to store minimum steps to reach each node (0 to 99999)
        vector<int> dist(100000, 1e9);
        dist[start] = 0;
        
        while (!q.empty()) {
            auto it = q.front();
            q.pop();
            
            int node = it.first;
            int steps = it.second;
            
            // Try multiplying the current node with each element in arr
            for (int itm : arr) {
                int num = (1LL * node * itm) % mod;
                
                // If we found a shorter path to 'num'
                if (steps + 1 < dist[num]) {
                    dist[num] = steps + 1;
                    
                    // If we reached the target end node, return the steps + 1
                    if (num == end) {
                        return steps + 1;
                    }
                    
                    q.push({num, steps + 1});
                }
            }
        }
        
        // If the queue is exhausted and end was never reached
        return -1;
    }
};

7. Complexity Analysis

8. Summary and Key Takeaways

• Problem Nature: Finding minimum steps with uniform edge weights ($+1$ per operation) maps directly to Breadth-First Search (BFS).
• State Space Bound: Because of the modulo operation (% 10^5), the maximum number of unique states is strictly bounded between 0 and 99,999.
• Queue Efficiency: Since step costs increase monotonically ($+1$ at each level), a standard FIFO queue is sufficient; a priority queue is unnecessary.
• Early Exit Optimization: Checking whether num == end immediately upon calculating a new valid state allows us to terminate early without processing the entire queue.
• Modulo Safety: Always cast multiplications to long long (1LL * node * itm) before applying modulo to prevent potential integer overflow issues in compiled languages.