Problem Description

You have two integers n and m that have the same number of digits. Your goal is to transform n into m through a series of operations while maintaining certain constraints.

Allowed Operations:

• Pick any digit in n that is not 9 and increase it by 1
• Pick any digit in n that is not 0 and decrease it by 1

Key Constraint:

• At no point during the transformation (including the initial value and after each operation) can n be a prime number

Cost Calculation: The cost of the entire transformation is the sum of all values that n takes during the process. This includes:

• The initial value of n
• Every intermediate value after each operation
• The final value m

Objective: Find the minimum cost to transform n into m. If it's impossible to do so without encountering a prime number, return -1.

Example Walkthrough: If we need to transform 12 to 15:

• 12 (starting value, not prime)
• 13 (increase second digit by 1, but 13 is prime - not allowed!)
• We'd need to find an alternative path that avoids prime numbers

The solution uses:

1. Sieve of Eratosthenes to precompute all prime numbers up to 99999
2. Dijkstra's algorithm (priority queue) to find the minimum cost path from n to m
3. Each state in the graph represents a number, and edges represent valid digit modifications that don't result in prime numbers
4. The priority queue ensures we explore paths with lower cumulative costs first

Intuition

The problem asks us to find the minimum cost path from n to m, where each step has a cost (the value of the number at that step). This immediately suggests a shortest path problem in a graph.

Think of each possible number as a node in a graph. Two nodes are connected if we can transform one into the other by changing a single digit (increase by 1 or decrease by 1). However, we can't visit any node that represents a prime number - these nodes are essentially "blocked" or removed from our graph.

Since we want to minimize the total sum of all numbers we visit (not just the number of steps), we need a weighted shortest path algorithm. Each edge has a weight equal to the destination node's value. This is why Dijkstra's algorithm is perfect here - it finds the path with minimum total weight.

The key insights are:

1. Graph representation: Each valid (non-prime) number is a node, and valid single-digit changes create edges between nodes
2. Edge weights: The weight of moving to a number is that number itself (since we add it to our cost)
3. Precomputation: Since we'll be checking many numbers for primality, it's efficient to precompute all primes up to 99999 using the Sieve of Eratosthenes
4. State exploration: From any current number, we try modifying each digit (up or down) to generate neighboring states, but only add them to our queue if they're not prime

The priority queue in Dijkstra's ensures we always process the path with the lowest cumulative cost first, guaranteeing we find the optimal solution when we reach m. If we exhaust all possibilities without reaching m, it means there's no valid path avoiding prime numbers, so we return -1.

Learn more about Graph, Math, Shortest Path and Heap (Priority Queue) patterns.

Solution Approach

The implementation consists of three main components: prime number precomputation, graph traversal using Dijkstra's algorithm, and neighbor generation through digit manipulation.

1. Prime Number Precomputation (run_sieve)

We use the Sieve of Eratosthenes to identify all prime numbers up to 99999:

self.sieve = [True] * 100000
self.sieve[0], self.sieve[1] = False, False
for i in range(2, 100000):
    if self.sieve[i]:
        for j in range(2 * i, 100000, i):
            self.sieve[j] = False

This creates a boolean array where sieve[i] = True means i is prime. We mark 0 and 1 as non-prime, then for each prime i, mark all its multiples as non-prime.

2. Main Search Algorithm (solve)

We implement Dijkstra's algorithm using a min-heap priority queue:

pq = []
heapq.heappush(pq, (n, n))  # (cumulative_cost, current_number)
visited = set()

The queue stores tuples of (cumulative_cost, current_number). The heap ensures we always process the path with minimum cost first.

3. Neighbor Generation

For each number, we generate neighbors by modifying each digit:

s = list(str(cur))  # Convert number to list of digit characters
for i in range(len(s)):
    c = s[i]  # Store original digit for restoration
  
    # Try increasing digit by 1 (if not 9)
    if s[i] < '9':
        s[i] = chr(ord(s[i]) + 1)
        next_ = int(''.join(s))
        if not self.sieve[next_] and next_ not in visited:
            heapq.heappush(pq, (sum_ + next_, next_))
        s[i] = c  # Restore original digit
  
    # Try decreasing digit by 1 (if not 0, and won't create leading zero)
    if s[i] > '0' and not (i == 0 and s[i] == '1'):
        s[i] = chr(ord(s[i]) - 1)
        next_ = int(''.join(s))
        if not self.sieve[next_] and next_ not in visited:
            heapq.heappush(pq, (sum_ + next_, next_))
        s[i] = c  # Restore original digit

Key checks during neighbor generation:

• Don't increase a digit if it's already 9
• Don't decrease a digit if it's 0
• Don't decrease the first digit if it's 1 (would create a number with fewer digits)
• Only add non-prime, unvisited numbers to the queue

4. Main Entry Point (minOperations)

The function first checks if either n or m is prime - if so, the transformation is impossible:

if self.sieve[n] or self.sieve[m]:
    return -1
return self.solve(n, m)

The algorithm continues until either:

• We reach m (return the cumulative cost)
• The priority queue is empty (no path exists, return -1)

The time complexity is O(V log V) where V is the number of reachable non-prime numbers, and space complexity is O(V) for the visited set and priority queue.

Example Walkthrough

Let's transform 10 to 14 and find the minimum cost path.

Step 1: Check Initial Constraints

• Is 10 prime? No (10 = 2 × 5)
• Is 14 prime? No (14 = 2 × 7)
• Both are non-prime, so transformation is possible

Step 2: Initialize Dijkstra's Algorithm

• Priority Queue: [(10, 10)] (cost, current_number)
• Visited: {}
• Initial cost includes starting value: 10

Step 3: Process States

Iteration 1: Pop (10, 10)

• Mark 10 as visited
• Generate neighbors by modifying digits:
  • First digit (1): Can't decrease (would make 00), increase to 2 → 20 (not prime)
  • Second digit (0): Can't decrease (already 0), increase to 1 → 11 (prime - skip!)
• Add to queue: [(30, 20)] (cost = 10 + 20 = 30)

Iteration 2: Pop (30, 20)

• Mark 20 as visited
• Generate neighbors:
  • First digit (2): Decrease to 1 → 10 (already visited - skip!) Increase to 3 → 30 (not prime)
  • Second digit (0): Increase to 1 → 21 (not prime)
• Add to queue: [(60, 30), (51, 21)] → Priority queue reorders to [(51, 21), (60, 30)]

Iteration 3: Pop (51, 21)

• Mark 21 as visited
• Generate neighbors:
  • First digit (2): Decrease to 1 → 11 (prime - skip!) Increase to 3 → 31 (prime - skip!)
  • Second digit (1): Decrease to 0 → 20 (already visited - skip!) Increase to 2 → 22 (not prime)
• Add to queue: [(60, 30), (73, 22)] → Reorders to [(60, 30), (73, 22)]

Iteration 4: Pop (60, 30)

• Mark 30 as visited
• Generate neighbors:
  • First digit (3): Decrease to 2 → 20 (already visited - skip!) Increase to 4 → 40 (not prime)
  • Second digit (0): Increase to 1 → 31 (prime - skip!)
• Add to queue: [(73, 22), (100, 40)]

Iteration 5: Pop (73, 22)

• Mark 22 as visited
• Generate neighbors:
  • First digit (2): Decrease to 1 → 12 (not prime) Increase to 3 → 32 (not prime)
  • Second digit (2): Decrease to 1 → 21 (already visited - skip!) Increase to 3 → 23 (prime - skip!)
• Add to queue: [(85, 12), (100, 40), (105, 32)]

Iteration 6: Pop (85, 12)

• Mark 12 as visited
• Generate neighbors:
  • First digit (1): Increase to 2 → 22 (already visited - skip!)
  • Second digit (2): Decrease to 1 → 11 (prime - skip!) Increase to 3 → 13 (prime - skip!) Increase to 4 → 14 (TARGET FOUND!)
• Cost to reach 14 = 85 + 14 = 99

Final Answer: The minimum cost is 99

The path taken was: 10 → 20 → 21 → 22 → 12 → 14

• Cost = 10 + 20 + 21 + 22 + 12 + 14 = 99

This example demonstrates how:

1. We avoid prime numbers (11, 13, 23, 31)
2. We don't revisit already processed numbers
3. The priority queue ensures we explore lower-cost paths first
4. We accumulate the cost by adding each number we visit

Solution Implementation

Time and Space Complexity

Time Complexity: O(V * D * log(V)) where V is the number of valid non-prime numbers in the range and D is the number of digits.

• The sieve of Eratosthenes takes O(N * log(log(N))) where N = 100000
• In the worst case, we visit all valid non-prime numbers between 1 and 100000
• For each number visited, we perform:
  • String conversion: O(D) where D is the number of digits (max 5 for numbers < 100000)
  • For each digit position, we try at most 2 operations (increment and decrement): O(D)
  • Each heap push operation: O(log(V)) where V is the size of the heap
• The heap can contain at most all valid non-prime numbers
• Overall: O(V * D * log(V)) for the main algorithm plus O(N * log(log(N))) for the sieve

Space Complexity: O(N + V)

• Sieve array: O(100000) = O(N)
• Heap: Can contain at most O(V) elements where V is the number of valid non-prime numbers
• Visited set: Can contain at most O(V) elements
• String manipulation uses O(D) temporary space where D is the number of digits (max 5)
• Overall: O(N + V) where N = 100000

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrect Handling of Leading Zeros

One of the most common mistakes is allowing the first digit to be decremented to 0, which would create an invalid number or a number with fewer digits.

Pitfall Example:

# WRONG: This allows 1234 to become 0234, which is interpreted as 234
if digits[i] > '0':
    digits[i] = chr(ord(digits[i]) - 1)
    next_number = int(''.join(digits))  # 1234 -> "0234" -> 234 (wrong!)

Solution: Always check if we're modifying the first digit and it's currently '1':

# CORRECT: Prevent first digit from becoming 0
if digits[i] > '0' and not (i == 0 and digits[i] == '1'):
    digits[i] = chr(ord(digits[i]) - 1)
    next_number = int(''.join(digits))

2. Not Restoring the Original Digit After Each Modification

When trying different digit modifications, forgetting to restore the original digit leads to cumulative changes that produce incorrect neighbors.

Pitfall Example:

# WRONG: Digit not restored, affecting subsequent iterations
for i in range(len(digits)):
    if digits[i] < '9':
        digits[i] = chr(ord(digits[i]) + 1)
        # Process next_number...
        # Missing: digits[i] = original_digit
  
    if digits[i] > '0':  # This now operates on the modified digit!
        digits[i] = chr(ord(digits[i]) - 1)

Solution: Store and restore the original digit after each operation:

# CORRECT: Save and restore original digit
for i in range(len(digits)):
    original_digit = digits[i]  # Save original
  
    if digits[i] < '9':
        digits[i] = chr(ord(digits[i]) + 1)
        # Process next_number...
        digits[i] = original_digit  # Restore
  
    if digits[i] > '0' and not (i == 0 and digits[i] == '1'):
        digits[i] = chr(ord(digits[i]) - 1)
        # Process next_number...
        digits[i] = original_digit  # Restore

3. Processing Already Visited Nodes Multiple Times

Without proper visited state checking, the same number might be processed multiple times with different cumulative costs, leading to incorrect results or infinite loops.

Pitfall Example:

# WRONG: Adding to visited set after processing neighbors
while priority_queue:
    cumulative_sum, current_number = heapq.heappop(priority_queue)
  
    if current_number == m:
        return cumulative_sum
  
    # Generate and add neighbors...
  
    visited.add(current_number)  # Too late! Might process same number again

Solution: Check and mark visited status immediately after popping from the queue:

# CORRECT: Check visited status right after popping
while priority_queue:
    cumulative_sum, current_number = heapq.heappop(priority_queue)
  
    if current_number in visited:
        continue  # Skip if already processed
    visited.add(current_number)  # Mark as visited immediately
  
    if current_number == m:
        return cumulative_sum

4. Incorrect Priority Queue Tuple Order

Using the wrong order in the priority queue tuple causes the algorithm to not prioritize minimum cost paths.

Pitfall Example:

# WRONG: Current number as first element makes heap sort by number value
heapq.heappush(priority_queue, (current_number, cumulative_sum))

Solution: Always put the cumulative sum first for proper minimum cost prioritization:

# CORRECT: Cumulative sum first for minimum cost priority
heapq.heappush(priority_queue, (cumulative_sum, current_number))

5. Off-by-One Errors in Sieve Implementation

Incorrectly implementing the sieve bounds or loop ranges can mark wrong numbers as prime/non-prime.

Pitfall Example:

# WRONG: Starting from i instead of 2*i marks prime itself as non-prime
for i in range(2, 100000):
    if self.sieve[i]:
        for j in range(i, 100000, i):  # Should start from 2*i
            self.sieve[j] = False

Solution: Start marking multiples from 2*i, not i itself:

# CORRECT: Start from 2*i to preserve the prime number itself
for i in range(2, 100000):
    if self.sieve[i]:
        for j in range(2 * i, 100000, i):
            self.sieve[j] = False