Introduction

In this blog post, we will explore how to solve the LeetCode problem "3377" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are given two integers n and m that consist of the same number of digits. You can perform the following operations any number of times: Choose any digit from n that is not 9 and increase it by 1. Choose any digit from n that is not 0 and decrease it by 1. The integer n must not be a prime number at any point, including its original value and after each operation. The cost of a transformation is the sum of all values that n takes throughout the operations performed. Return the minimum cost to transform n into m. If it is impossible, return -1. Example 1: Input: n = 10, m = 12 Output: 85 Explanation: We perform the following operations: Increase the first digit, now n = 20. Increase the second digit, now n = 21. Increase the second digit, now n = 22. Decrease the first digit, now n = 12. Example 2: Input: n = 4, m = 8 Output: -1 Explanation: It is impossible to make n equal to m. Example 3: Input: n = 6, m = 2 Output: -1 Explanation: Since 2 is already a prime, we can't make n equal to m. Constraints: 1 <= n, m < 104 n and m consist of the same number of digits.

Explanation

Here's the breakdown of the solution:

• Graph Representation: Model the problem as a graph where each node represents a number. An edge exists between two nodes if one can be reached from the other by incrementing/decrementing a single digit, ensuring that the intermediate numbers are never prime.
• Breadth-First Search (BFS): Use BFS to find the shortest path (minimum number of operations) from the initial number 'n' to the target number 'm'. Keep track of the cumulative cost along each path.

• Prime Check Optimization: Precompute a list of primes up to a certain limit (10000) using the Sieve of Eratosthenes for efficient prime checking during graph construction and BFS.

• Runtime Complexity: O(V + E), where V is the number of possible states (numbers between 1000 and 9999 for 4-digit numbers) and E is the number of possible transitions between states. The Sieve of Eratosthenes takes O(N log log N) time, which can be considered O(N) for our constraints. Storage Complexity: O(V) to store the graph, visited nodes, and cost.

import collections

def solve():
    def is_prime(num, primes):
        return num in primes

    def get_neighbors(num, primes):
        s_num = str(num)
        neighbors = []
        for i in range(len(s_num)):
            digit = int(s_num[i])

            # Increment
            if digit < 9:
                new_digit = digit + 1
                new_num_str = s_num[:i] + str(new_digit) + s_num[i+1:]
                new_num = int(new_num_str)
                if not is_prime(new_num, primes):
                    neighbors.append(new_num)

            # Decrement
            if digit > 0:
                new_digit = digit - 1
                new_num_str = s_num[:i] + str(new_digit) + s_num[i+1:]
                new_num = int(new_num_str)
                if not is_prime(new_num, primes):
                    neighbors.append(new_num)
        return neighbors

    def bfs(start, end, primes, n_digits):
        q = collections.deque([(start, 0, start)])  # (node, cost, total_cost)
        visited = {start}

        while q:
            node, cost, total_cost = q.popleft()

            if node == end:
                return total_cost

            neighbors = get_neighbors(node, primes)
            for neighbor in neighbors:
                if neighbor not in visited:
                    visited.add(neighbor)
                    q.append((neighbor, cost + neighbor, total_cost + neighbor))
        return -1

    n = int(input())
    m = int(input())

    n_digits = len(str(n))
    limit = 10 ** n_digits
    lower_bound = 10**(n_digits-1)

    # Sieve of Eratosthenes
    primes = set()
    is_composite = [False] * limit
    for i in range(2, limit):
        if not is_composite[i]:
            if i >= lower_bound:
                primes.add(i)
            for j in range(i * i, limit, i):
                is_composite[j] = True

    if is_prime(m, primes):
        print(-1)
        return

    print(bfs(n, m, primes, n_digits))


    # Code
    ```python
    import olletions
def solve():
    def is_prime(num, primes):
        return num in primes

    def get_neighbors(num, primes):
        s_num = str(num)
        neighbors = []
        for i in range(len(s_num)):
            digit = int(s_num[i])

            # Increment
            if digit < 9:
                new_digit = digit + 1
                new_num_str = s_num[:i] + str(new_digit) + s_num[i+1:]
                new_num = int(new_num_str)
                if not is_prime(new_num, primes):
                    neighbors.append(new_num)

            # Decrement
            if digit > 0:
                new_digit = digit - 1
                new_num_str = s_num[:i] + str(new_digit) + s_num[i+1:]
                new_num = int(new_num_str)
                if not is_prime(new_num, primes):
                    neighbors.append(new_num)
        return neighbors

    def bfs(start, end, primes, n_digits):
        q = collections.deque([(start, 0, start)])  # (node, cost, total_cost)
        visited = {start}

        while q:
            node, cost, total_cost = q.popleft()

            if node == end:
                return total_cost

            neighbors = get_neighbors(node, primes)
            for neighbor in neighbors:
                if neighbor not in visited:
                    visited.add(neighbor)
                    q.append((neighbor, cost + neighbor, total_cost + neighbor))
        return -1

    n = int(input())
    m = int(input())

    n_digits = len(str(n))
    limit = 10 ** n_digits
    lower_bound = 10**(n_digits-1)

    # Sieve of Eratosthenes
    primes = set()
    is_composite = [False] * limit
    for i in range(2, limit):
        if not is_composite[i]:
            if i >= lower_bound:
                primes.add(i)
            for j in range(i * i, limit, i):
                is_composite[j] = True

    if is_prime(m, primes):
        print(-1)
        return

    print(bfs(n, m, primes, n_digits))

solve()