Solution Explanation

This problem asks to find the smallest integer greater than k that is a multiple of k and consists only of digits digit1 and digit2. The solution uses a breadth-first search (BFS) approach to explore possible numbers.

Algorithm:

1. Handle Trivial Cases: If both digit1 and digit2 are 0, no such integer exists, so return -1. If digit1 is greater than digit2, swap them to maintain consistency. This optimization avoids redundant calculations.

2. Breadth-First Search (BFS): A queue (q) is initialized with 0. The BFS explores all possible numbers built from digit1 and digit2. In each iteration:

   • Dequeue a number (x) from the queue.
   • If x exceeds the maximum 32-bit integer, return -1.
   • If x is greater than k and is a multiple of k ( x > k && x % k == 0), it's the solution, so return x.
   • Enqueue two new numbers: x * 10 + digit1 and x * 10 + digit2 (unless digit1 and digit2 are the same). These represent extending the current number with either digit.

Time Complexity: The time complexity is difficult to express precisely because the number of integers explored depends on the values of k, digit1, and digit2. In the worst case, it could be exponential, particularly if the smallest such integer is large and many numbers need to be explored before finding it. However, in many practical scenarios, it would complete quickly. A tight bound is difficult to provide.

Space Complexity: The space complexity is O(B), where B is the maximum length of the numbers generated during the BFS. In the worst-case scenario, B would be proportional to the logarithm of the maximum 32-bit integer, making space complexity approximately O(log(231-1)). This is because the queue stores integers, and the number of digits in the largest integer considered is logarithmic with respect to the integer's value.

Code Explanation (Python Example):

class Solution:
    def findInteger(self, k: int, digit1: int, digit2: int) -> int:
        if digit1 == 0 and digit2 == 0:
            return -1
        if digit1 > digit2:
            return self.findInteger(k, digit2, digit1)
        q = deque([0]) # Initialize queue with 0
        while 1: #Infinite loop, will break when solution is found or limit is exceeded.
            x = q.popleft()
            if x > 2**31 - 1:
                return -1
            if x > k and x % k == 0:
                return x
            q.append(x * 10 + digit1)
            if digit1 != digit2:
                q.append(x * 10 + digit2)
 

The other language examples follow the same logic, adapting the data structures (queues) and syntax to their respective languages. The core BFS algorithm remains consistent across all implementations.