Concatenated Divisibility

Concatenated Divisibility

1. What is this problem about?

The Concatenated Divisibility coding problem is a complex mathematical and algorithmic challenge. You are given an array of integers and a target value k. Your goal is to find the number of pairs $(i, j)$ such that the concatenation of array[i] and array[j] is divisible by k. For example, if the numbers are 12 and 345, their concatenation is 12345. The brute-force approach involves checking every possible pair, but for large arrays, this becomes computationally expensive.

2. Why is this asked in interviews?

Meta often uses the Concatenated Divisibility interview question to test a candidate's mastery of Number Theory and Modular Arithmetic. It requires transforming a string-like operation (concatenation) into a mathematical formula: Concat(A, B) = A * 10^(length of B) + B. This tests if you can move beyond simple simulation to find an optimized solution using Frequency Maps and the properties of remainders.

3. Algorithmic pattern used

This problem utilizes Modular Arithmetic and Pre-computation with Hash Maps.

1. The condition (A * 10^L + B) % k == 0 can be rewritten as (A * 10^L) % k == -B % k.
2. For each possible length L (from 1 to 10), pre-calculate the remainder of (A * 10^L) % k for all numbers in the array.
3. Store these remainders in a frequency map for each L.
4. Then, for each number B in the array, find its length L_B and its remainder r = B % k. Look up how many numbers A satisfy the condition using the map for L_B.

4. Example explanation

Suppose array = [12, 3], k = 3.

• Concatenate 12 and 3: 123. 123 % 3 = 0. (Match!)
• Concatenate 3 and 12: 312. 312 % 3 = 0. (Match!) The formula approach:
• 12 * 10^1 + 3 = 120 + 3 = 123.
• 3 * 10^2 + 12 = 300 + 12 = 312. By checking remainders against a map, we can count these without actually forming the large concatenated numbers.

5. Common mistakes candidates make

• String Conversion: Converting integers to strings to concatenate them is extremely slow and uses significant memory.
• Modulo Pitfalls: Forgetting that -B % k should be handled as (k - (B % k)) % k to ensure a positive remainder in languages like Java or C++.
• Overflow: Not using long or 64-bit integers for the intermediate product A * 10^L, which can easily exceed the 32-bit integer limit.

6. Interview preparation tip

Master the distributive property of the modulo operator: (A + B) % k = ((A % k) + (B % k)) % k. This is the foundation for optimizing many "Hard" level math problems in coding interviews.