Smallest Divisible Digit Product I

1. Define a constant LIMIT (e.g., 200) as the upper bound for pre-computation.\n2. Create a static array digitProducts of size LIMIT + 1.\n3. In a static block, iterate from i = 1 to LIMIT.\n4. For each i, calculate its digit product and store it in digitProducts[i].\n5. In the main function, iterate with i from n to LIMIT.\n6. Check if digitProducts[i] is divisible by t.\n7. If it is, return i as the result.\n8. If the loop finishes without finding an answer (unlikely), implement a fallback to check numbers greater than LIMIT.

Given the small constraints (n <= 100), the answer is expected to be a relatively small number. We can establish a fixed upper limit, for instance, 200, which should comfortably contain the answer for any valid input.\nAn array, let's call it digitProducts, is used to store the product of digits for each number from 1 up to this limit.\nThis array is populated once, possibly in a static initializer block. For each number i in the range, its digit product is calculated and stored at digitProducts[i].\nThe smallestDivisibleDigitProduct function then iterates from n up to the limit. For each number i, it retrieves the pre-computed product digitProducts[i] and checks if it's divisible by t.\nThe first number i that satisfies digitProducts[i] % t == 0 is the smallest such number greater than or equal to n, so it is returned.\nA fallback mechanism for numbers beyond the limit can be included for robustness, although it's unlikely to be triggered with the given constraints.\n

1java\nclass Solution {\n    private static final int LIMIT = 200;\n    private static final long[] digitProducts = new long[LIMIT + 1];\n\n    static {\n        digitProducts[0] = 0;\n        for (int i = 1; i <= LIMIT; i++) {\n            digitProducts[i] = calculateDigitProduct(i);\n        }\n    }\n\n    private static long calculateDigitProduct(int n) {\n        if (n == 0) return 0;\n        long product = 1;\n        int temp = n;\n        while (temp > 0) {\n            int digit = temp % 10;\n            if (digit == 0) return 0; // Optimization\n            product *= digit;\n            temp /= 10;\n        }\n        return product;\n    }\n\n    public long smallestDivisibleDigitProduct(int n, int t) {\n        for (int i = n; i <= LIMIT; i++) {\n            if (digitProducts[i] % t == 0) {\n                return i;\n            }\n        }\n        // Fallback for cases where answer > LIMIT\n        long num = LIMIT + 1;\n        while (true) {\n            if (calculateDigitProduct((int)num) % t == 0) {\n                return num;\n            }\n            num++;\n        }\n    }\n}\n

1. Initialize a variable num with the input n.\n2. Enter an infinite while loop.\n3. Inside the loop, calculate the product of the digits of num.\n a. Handle the case where num has a digit 0; the product is 0.\n b. Otherwise, iterate through the digits of num by repeatedly using modulo 10 and division by 10, multiplying them into a running product.\n4. Check if the calculated digit product is divisible by t.\n5. If product % t == 0, then num is the answer. Return num.\n6. If not, increment num by 1 and continue the loop.

The core idea is to perform a linear search for the answer. We start with the number n and check if it satisfies the condition. If not, we check n+1, then n+2, and so on, until we find a valid number.\nThe algorithm is implemented with a while loop that starts with num = n.\nInside the loop, a helper function getDigitProduct(num) is called. This function computes the product of the digits of num.\nThe getDigitProduct function is implemented by repeatedly taking the number modulo 10 to get the last digit and then dividing by 10. An important optimization is to recognize that if any digit is 0, the product is 0.\nThe calculated product is then checked for divisibility by t using the modulo operator (product % t == 0).\nIf the condition is met, the loop terminates, and the current num is returned. Otherwise, num is incremented, and the process repeats.\nGiven the small constraints on n and t, the answer is always found after a very small number of iterations, making this approach extremely fast in practice.\n

1java\nclass Solution {\n    private long getDigitProduct(long n) {\n        if (n == 0) {\n            return 0;\n        }\n        long product = 1;\n        long temp = n;\n        while (temp > 0) {\n            long digit = temp % 10;\n            if (digit == 0) {\n                // If any digit is 0, the product is 0.\n                return 0;\n            }\n            product *= digit;\n            temp /= 10;\n        }\n        return product;\n    }\n\n    public long smallestDivisibleDigitProduct(int n, int t) {\n        long num = n;\n        while (true) {\n            if (getDigitProduct(num) % t == 0) {\n                return num;\n            }\n            num++;\n        }\n    }\n}\n