From N! to N^2 (Dynamic Programming) - Part II

I tried this problem using brute-force but it was clear that it wouldn't work with an input of num=3000. Complexity for the brute-force was 3000 factorial.

Instead, this was a problem similar to counting coins or solving Diophantine equations: if you manage to have all the solutions from 0..num-1, you can find the solution for num. Notice that the solution for num will require an extra loop from i=1..num where you're going to compare the current solution for num dp[num] to all solutions dp[num-i], assuming that i ends with the digit k. The two loops together lead to an num^2 solution, which runs very quick given the small num. Solution is down below, Merci, ACC.

Sum of Numbers With Units Digit K - LeetCode

2310. Sum of Numbers With Units Digit K

Medium

Given two integers num and k, consider a set of positive integers with the following properties:

• The units digit of each integer is k.
• The sum of the integers is num.

Return the minimum possible size of such a set, or -1 if no such set exists.

Note:

• The set can contain multiple instances of the same integer, and the sum of an empty set is considered 0.
• The units digit of a number is the rightmost digit of the number.

 

Example 1:

Input: num = 58, k = 9
Output: 2
Explanation:
One valid set is [9,49], as the sum is 58 and each integer has a units digit of 9.
Another valid set is [19,39].
It can be shown that 2 is the minimum possible size of a valid set.

Example 2:

Input: num = 37, k = 2
Output: -1
Explanation: It is not possible to obtain a sum of 37 using only integers that have a units digit of 2.

Example 3:

Input: num = 0, k = 7
Output: 0
Explanation: The sum of an empty set is considered 0.

 

Constraints:

• 0 <= num <= 3000
• 0 <= k <= 9

Accepted

13,189

Submissions

56,229

public int MinimumNumbers(int num, int k)
{
    int[] dp = new int[num + 1];
    dp[0] = 0;
    for (int i = 1; i < dp.Length; i++) dp[i] = -1;

    for (int i = 1; i <= num; i++)
    {
        for (int j = 1; j <= i; j++)
        {
            if (j % 10 == k && dp[i - j] != -1)
            {
                dp[i] = (dp[i] == -1) ? dp[i - j] + 1 : Math.Min(dp[i], dp[i - j] + 1);
            }
        }
    }

    return dp[num];
}