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Formatted question description: https://leetcode.ca/all/2310.html

2310. Sum of Numbers With Units Digit K

  • Difficulty: Medium.
  • Related Topics: Math, Dynamic Programming, Greedy, Enumeration.
  • Similar Questions: Digit Count in Range, Count Integers With Even Digit Sum.

Problem

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

Solution (Java, C++, Python)

  • class Solution {
    public int minimumNumbers(int nums, int k) {
        // Base Case Check
        if (nums == 0) {
            return 0;
        }
        int x = nums % 10;
        for (int i = 1; i <= 10; i++) {
            // check if the unit digits are equal for any case and if n>k*i
            if ((k * i) % 10 == x && nums >= k * i) {
                return i;
            }
        }
        // in case nothing matches
        return -1;
    }
}

  • Todo

  • class Solution:
    def minimumNumbers(self, num: int, k: int) -> int:
        if num == 0:
            return 0
        for i in range(1, num + 1):
            if (t := num - k * i) >= 0 and t % 10 == 0:
                return i
        return -1

############

# 2310. Sum of Numbers With Units Digit K
# https://leetcode.com/problems/sum-of-numbers-with-units-digit-k/

class Solution:
    def minimumNumbers(self, num: int, k: int) -> int:
        if num == 0: return 0
        
        A = []
        INF = 100000
        
        for x in range(k, num + 10, 10):
            A.append(x)

        A.reverse()
        
        @cache
        def go(index, total):
            if index >= len(A):
                return INF
            
            if total > num:
                return INF
            
            if total == num:
                return 0
            
            # skip current
            res = go(index + 1, total)
            
            # take current
            if A[index] > 0:
                res = min(res, 1 + go(index, total + A[index]))
            
            return res
        
        ans = go(0, 0)
        if ans == 100000:
            return -1
        
        return ans

Explain:

nope.

Complexity:

  • Time complexity : O(n).
  • Space complexity : O(n).

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