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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)

• Java
• Python
• C++
• Go
• TypeScript

• 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;
      }
  }
  
  ############
  
  class Solution {
      public int minimumNumbers(int num, int k) {
          if (num == 0) {
              return 0;
          }
          for (int i = 1; i <= num; ++i) {
              int t = num - k * i;
              if (t >= 0 && t % 10 == 0) {
                  return i;
              }
          }
          return -1;
      }
  }
  

• 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
              
                  
  
  

• class Solution {
  public:
      int minimumNumbers(int num, int k) {
          if (num == 0) return 0;
          for (int i = 1; i <= num; ++i) {
              int t = num - k * i;
              if (t >= 0 && t % 10 == 0) return i;
          }
          return -1;
      }
  };
  

• func minimumNumbers(num int, k int) int {
  	if num == 0 {
  		return 0
  	}
  	for i := 1; i <= num; i++ {
  		t := num - k*i
  		if t >= 0 && t%10 == 0 {
  			return i
  		}
  	}
  	return -1
  }
  

• function minimumNumbers(num: number, k: number): number {
      if (!num) return 0;
      let digit = num % 10;
      for (let i = 1; i < 11; i++) {
          let target = i * k;
          if (target <= num && target % 10 == digit) return i;
      }
      return -1;
  }
  
  

Explain:

nope.

Complexity:

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

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