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2310. Sum of Numbers With Units Digit K

Description

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

Solutions

Solution 1: Math + Enumeration

Each number that meets the splitting condition can be represented as $10x_i + k$. If there are $n$ such numbers, then $\textit{num} - n \times k$ must be a multiple of $10$.

We enumerate $n$ from small to large, and find the first $n$ that satisfies $\textit{num} - n \times k$ being a multiple of $10$. Since $n$ cannot exceed $\textit{num}$, the maximum value of $n$ is $\textit{num}$.

We can also only consider the units digit. If the units digit satisfies the condition, the higher digits can be arbitrary.

The time complexity is $O(n)$, where $n$ is the size of $\textit{num}$. The space complexity is $O(1)$.

Solution 2

This implementation follows the required operations directly. It traverses the relevant values and updates its state as each value is processed. After all required states have been considered, the maintained result is returned.

Solution 3

This implementation uses depth-first search, followed by memoization. It traverses the relevant values and updates its state as each value is processed. Previously computed states are retained so the same subproblem does not need to be solved again. After all required states have been considered, the maintained result is returned.

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

• 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;
      }
  }
  
  
  // Solution 2
  class Solution {
      public int minimumNumbers(int num, int k) {
          if (num == 0) {
              return 0;
          }
          for (int i = 1; i <= 10; ++i) {
              if ((k * i) % 10 == num % 10 && k * i <= num) {
                  return i;
              }
          }
          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;
      }
  };
  
  
  // Solution 2
  class Solution {
  public:
      int minimumNumbers(int num, int k) {
          if (!num) return 0;
          for (int i = 1; i <= 10; ++i)
              if ((k * i) % 10 == num % 10 && k * i <= num)
                  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
  
  
  # Solution 2
  class Solution:
      def minimumNumbers(self, num: int, k: int) -> int:
          if num == 0:
              return 0
          for i in range(1, 11):
              if (k * i) % 10 == num % 10 and k * i <= num:
                  return i
          return -1
  
  
  
  # Solution 3
  class Solution:
      def minimumNumbers(self, num: int, k: int) -> int:
          @cache
          def dfs(v):
              if v == 0:
                  return 0
              if v < 10 and v % k:
                  return inf
              i = 0
              t = inf
              while (x := i * 10 + k) <= v:
                  t = min(t, dfs(v - x))
                  i += 1
              return t + 1
  
          if num == 0:
              return 0
          if k == 0:
              return -1 if num % 10 else 1
          ans = dfs(num)
          return -1 if ans >= inf else ans
  
  

• 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
  }
  
  
  // Solution 2
  func minimumNumbers(num int, k int) int {
  	if num == 0 {
  		return 0
  	}
  	for i := 1; i <= 10; i++ {
  		if (k*i)%10 == num%10 && k*i <= num {
  			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;
  }
  
  

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