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2310. Sum of Numbers With Units Digit K

  • Time: $O(1)$
  • Space: $O(1)$
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class Solution {
 public:
  int minimumNumbers(int num, int k) {
    if (num == 0)
      return 0;

    // Assume the size of the set is n, and the numbers in the set are X1, X2,
    // ..., Xn. Since the units digit of each number is k, X1 + X2 + ... + Xn =
    // N * k + 10 * (x1 + x2 + ... + xn) = num. Therefore, the goal is to find
    // the n s.t. n * k % 10 = num % 10
    for (int i = 1; i <= 10 && i * k <= num; ++i)
      if (i * k % 10 == num % 10)
        return i;

    return -1;
  }
};

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class Solution {
  public int minimumNumbers(int num, int k) {
    if (num == 0)
      return 0;

    // Assume the size of the set is n, and the numbers in the set are X1, X2,
    // ..., Xn. Since the units digit of each number is k, X1 + X2 + ... + Xn =
    // N * k + 10 * (x1 + x2 + ... + xn) = num. Therefore, the goal is to find
    // the n s.t. n * k % 10 = num % 10
    for (int i = 1; i <= 10 && i * k <= num; ++i)
      if (i * k % 10 == num % 10)
        return i;

    return -1;
  }
}

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class Solution:
  def minimumNumbers(self, num: int, k: int) -> int:
    if num == 0:
      return 0

    # Assume the size of the set is n, and the numbers in the set are X1, X2,
    # ..., Xn. Since the units digit of each number is k, X1 + X2 + ... + Xn =
    # N * k + 10 * (x1 + x2 + ... + xn) = num. Therefore, the goal is to find
    # the n s.t. n * k % 10 = num % 10
    for i in range(1, 11):
      if i * k > num + 1:
        break
      if i * k % 10 == num % 10:
        return i

    return -1