leetcode 2310. Sum of Numbers With Units Digit K （python）

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describe

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.

Note:

0 <= num <= 3000
0 <= k <= 9

analysis

According to the meaning , Given two integers num and k , Consider a set of positive integers with the following properties ：

• The bits of each integer are k
• The sum of integers is num

Returns the smallest possible size of such a collection , If no such collection exists , Then return to -1 . It should be noted that , A collection can contain multiple instances of the same integer , The sum of empty sets is considered to be 0 .

This problem can be counted as a mathematical problem , Let's assume that there is now a in the set n It's an integer , Then according to the above two conditions , We can know the following formula ：

• n₁ + n₂ + ... + n_n = num

Because every bit of a positive number is k , Then it can become ：

• (k+10*a₁) + (k+10*a₂) + ... + (k+10*a_n) = n*k + 10*(a₁+a₂+...+a_n) = num

Get the above formula , You can get this num-n*k Must be 10 Multiple , So we traverse from 1 To 10 , If there is, the answer will be found , Otherwise go straight back to -1 .

Why do you know that if there is any result, it will be in 1 To 10 Between , Because we are right about 10 To take the mold .

The time complexity is O(1) , The space complexity is O(1) .

answer

class Solution(object):
def minimumNumbers(self, num, k):
"""
:type num: int
:type k: int
:rtype: int
"""
if num == 0 :return 0
for i in range(1, 11):
if (num-i*k) % 10 == 0 and i*k <= num:
return i
return -1

Running results

298 / 298 test cases passed.
Status: Accepted
Runtime: 43 ms
Memory Usage: 13.3 MB

Original link

leetcode.com/contest/wee…

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