LeetCode 2310. Sum of Numbers With Units Digit K Solution in Java, C++, Python & More | Explanation + Code

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 num - n × k must be a multiple of 10.

We enumerate n from small to large, and find the first n that satisfies num - n × k being a multiple of 10. Since n cannot exceed num, the maximum value of n is 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 num. The space complexity is O(1).

PythonJavaC++GoTypeScript

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(code-box)

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; } }(code-box)

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; } };(code-box)

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 }(code-box)

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; }(code-box)

Solution 2

PythonJavaC++Go

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(code-box)

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; } }(code-box)

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; } };(code-box)

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 }(code-box)

Solution 3

Python

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(code-box)