1689. Partitioning Into Minimum Number Of Deci-Binary Numbers

#### QUESTION:

A decimal number is called deci-binary if each of its digits is either 0 or 1 without any leading zeros. For example, 101 and 1100 are deci-binary, while 112 and 3001 are not.

Given a string n that represents a positive decimal integer, return the minimum number of positive deci-binary numbers needed so that they sum up to n.

Example 1:

``````Input: n = "32"
Output: 3
Explanation: 10 + 11 + 11 = 32
``````

Example 2:

``````Input: n = "82734"
Output: 8
``````

Example 3:

``````Input: n = "27346209830709182346"
Output: 9
``````

Constraints:

``````1 <= n.length <= 105
n consists of only digits.
n does not contain any leading zeros and represents a positive integer.
``````

#### EXPLANATION:

1. 首先创建流式操作,获取到max的值
2. 将max的值转换成int值,也就是对应字符的ascii值
3. 将这个值减去’0’的值,就可以获取到对应的值了.

#### SOLUTION:

``````class Solution {
fun minPartitions(n: String): Int {
var result = 0
n.toCharArray().forEach { c ->
result = Math.max(result,Integer.parseInt(c.toString()))
}
return result
}
}
// 一行代码
fun minPartitions(n: String): Int {
return n.chars().max().asInt - 48
}
``````
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