1689. Partitioning Into Minimum Number Of Deci-Binary Numbers

package LeetCode_1689

/**
 * 1689. Partitioning Into Minimum Number Of Deci-Binary Numbers
 * https://leetcode.com/problems/partitioning-into-minimum-number-of-deci-binary-numbers/
 * 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. 1 <= n.length <= 105
2. n consists of only digits.
3. n does not contain any leading zeros and represents a positive integer.
 * */
class Solution {
    /*
    * solution: greedy, get the max digit, Time:O(L), Space:O(1);
    * for example: 135:
    * init 5 deci-binary number with length is 3:
    * 000
    * 000
    * 000
    * 000
    * 000
    * then we can fill like below:
    * 111
    * 011
    * 011
    * 001
    * 001
    * 111+11+11+1+1 = 135
    * */
    fun minPartitions(n: String): Int {
        var max = 0
        for (c in n) {
            max = Math.max(max, c - '0')
        }
        return max
    }
}