1689. Partitioning Into Minimum Number Of Deci-Binary Numbers

Problem

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 <= 10^5

• n consists of only digits.

• n does not contain any leading zeros and represents a positive integer.

Solution (Java)

class Solution {
    public int minPartitions(String n) {
        char[] tempArray = n.toCharArray();
        int result = 0;
        for (int i = 0; i < n.length(); i++) {
            result = Math.max(result, tempArray[i] - 48);
        }
        return result;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).