# Question

Formatted question description: https://leetcode.ca/all/1689.html

 1689. 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 <= n.length <= 105
n consists of only digits.
n does not contain any leading zeros and represents a positive integer.



# Algorithm

Find the largest single digit… More of math proof about it…

# Code

• class Solution {
public int minPartitions(String n) {
int max = 0;
int length = n.length();
for (int i = 0; i < length; i++) {
int digit = n.charAt(i) - '0';
max = Math.max(max, digit);
}
return max;
}
}

• // OJ: https://leetcode.com/problems/partitioning-into-minimum-number-of-deci-binary-numbers/
// Time: O(N)
// Space: O(1)
class Solution {
public:
int minPartitions(string n) {
return *max_element(begin(n), end(n)) - '0';
}
};

• class Solution:
def minPartitions(self, n: str) -> int:
return int(max(n))

############

# 1689. Partitioning Into Minimum Number Of Deci-Binary Numbers
# https://leetcode.com/problems/partitioning-into-minimum-number-of-deci-binary-numbers/

class Solution:
def minPartitions(self, n: str) -> int:
res = float('-inf')

for c in n:
res = max(int(c), res)

return res