# 1689. Partitioning Into Minimum Number Of Deci-Binary Numbers

## Description

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.

## Solutions

Solution 1: Quick Thinking

The problem is equivalent to finding the maximum number in the string.

The time complexity is $O(n)$, where $n$ is the length of the string. The space complexity is $O(1)$.

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

• class Solution {
public:
int minPartitions(string n) {
int ans = 0;
for (char& c : n) ans = max(ans, c - '0');
return ans;
}
};

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


• func minPartitions(n string) (ans int) {
for _, c := range n {
if t := int(c - '0'); ans < t {
ans = t
}
}
return
}

• function minPartitions(n: string): number {
let nums = n.split('').map(d => parseInt(d));
let ans = Math.max(...nums);
return ans;
}


• impl Solution {
pub fn min_partitions(n: String) -> i32 {
let mut ans = 0;
for c in n.as_bytes() {
ans = ans.max((c - b'0') as i32);
}
ans
}
}