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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

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