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

Java

    public class Solution {
        public int minPartitions(String n) {
            int num = 0;
            for (char digit : n.toCharArray()) {
                num = Math.max(num, digit);
                if (num == '9') {
                    break;
                }
            }
            return num - '0';
        }
    }

Java

  • 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';
        }
    };
    
  • # 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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