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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
• C++
• Python
• Go
• TypeScript
• RenderScript

• 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;
      }
  }
  
  ############
  
  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;
      }
  }
  

• // 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
  

• 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
      }
  }
  
  

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