1689. Partitioning Into Minimum Number Of Deci-Binary Numbers

1689. Partitioning Into Minimum Number Of Deci-Binary Numbers (Medium)

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.

Related Topics:
Greedy

Solution 1.

Simply return the maximum digit in n.

We can express the n in this way -- turn each digit into a column of 1s.

For example n = 212301

2 1 2 3 0 1

0 0 0 1 0 0
1 0 1 1 0 0
1 1 1 1 0 1

The rows are the deci-binary numbers we use to partition n.

// OJ: https://leetcode.com/problems/partitioning-into-minimum-number-of-deci-binary-numbers/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
    int minPartitions(string n) {
        return *max_element(begin(n), end(n)) - '0';
    }
};