Partitioning Into Minimum Number Of Deci-Binary Numbers#

Problem statement#

[1]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 <= 10^5.

  • n consists of only digits.

  • n does not contain any leading zeros and represents a positive integer.

Solution: Identify the maximum digit of n#

Any digit d can be obtained by summing the digit 1 d times.

The problem turns into identifying the maximum digit of n.

Example 2#

For n = "82734" the answer is 8 because:

  82734
= 11111 
+ 11111 
+ 10111 
+ 10101 
+ 10100
+ 10100
+ 10100
+ 10000

Code#

#include <iostream>
using namespace std;
int minPartitions(const string& n) {
    char maxDigit = '0';
    for (auto& d : n) {
        maxDigit = max(maxDigit, d);
    }
    return maxDigit - '0';
}
int main() {
    cout << minPartitions("32") << endl;
    cout << minPartitions("82734") << endl;
    cout << minPartitions("27346209830709182346") << endl;
}

Output:
3
8
9

Complexity#

  • Runtime: O(N), where N = n.length.

  • Extra space: O(1).

Conclusion#

This problem can be efficiently solved by identifying the maximum digit in the string. Since each deci-binary number can only contain digits from 0 to 9, the maximum digit determines the minimum number of deci-binary numbers needed.

By finding the maximum digit in the string and converting it to an integer, the solution effectively determines the minimum number of deci-binary numbers required.