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1689. Partitioning Into Minimum Number Of Deci-Binary Numbers

1689. Partitioning Into Minimum Number Of Deci-Binary Numbers

The math behind the problem

Nhut Nguyen's photo
Nhut Nguyen
·Oct 17, 2022·

2 min read

Problem statement

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(string n) {
    char maxDigit = '0';
    for (char& 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).

References

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