Partitioning Into Minimum Number Of Deci Binary Numbers - Leetcode Solution

Problem:

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.

Solution:

To solve this problem, we need to find the largest digit in the given number, which will determine the number of deci-binary numbers needed to sum up to the given number.

For example, if the given number is "32", the largest digit is 3, which means we need a minimum of three deci-binary numbers to represent 32.

Algorithm:

1. Initialize a variable max_digit to 0 and loop through each digit of the given number n.
2. Convert the current digit to an integer and compare it with max_digit. If it is greater than max_digit, update max_digit to the current digit.
3. Return the value of max_digit as the minimum number of deci-binary numbers needed to represent the given number.

Python Code:

class Solution: def minPartitions(self, n: str) -> int: max_digit = 0 # initialize max_digit to 0 for digit in n: digit = int(digit) if digit > max_digit: max_digit = digit return max_digit

Complexity Analysis:

Time complexity: O(n), where n is the length of the given number n.

Space complexity: O(1), as we are not using any extra data structures.

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