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The key to solving this problem is understanding the relationship between the digits of the number and the deci-binary numbers. The minimum number of deci-binary numbers needed is dictated by the maximum digit in the string representation of the number. This is because to construct the number using deci-binary numbers (which only contain 0 and 1), at least one of these numbers must have a '1' in each digit position where the given number is non-zero. Each deci-binary number can only add 1 to each digit position.
Time Complexity: O(n), where n is the length of the string.
Space Complexity: O(1), as we only use a constant amount of extra space.
1public class Solution {
2 public int minPartitions(String n) {
3 char maxDigit = '0';
4 for (char digit :
In this Java solution, we convert the string into an array of characters and iterate through each character to find the maximum. The result is the minimal required number of deci-binary numbers.
Instead of iterating through the string directly, an alternate way to solve the problem can be to map the string into an array of numbers first, then find the maximum in that array. However, this adds overhead without practical advantage, since the core of the solution relies on identifying the maximum digit in the string.
Time Complexity: O(n), n being the length of the string.
Space Complexity: O(1).
While this approach doesn't effectively differ in the outcome or complexity, it demonstrates the same maximum digit principle, emphasizing direct character handling in the number string.