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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.
1function minPartitions(n) {
2 let maxDigit = '0';
3 for (const digit of n) {
4 if (digit > maxDigit
In this JavaScript code, we iterate over the string 'n', updating the maxDigit whenever we find a larger digit. The maximum digit tells us how many deci-binary numbers are necessary.
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).
#include <string>
#include <algorithm>
int minPartitions(std::string n) {
return *std::max_element(n.begin(), n.end()) - '0';
}
int main() {
std::string n = "27346209830709182346";
std::cout << minPartitions(n) << "\n";
return 0;
}In this C++ solution, we use the STL function 'std::max_element' to find the maximum digit directly in the string. This method showcases how to leverage library functions for concise code.