LeetCode 1689: Partitioning Into Minimum Number Of Deci-Binary Numbers Solution
Master LeetCode problem 1689 (Partitioning Into Minimum Number Of Deci-Binary Numbers), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.
Problem Explanation
Explanation
To determine the minimum number of positive deci-binary numbers needed to sum up to a given positive decimal integer n, we need to find the maximum digit in the decimal representation of n. The reason is that each deci-binary number can only have digits 0 or 1, and the largest digit in n determines how many deci-binary numbers are needed to represent n.
Algorithm
- Iterate through the digits of
nto find the maximum digit. - The maximum digit found will be the minimum number of positive deci-binary numbers needed to sum up to
n.
Time Complexity
The time complexity of this algorithm is O(n), where n is the length of the input string n.
Space Complexity
The space complexity is O(1) as we are using only a constant amount of extra space.
Solution Code
class Solution {
public int minPartitions(String n) {
int maxDigit = 0;
for (int i = 0; i < n.length(); i++) {
int digit = n.charAt(i) - '0';
maxDigit = Math.max(maxDigit, digit);
}
return maxDigit;
}
}Try It Yourself
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What is the time complexity of LeetCode 1689 (Partitioning Into Minimum Number Of Deci-Binary Numbers)?
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