At first glance, this problem looks like a complex mathematical challenge involving partitions and sums. However, once you peek under the hood, you will find it is actually a clever logic puzzle that tests your ability to identify the "bottleneck" in a system.
Problem Summary
You're given:
A string $n$ representing a very large positive integer.
Your goal:
Find the minimum number of "deci-binary" numbers (numbers consisting only of digits 0 and 1) that sum up to $n$.
Intuition
To solve this, we need to think about how addition works digit by digit. A deci-binary number can only contribute a value of 0 or 1 to any specific decimal place.
If you have a digit 7 in the units place, you need at least 7 separate deci-binary numbers to provide enough 1s to sum up to that 7. Even if other digits in the number are smaller, like 2 or 3, they can be satisfied by using 0s in those specific positions for some of the deci-binary numbers.
Therefore, the "limiting factor" is simply the largest digit present in the string. If the string contains a 9, you need 9 deci-binary numbers. If the largest digit is 3, you only need 3. The logic is as simple as finding the maximum character in the string.
Walkthrough: Understanding the Examples
Example 1: $n = 32$
- The digits are 3 and 2.
- To get the 3 in the tens place, we need at least three 1s (10 + 10 + 10).
- To get the 2 in the units place, we only need two 1s.
- We can use: $11 + 11 + 10 = 32$.
- Total numbers used: 3 (which is the maximum digit).
Example 2: $n = 82734$
- The digits are 8, 2, 7, 3, and 4.
- The largest digit is 8.
- To satisfy the 8, we must have 8 separate numbers. All other positions (2, 7, 3, 4) can be formed using 8 or fewer 1s.
- Result: 8.
Code Blocks
C++
class Solution {
public:
int minPartitions(string n) {
int maxi = 0;
for(char c : n) {
// Convert character to integer and track the maximum
if(maxi < c - '0') {
maxi = c - '0';
}
}
return maxi;
}
};
Python
class Solution:
def minPartitions(self, n: str) -> int:
# The result is simply the maximum digit in the string
# We convert the characters to integers to find the max
return int(max(n))
JavaScript
/**
* @param {string} n
* @return {number}
*/
var minPartitions = function(n) {
let maxi = 0;
for (let i = 0; i < n.length; i++) {
// Find the highest digit in the string
let digit = Number(n[i]);
if (digit > maxi) {
maxi = digit;
}
// Optimization: if we find a 9, we can stop early
if (maxi === 9) return 9;
}
return maxi;
};
Key Takeaways
- Greedy Logic: Often, the most extreme value in a dataset (the maximum) dictates the requirements for the entire set.
- String Manipulation: When dealing with massive numbers that exceed standard integer limits ($10^{100000}$), treating the number as a string is the only way to process it.
- Complexity: This solution runs in $O(n)$ time where $n$ is the length of the string, and $O(1)$ extra space, making it highly efficient.
Final Thoughts
This problem is a classic "Aha!" moment in technical interviews. It looks like a dynamic programming or partition problem, but it is actually a test of observation. In real-world software engineering, this mirrors resource allocation. If you are building a system to process parallel tasks, your total execution time is often limited by the single longest task in the queue, just as our sum was limited by the largest digit.
Top comments (2)
Clean walkthrough — the key insight that each digit position is independent and the bottleneck is just max(digits) makes this way more intuitive than trying to think about partition sums.
Thanks sir, Glad you liked it !