1689. Partitioning Into Minimum Number Of Deci-Binary Numbers

MediumGreedyString

Problem Description

This LeetCode problem defines a term "deci-binary" which refers to a number that comprises only 0 and 1 digits. A deci-binary number does not contain any digits other than 0 and 1 and should not have leading zeros. The task is to determine the minimum quantity of deci-binary numbers that can be summed together to equal a given positive decimal number n. The input n is provided as a string representation of a decimal integer, and the expected output is an integer representing the minimum number of deci-binary numbers required to sum up to n.

Intuition

The intuition behind the solution is that the minimum number of deci-binary numbers required is determined by the largest digit in the original number n. Since deci-binary numbers can only have the digits 0 or 1, and no leading zeroes are allowed, each deci-binary number can only contribute a 1 towards a single decimal place in the sum.

To build the original number n using deci-binary numbers, you would start by placing a 1 in the position of the largest digit of n, and then subtract 1 from this digit. You would repeat this process, placing a 1 in the position of the current largest digit after subtraction, until all the digits in the original number are reduced to 0.

For example, if n = "123", we would need a 1 in the hundreds place, tens place, and ones place to start, and then continue placing 1s in the tens and ones place until the number is exhausted. It would take three deci-binary numbers to reduce the hundreds place, two to reduce the tens, and three to reduce the ones place, resulting in a total of three deci-binary numbers (3 being the largest digit in n).

Therefore, the solution is as simple as finding the maximum digit in the string n, as this will dictate how many times a 1 needs to be placed in the position of this digit across all deci-binary numbers. The Python code int(max(n)) efficiently finds the largest digit and converts it to an integer, which is the minimum number of deci-binary numbers needed.

Learn more about Greedy patterns.

Solution Approach

The implementation of the solution is remarkably straightforward and doesn't require complex algorithms or data structures. It also doesn't involve a pattern recognition exercise typical for many combinatorial problems; rather, it simply relies on a characteristic of the given decimal number—the value of its highest digit.

The core of the solution approach lies in understanding that, since we can only use deci-binary numbers (numbers with digits 0 or 1 only), the maximum number one such number can contribute to any digit place of the original number n is 1. That's because having any digit greater than 1 would violate the definition of deci-binary.

The algorithm is as follows:

Iterate through each character in the string n. This represents each digit of the decimal number.
Convert each character to an integer.
Keep track of the maximum integer value found during the iteration.

Since we are given the number in the form of a string, no conversion to a number is needed to find the maximum digit value. This is because the characters '0' to '9' have increasing ASCII values, so comparing them as characters yields the same result as comparing their numerical values.

In terms of the solution code provided:

class Solution:
    def minPartitions(self, n: str) -> int:
        return int(max(n))

max(n) gets the character with the highest ASCII value in the string n, which corresponds to the highest digit in the decimal number. Converting this character back to an integer with int() gives us the minimum number of deci-binary numbers needed. The reason for this conversion is that the max() function would give us the maximum character (i.e., digit in the form of a string), but the problem requires us to return an integer.

No additional data structures are required because we only need to determine the maximum digit, which is a single value, and no additional computation or storage is necessary. The algorithm's time complexity is O(m), where m is the number of digits in the string representation of the number n, because it requires a single pass through all the digits to find the maximum one. The space complexity is O(1) as it only uses a constant amount of additional memory.

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Example Walkthrough

Let's illustrate the solution approach using the example n = "82734". We want to find the minimum number of deci-binary numbers that sum up to this number.

The first step is to examine each digit of n. We check 8, 2, 7, 3, and 4.
We find that the largest digit in this number is 8.
Knowing that a deci-binary number only consists of 0s and 1s, the largest number it can contribute to any single digit is 1.
Therefore, we would need at least eight deci-binary numbers to contribute to the digit 8 in the thousands place.
No other digit in the number 82734 is larger than 8, so no more than eight deci-binary numbers will be needed for any other digit.
Hence, the answer is 8, which indicates that eight deci-binary numbers would be enough to add up to 82734.

To visualize this, we can represent the deci-binary numbers and their sum like so:

Deci-binary numbers that sum to 82734:
11111
11111
11111
11111
11111
11111
11111
10000
-------
82734

As you can see, each deci-binary number contributes a 1 to every position but in the last deci-binary number, we only need to contribute to the thousands place to make the sum equal to 82734. It takes eight such numbers to match the maximum digit 8 in the original number.

By executing the line int(max(n)), we take the maximum digit character, which is '8', and convert it into an integer, giving us the final answer, 8.

Solution Implementation

1class Solution:
2    def minPartitions(self, n: str) -> int:
3        # Find the maximum digit in the string representation of the number
4        max_digit = max(n)
5      
6        # Convert the maximum digit from string to integer
7        min_partitions = int(max_digit)
8      
9        # Return the minimum number of partitions required
10        return min_partitions
11
12# Explanation: The provided method minPartitions is used to find the minimum number of 
13# decimal digits needed to write down the number n in such a way that each digit is 
14# used only once. The input is a string representation of a non-negative integer n, 
15# and the method returns an integer representing the answer. The logic simply finds 
16# the maximum digit in the string, as this will be the minimum number needed. For 
17# example, for input '82734', the maximum digit is '8', so you would need at least 
18# 8 partitions (since the number must be decomposable into a sum of numbers containing 
19# each digit at most once).
20

1class Solution {
2    // Method to find the minimum number of partitions required
3    public int minPartitions(String n) {
4        // Initialize the variable to store the maximum digit in the string
5        int maxDigit = 0;
6
7        // Loop through each character of the string representing the number
8        for (int i = 0; i < n.length(); ++i) {
9            // Find the numeric value of the current digit
10            int currentDigit = n.charAt(i) - '0';
11
12            // Update maxDigit if the current digit is greater than the maxDigit so far
13            maxDigit = Math.max(maxDigit, currentDigit);
14
15            // If the maximum digit is 9, we can return it immediately as it's the highest possible digit
16            if (maxDigit == 9) {
17                break;
18            }
19        }
20
21        // Return the maximum digit as the minimum number of partitions required
22        return maxDigit;
23    }
24}
25

1class Solution {
2public:
3    // Function to calculate the minimum number of decimal digits 
4    // one must add to decompose the given string of digits
5    int minPartitions(string n) {
6        // Initialize the answer to zero
7        int maxDigit = 0;
8      
9        // Iterate over each character in the string
10        for (const char& digitChar : n) {
11            // Convert the character to the corresponding integer digit
12            int digit = digitChar - '0';
13          
14            // Update the maximum digit found so far
15            maxDigit = std::max(maxDigit, digit);
16        }
17      
18        // Return the highest digit as the minimum number of partitions needed
19        return maxDigit;
20    }
21};
22

1// Function to find the minimum number of deci-binary numbers
2// needed such that they sum up to the string n.
3// A deci-binary number is a number base-10 that each of its digits
4// is either 0 or 1 without any leading zeros.
5function minPartitions(n: string): number {
6    // Split the input string into an array of its characters,
7    // then map each character to its integer representation
8    let digits = n.split('').map(digit => parseInt(digit));
9  
10    // Find the maximum digit in the array, as this will be the
11    // minimum number of deci-binary numbers needed
12    let maxDigit = Math.max(...digits);
13  
14    // Return the maximum digit, which represents the answer
15    return maxDigit;
16}
17

Time and Space Complexity

Time Complexity

The time complexity of the function is determined by finding the maximum digit in the string n. Since the max() function iterates through each character in the string once, the time complexity is O(d), where d is the length of the input string n.

Space Complexity

The space complexity of the function is O(1). This is because it only uses a fixed amount of additional space to store the maximum digit found in the string, regardless of the length of the input.

Learn more about how to find time and space complexity quickly using problem constraints.

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Question 1 out of 10

You are given an array of intervals where intervals[i] = [start_i, end_i] represent the start and end of the ith interval. You need to merge all overlapping intervals and return an array of the non-overlapping intervals that cover all the intervals in the input.

Sort the intervals by their start time, then iterate through the sorted intervals, merging any overlapping intervals.

Brute force through all interval combinations and find the answer with the least total intervals.

Iterate through the intervals in their original order, merging any overlapping intervals.

Use a priority queue to keep track of the intervals, merging any overlapping intervals.

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