LeetCode 3697 - Compute Decimal Representation

This problem asks us to decompose a positive integer n into the fewest possible base-10 components, where a base-10 component is defined as a single digit (1-9) multiplied by a power of 10. Essentially, we are breaking n down into its decimal place contributions.

leetcodeeasyarraymath

LeetCode Problem 3697

Difficulty: 🟢 Easy
Topics: Array, Math

Solution

Problem Understanding

This problem asks us to decompose a positive integer n into the fewest possible base-10 components, where a base-10 component is defined as a single digit (1-9) multiplied by a power of 10. Essentially, we are breaking n down into its decimal place contributions. For example, for n = 537, the hundreds place contributes 500, the tens place 30, and the ones place 7. These are exactly the base-10 components.

The input is guaranteed to be a positive integer up to 1 billion (1 <= n <= 10^9), which means performance must be linear or better with respect to the number of digits in n. The expected output is an array of integers representing the base-10 components in descending order, which is the natural order if we process the digits from left to right.

Important edge cases include numbers that have zeros in some positions, such as 102 (to avoid including zero as a component), single-digit numbers (6), and very large numbers close to the upper limit (10^9).

Approaches

A brute-force approach would attempt to generate all possible combinations of base-10 components that sum to n and select the combination with the fewest elements. This would involve checking all partitions of n in terms of powers of 10 and digits 1-9. While this is theoretically correct, it is highly inefficient, because the number of combinations grows exponentially with the number of digits. For large n, this approach is computationally infeasible.

The key insight for an optimal solution comes from recognizing that each digit of n naturally represents a base-10 component multiplied by its positional power of 10. By iterating through each digit and constructing the component only if it is non-zero, we automatically obtain the minimal set of base-10 components. This observation leverages the decimal system directly and avoids unnecessary computation. The resulting components are naturally in descending order if we process digits from the highest to the lowest position.

Approach Time Complexity Space Complexity Notes
Brute Force O(2^d) O(d) Generate all partitions using digits and powers of 10, where d is number of digits
Optimal O(d) O(d) Process each digit once, constructing non-zero components

Algorithm Walkthrough

  1. Convert the integer n to its string representation so we can process each digit individually from left to right. This allows easy access to each decimal place.
  2. Initialize an empty list components to store the base-10 components.
  3. Iterate through the string representation of n, tracking the current index i and digit character ch.
  4. For each digit, convert it back to an integer digit_value. Calculate the positional multiplier as 10^(length_of_n - i - 1) to determine the actual value of this digit in n.
  5. If the digit is non-zero, multiply it by the positional multiplier and append the result to components. Zero digits are skipped since they do not contribute to the sum.
  6. After processing all digits, return components, which will be in descending order.

Why it works: By processing digits from highest to lowest, we ensure that each non-zero digit contributes exactly once to the decomposition. Skipping zeros avoids unnecessary components. This guarantees that the sum of the components equals n and that the number of components is minimal, because each digit represents the maximal contribution at its place.

Python Solution

from typing import List

class Solution:
    def decimalRepresentation(self, n: int) -> List[int]:
        n_str = str(n)
        components = []
        length = len(n_str)
        
        for i, ch in enumerate(n_str):
            digit_value = int(ch)
            if digit_value != 0:
                component = digit_value * (10 ** (length - i - 1))
                components.append(component)
        
        return components

In this implementation, we first convert the integer to a string to easily iterate through each digit. We then calculate the positional value for each non-zero digit by multiplying it with the appropriate power of 10. This directly constructs each base-10 component in descending order, matching the problem requirement.

Go Solution

func decimalRepresentation(n int) []int {
    var components []int
    divisor := 1
    
    // Find the highest power of 10 less than or equal to n
    for temp := n; temp >= 10; temp /= 10 {
        divisor *= 10
    }
    
    for divisor > 0 {
        digit := n / divisor
        if digit != 0 {
            components = append(components, digit*divisor)
        }
        n %= divisor
        divisor /= 10
    }
    
    return components
}

The Go implementation avoids converting the integer to a string. Instead, it computes the largest power of 10 less than n and iterates by dividing and taking modulo to extract each digit. This approach is more idiomatic for Go and efficiently constructs the result slice.

Worked Examples

Example 1: n = 537

Step Digit Multiplier Component Added Components
1 5 100 500 [500]
2 3 10 30 [500, 30]
3 7 1 7 [500, 30, 7]

Output: [500, 30, 7]

Example 2: n = 102

Step Digit Multiplier Component Added Components
1 1 100 100 [100]
2 0 10 0 skipped [100]
3 2 1 2 [100, 2]

Output: [100, 2]

Example 3: n = 6

Step Digit Multiplier Component Added Components
1 6 1 6 [6]

Output: [6]

Complexity Analysis

Measure Complexity Explanation
Time O(d) Each digit is processed once, where d is the number of digits in n
Space O(d) We store one component per non-zero digit

Since d <= 10 for n <= 10^9, this solution is extremely efficient.

Test Cases

# Provided examples
assert Solution().decimalRepresentation(537) == [500,30,7]  # Example 1
assert Solution().decimalRepresentation(102) == [100,2]     # Example 2
assert Solution().decimalRepresentation(6) == [6]           # Example 3

# Edge cases
assert Solution().decimalRepresentation(1) == [1]           # smallest number
assert Solution().decimalRepresentation(10) == [10]         # single zero digit
assert Solution().decimalRepresentation(1000000000) == [1000000000]  # largest number

# Random cases
assert Solution().decimalRepresentation(90705) == [90000,700,5] # mix of zeros
assert Solution().decimalRepresentation(111) == [100,10,1]       # repeated digits
Test Why
537 Normal multi-digit number with all non-zero digits
102 Zero in the middle position
6 Single-digit number
1 Minimal input value
10 Zero in the units place
1000000000 Maximal input value, large power of 10
90705 Sparse non-zero digits across multiple places
111 Repeated digits

Edge Cases

  1. Single-digit numbers: These numbers are already base-10 components. The algorithm correctly identifies that no decomposition is necessary and returns the number itself.
  2. Numbers with zero digits: For example, 102 has a zero in the tens place. If the algorithm were to include zero as a component, it would violate the problem requirement. Skipping zero ensures correctness.
  3. Large numbers at the upper constraint: n = 10^9 tests whether the solution handles integer bounds and large powers of 10. The algorithm scales linearly with the number of digits, not the magnitude, so it handles this efficiently.

These edge cases could trip up a naive implementation, but the digit-by-digit processing strategy safely handles them.