LeetCode 3300 - Minimum Element After Replacement With Digit Sum

Difficulty: 🟢 Easy
Topics: Array, Math

Solution

Problem Understanding

The problem asks us to transform an array of integers nums by replacing each element with the sum of its digits. After this transformation, we need to determine the minimum element in the resulting array. The input array nums consists of 1 to 100 elements, and each element ranges from 1 to 10,000. The constraints ensure that the array is small enough to iterate over entirely, and the digit sum calculation for each number is manageable.

The task is straightforward conceptually, but it is important to correctly compute the sum of digits for numbers with multiple digits, particularly edge cases like numbers with repeated nines or very small numbers. The expected output is a single integer, the minimum of all digit sums after transformation.

Approaches

The brute-force approach is also optimal in this case, as the input size is small. We iterate through each number in the array, compute the sum of its digits, and track the minimum sum found. This approach is simple, easy to implement, and efficient given the constraints. There is no need for further optimization or complex data structures.

The key insight is that the sum of digits is independent for each number, and we only care about the minimum value after transformation. Therefore, we do not need to store the entire transformed array; maintaining a running minimum is sufficient.

Approach	Time Complexity	Space Complexity	Notes
Brute Force	O(n * log10(maxNum))	O(1)	Iterate over the array, sum digits for each number, track minimum. Optimal for given constraints
Optimal	O(n * log10(maxNum))	O(1)	Same as brute force; further optimization is unnecessary

Algorithm Walkthrough

Initialize a variable min_val with a large value, e.g., infinity, to track the minimum digit sum found.
Iterate through each number num in the input array nums.
For each num, compute the sum of its digits. This can be done by repeatedly taking num % 10 to extract the last digit, adding it to a running sum, and then updating num //= 10 to remove the last digit.
Compare the computed digit sum with min_val and update min_val if the current sum is smaller.
After processing all numbers, return min_val.

Why it works: The algorithm processes every number exactly once, correctly computes the sum of digits, and maintains a running minimum. This guarantees that at the end, min_val holds the smallest digit sum in the array.

Python Solution

from typing import List

class Solution:
    def minElement(self, nums: List[int]) -> int:
        min_val = float('inf')
        for num in nums:
            digit_sum = 0
            n = num
            while n > 0:
                digit_sum += n % 10
                n //= 10
            min_val = min(min_val, digit_sum)
        return min_val

Implementation Explanation: The code initializes min_val to infinity. For each number, it computes the sum of digits using a simple while loop. The minimum is updated dynamically. The code correctly handles all input constraints, including single-digit numbers and numbers with multiple digits.

Go Solution

func minElement(nums []int) int {
    minVal := 1<<31 - 1 // initialize with max int
    for _, num := range nums {
        n := num
        digitSum := 0
        for n > 0 {
            digitSum += n % 10
            n /= 10
        }
        if digitSum < minVal {
            minVal = digitSum
        }
    }
    return minVal
}

Go-specific Notes: In Go, we initialize minVal with math.MaxInt32 (or 1<<31 - 1) to ensure any digit sum will be smaller. The iteration and digit sum computation mirror the Python logic. Go uses integer division and modulo, similar to Python.

Worked Examples

Example 1: nums = [10,12,13,14]

num	Digit Sum	Running Min
10	1+0=1	1
12	1+2=3	1
13	1+3=4	1
14	1+4=5	1

Output: 1

Example 2: nums = [999,19,199]

num	Digit Sum	Running Min
999	9+9+9=27	27
19	1+9=10	10
199	1+9+9=19	10

Output: 10

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n * log10(maxNum))	Each number requires computing its digit sum, which is proportional to the number of digits (~log10(num))
Space	O(1)	Only constant space is used for variables; no additional data structures required

The algorithm efficiently handles the maximum constraints, as even the largest number 10,000 has only 5 digits.

Test Cases

# Provided examples
assert Solution().minElement([10,12,13,14]) == 1  # minimum digit sum is 1
assert Solution().minElement([1,2,3,4]) == 1      # minimum digit sum is 1
assert Solution().minElement([999,19,199]) == 10 # minimum digit sum is 10

# Edge and boundary cases
assert Solution().minElement([5]) == 5           # single-element array
assert Solution().minElement([10000, 9999]) == 1 # large numbers
assert Solution().minElement([11, 22, 33]) == 2  # all two-digit numbers
assert Solution().minElement([123, 321, 213]) == 6 # digit sum equals 6 for all
assert Solution().minElement([1,10,100,1000]) == 1 # mixed powers of 10

Test	Why
[10,12,13,14]	Provided example with small numbers
[1,2,3,4]	Provided example, all single-digit numbers
[999,19,199]	Provided example with large sums
[5]	Single-element array, minimal edge case
[10000, 9999]	Large numbers to check digit sum calculation
[11, 22, 33]	Two-digit numbers to test repeated digits
[123,321,213]	Mixed permutations of digits to ensure sum is correct
[1,10,100,1000]	Powers of 10 to confirm handling of zeros

Edge Cases

Single-element array: If nums has only one element, the function must correctly return the digit sum of that element. The algorithm handles this naturally by iterating once.

Numbers with zeros: Numbers like 10, 100, or 1000 contain zeros, which should not affect the digit sum. The algorithm processes zeros correctly since n % 10 yields 0 and n //= 10 reduces the number.

Large numbers: Numbers up to 10,000 can appear, which have up to 5 digits. The sum-of-digits computation must handle multiple digits correctly. The algorithm does this in a loop that processes each digit sequentially.