#3300

Easy

LeetCode Problem Workspace

Minimum Element After Replacement With Digit Sum

Replace each number with its digit sum and return the smallest resulting value, using array plus math techniques efficiently.

array math

Problem Statement

Given an array of integers, replace every number with the sum of its digits. After performing this replacement on all elements, determine and return the minimum value present in the array. Each digit sum operation reduces numbers to smaller values but can vary depending on the original integer composition.

For example, if nums = [10,12,13,14], replacing each element with its digit sum results in [1, 3, 4, 5], and the minimum element is 1. Ensure your solution handles arrays of length up to 100 and integers up to 10,000 efficiently using direct computation of digit sums.

Examples

Example 1

Input: nums = [10,12,13,14]

Output: 1

nums becomes [1, 3, 4, 5] after all replacements, with minimum element 1.

Example 2

Input: nums = [1,2,3,4]

Output: 1

nums becomes [1, 2, 3, 4] after all replacements, with minimum element 1.

Example 3

Input: nums = [999,19,199]

Output: 10

nums becomes [27, 10, 19] after all replacements, with minimum element 10.

Constraints

1 <= nums.length <= 100
1 <= nums[i] <= 104

Solution Approach

Iterate and Sum Digits Directly

Loop through each element in the array and calculate its digit sum by repeatedly dividing by 10 and summing the remainders. Keep track of the minimum value while traversing to avoid an extra pass.

Convert to String for Digit Extraction

Transform each integer into a string, then map each character back to an integer to sum digits. This simplifies the digit sum logic and aligns with array plus math patterns for readability and correctness.

Single Pass Minimum Tracking

Instead of storing the transformed array, compute the digit sum and compare it immediately with a running minimum. This reduces space usage and ensures you never miss the smallest transformed value.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

Time complexity is O(n * k), where n is the array length and k is the number of digits in the largest number. Space complexity can be O(1) if using a running minimum without storing the transformed array.

What Interviewers Usually Probe

They may ask how to efficiently compute the digit sum without converting to string.
Expect questions on handling edge cases like single-digit numbers or maximum constraints.
They might test optimization for in-place minimum calculation versus storing intermediate results.

Common Pitfalls or Variants

Common pitfalls

Forgetting to replace all numbers before computing the minimum can yield incorrect results.
Incorrectly summing digits by assuming fixed-length numbers instead of iterating through all digits.
Using extra arrays unnecessarily when a running minimum suffices, increasing space complexity.

Follow-up variants

Instead of the minimum, return the maximum element after digit sum replacement.
Apply the digit sum replacement repeatedly until all numbers are single-digit, then return the minimum.
Replace only numbers above a certain threshold with their digit sum and return the smallest value.

FAQ

How do you calculate the minimum element after replacing numbers with digit sums?

Iterate through the array, compute each number's digit sum, and track the minimum value encountered.

Can this approach handle single-digit numbers or large integers efficiently?

Yes, single-digit numbers remain unchanged, and digit sums for large numbers can be computed efficiently with division or string conversion.

Does the order of elements affect the final minimum?

No, since all elements are transformed independently, the order does not affect the resulting minimum value.

Is it necessary to store the transformed array?

No, you can compute each digit sum and compare it with a running minimum to reduce space usage.

What is the main pattern behind Minimum Element After Replacement With Digit Sum?

The key pattern is array plus math: transform each array element by digit sum calculation, then identify the smallest result.

terminal

Solution

Solution 1: Simulation

We can traverse the array $\textit{nums}$. For each number $x$, we calculate the sum of its digits $y$. The minimum value among all $y$ is the answer.

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class Solution:
    def minElement(self, nums: List[int]) -> int:
        return min(sum(int(b) for b in str(x)) for x in nums)

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