3300. Minimum Element After Replacement With Digit Sum

Time: O(n * d)

Space: O(1)

Problem Overview

Each element is independently replaced by the sum of its digits, then we need the minimum among transformed values.

Intuition

Each element is independently replaced by the sum of its digits, then we need the minimum among transformed values. Since replacement does not depend on other elements, a single pass with a running minimum is optimal.

Algorithm

1Initialise ans = +infinity.
2For each number num in nums:
3 Compute digitSum(num) by repeatedly taking num % 10 and num /= 10.
4 Update ans = min(ans, digitSum(num)).
5Return ans after processing all elements.

Example Walkthrough

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

1.digitSum(10) = 1, ans = 1.
2.digitSum(12) = 3, ans stays 1.
3.digitSum(13) = 4, digitSum(14) = 5, ans stays 1.

Output: 1

Common Pitfalls

•Remember that digit sum for 0 is 0; guard if inputs can include 0.
•Do not sort just to find a minimum transformed value; one pass is enough.
•Digit-sum computation should use integer operations, not string conversion, for efficiency and simplicity.

3300.cs

// Approach: Replace each number by its digit sum and track the minimum digit sum seen.
// Compute digit sums in O(number of digits) per element and update a running minimum.
// Time: O(n * d) Space: O(1), where d is average digit count.

public class Solution
{
    public int MinElement(int[] nums)
    {
        int ans = int.MaxValue;
        foreach (int num in nums)
            ans = Math.Min(ans, GetDigitSum(num));
        return ans;
    }

    private int GetDigitSum(int num)
    {
        int digitSum = 0;
        while (num > 0)
        {
            digitSum += num % 10;
            num /= 10;
        }
        return digitSum;
    }
}

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