README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Easy	https://github.com/doocs/leetcode/edit/main/solution/3300-3399/3300.Minimum%20Element%20After%20Replacement%20With%20Digit%20Sum/README_EN.md	1181	Biweekly Contest 140 Q1	Array Math

3300. Minimum Element After Replacement With Digit Sum

中文文档

Description

You are given an integer array nums.

You replace each element in nums with the sum of its digits.

Return the minimum element in nums after all replacements.

 

Example 1:

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

Output: 1

Explanation:

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

Example 2:

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

Output: 1

Explanation:

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

Example 3:

Input: nums = [999,19,199]

Output: 10

Explanation:

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

 

Constraints:

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

Solutions

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.

The time complexity is $O(n \times \log M)$, where $n$ and $M$ are the length of the array $\textit{nums}$ and the maximum value in the array, respectively. The space complexity is $O(1)$.

Python3

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

Java

class Solution {
    public int minElement(int[] nums) {
        int ans = 100;
        for (int x : nums) {
            int y = 0;
            for (; x > 0; x /= 10) {
                y += x % 10;
            }
            ans = Math.min(ans, y);
        }
        return ans;
    }
}

C++

class Solution {
public:
    int minElement(vector<int>& nums) {
        int ans = 100;
        for (int x : nums) {
            int y = 0;
            for (; x > 0; x /= 10) {
                y += x % 10;
            }
            ans = min(ans, y);
        }
        return ans;
    }
};

Go

func minElement(nums []int) int {
	ans := 100
	for _, x := range nums {
		y := 0
		for ; x > 0; x /= 10 {
			y += x % 10
		}
		ans = min(ans, y)
	}
	return ans
}

TypeScript

function minElement(nums: number[]): number {
    let ans: number = 100;
    for (let x of nums) {
        let y = 0;
        for (; x; x = Math.floor(x / 10)) {
            y += x % 10;
        }
        ans = Math.min(ans, y);
    }
    return ans;
}