3300. Minimum Element After Replacement With Digit Sum
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 <= 1001 <= 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;
}