# 3190. Find Minimum Operations to Make All Elements Divisible by Three

## Description

You are given an integer array nums. In one operation, you can add or subtract 1 from any element of nums.

Return the minimum number of operations to make all elements of nums divisible by 3.

Example 1:

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

Output: 3

Explanation:

All array elements can be made divisible by 3 using 3 operations:

• Subtract 1 from 1.
• Subtract 1 from 4.

Example 2:

Input: nums = [3,6,9]

Output: 0

Constraints:

• 1 <= nums.length <= 50
• 1 <= nums[i] <= 50

## Solutions

### Solution 1: Mathematics

We directly iterate through the array $\text{nums}$. For each element $x$, we calculate the remainder of $x$ divided by 3, $x \bmod 3$. If the remainder is not 0, we need to make $x$ divisible by 3 with the minimum number of operations. Therefore, we can choose to either decrease $x$ by $x \bmod 3$ or increase $x$ by $3 - x \bmod 3$, and we accumulate the minimum of these two values to the answer.

The time complexity is $O(n)$, where $n$ is the length of the array $\text{nums}$. The space complexity is $O(1)$.

• class Solution {
public int minimumOperations(int[] nums) {
int ans = 0;
for (int x : nums) {
int mod = x % 3;
if (mod != 0) {
ans += Math.min(mod, 3 - mod);
}
}
return ans;
}
}

• class Solution {
public:
int minimumOperations(vector<int>& nums) {
int ans = 0;
for (int x : nums) {
int mod = x % 3;
if (mod) {
ans += min(mod, 3 - mod);
}
}
return ans;
}
};

• class Solution:
def minimumOperations(self, nums: List[int]) -> int:
ans = 0
for x in nums:
if mod := x % 3:
ans += min(mod, 3 - mod)
return ans


• func minimumOperations(nums []int) (ans int) {
for _, x := range nums {
if mod := x % 3; mod > 0 {
ans += min(mod, 3-mod)
}
}
return
}

• function minimumOperations(nums: number[]): number {
let ans = 0;
for (const x of nums) {
const mod = x % 3;
if (mod) {
ans += Math.min(mod, 3 - mod);
}
}
return ans;
}