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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.
  • Add 1 to 2.
  • 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 \(\textit{nums}\). 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 \(\textit{nums}\). The space complexity is \(O(1)\).

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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

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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;
    }
}

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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;
    }
};

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

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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;
}

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