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

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

Difficulty: Easy

Topics: Array, Math, Biweekly Contest 133

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

Hint:

1. If x % 3 != 0 we can always increment or decrement x such that we only need 1 operation.
2. Add min(nums[i] % 3, 3 - (num[i] % 3)) to the count of operations.

Solution:

We need to make all elements divisible by 3 with minimum operations, where each operation is adding or subtracting 1.

For any number, there are three possible cases:

1. If it's already divisible by 3: 0 operations needed
2. If it's 1 more than a multiple of 3: we can subtract 1 (1 operation)
3. If it's 2 more than a multiple of 3: we can add 1 (1 operation)

So for each number, the minimum operations needed is simply the minimum distance to the nearest multiple of 3, which can be calculated as min(num % 3, 3 - (num % 3)).

Let's implement this solution in PHP: 3190. Find Minimum Operations to Make All Elements Divisible by Three

<?php
/**
 * @param Integer[] $nums
 * @return Integer
 */
function minimumOperations($nums) {
    ...
    ...
    ...
    /**
     * go to ./solution.php
     */
}

// Test cases
echo minimumOperations([1,2,3,4]) . "\n";   // Output: 3
echo minimumOperations([3,6,9]) . "\n";     // Output: 0
?>

Explanation:

1. Iterate through each number in the array
2. Calculate remainder when divided by 3 using $num % 3
3. Find minimum operations for current number:
   • If remainder is 0: both $remainder and 3 - $remainder equal 0
   • If remainder is 1: min(1, 2) = 1 (subtract 1)
   • If remainder is 2: min(2, 1) = 1 (add 1)
4. Sum up all operations and return the total

Time Complexity: O(n) - we iterate through the array once
Space Complexity: O(1) - we only use a constant amount of extra space

Contact Links

If you found this series helpful, please consider giving the repository a star on GitHub or sharing the post on your favorite social networks 😍. Your support would mean a lot to me!

If you want more helpful content like this, feel free to follow me:

• LinkedIn
• GitHub