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Find Minimum Operations to Make All Elements Divisible by Three

Difficulty: Easy

Problem 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.

Key Insights

  • The remainder of each number when divided by 3 can only be 0, 1, or 2.
  • To make a number divisible by 3:
    • If the remainder is 1, subtracting 1 will make it divisible (1 operation).
    • If the remainder is 2, adding 1 will make it divisible (1 operation).
  • Therefore, for each element in nums, the total operations required can be calculated based on its remainder when divided by 3.
  • The final count of operations is the sum of the necessary operations for each element.

Space and Time Complexity

Time Complexity: O(n), where n is the length of the array nums, since we need to iterate through each element once.

Space Complexity: O(1), as we are using a constant amount of space for counting operations.

Solution

To solve the problem, we will iterate through each element in the nums array, calculate its remainder when divided by 3, and determine the number of operations needed to make it divisible by 3. We will maintain a counter to sum these operations and return the total count.

Code Solutions

def min_operations_to_make_divisible_by_three(nums):
    operations = 0
    for num in nums:
        remainder = num % 3
        # Calculate operations needed based on remainder
        operations += remainder  # Either 0, 1, or 2 operations needed
    return operations

# Example usage
print(min_operations_to_make_divisible_by_three([1, 2, 3, 4]))  # Output: 3
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