How do you solve Minimum Number of Operations to Convert Time on LeetCode?

Minimum Number of Operations to Convert Time (LeetCode #2224) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(1) time complexity and O(1) space complexity. Key insight: Always prioritize larger increments to minimize the number of operations.

What is the time complexity of Minimum Number of Operations to Convert Time?

The optimal solution for Minimum Number of Operations to Convert Time runs in O(1) time and uses O(1) space. The complexity is O(1) because we only perform a fixed number of operations (4 possible increments) regardless of the input size.

What is the brute force solution for Minimum Number of Operations to Convert Time?

The brute force approach for Minimum Number of Operations to Convert Time is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves trying all possible combinations of operations to reach the correct time from the current time. This method is straightforward but inefficient as it explores many unnecessary paths. The optimal Optimal Solution approach improves this to O(1).

What algorithm or data structure is used to solve Minimum Number of Operations to Convert Time?

Minimum Number of Operations to Convert Time uses the following concepts: String, Greedy. The recommended patterns to study are: Greedy, Dynamic Programming.

What are the interview tips for Minimum Number of Operations to Convert Time?

Always clarify the problem constraints and edge cases with the interviewer. Explain your thought process as you work through the problem. Consider both brute-force and optimal solutions to demonstrate your understanding.

What are common mistakes when solving Minimum Number of Operations to Convert Time?

Not considering the greedy approach and trying to brute-force all combinations. Failing to convert time correctly from 'HH:MM' format to minutes.

#2224

Minimum Number of Operations to Convert Time

Easy

StringGreedyGreedyDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(1)
Space	O(1)	O(1)

💡

Intuition

Time O(1)Space O(1)

The optimal solution focuses on minimizing the number of operations by always taking the largest possible step first. This greedy approach ensures that we reduce the difference to zero in the fewest moves.

⚙️

Algorithm

3 steps

1Step 1: Convert both current and correct times from 'HH:MM' format to total minutes since midnight.
2Step 2: Calculate the difference in minutes between correct and current.
3Step 3: Use a greedy approach to subtract the largest possible increments (60, 15, 5, 1) from the difference until it reaches zero, counting the number of operations.

solution.py18 lines

1# Full working Python code
2
3def minOperations(current: str, correct: str) -> int:
4    def to_minutes(time):
5        h, m = map(int, time.split(':'))
6        return h * 60 + m
7
8    current_minutes = to_minutes(current)
9    correct_minutes = to_minutes(correct)
10    diff = correct_minutes - current_minutes
11    operations = 0
12
13    for add in [60, 15, 5, 1]:
14        operations += diff // add
15        diff %= add
16
17    return operations
18

ℹ

Complexity note: The complexity is O(1) because we only perform a fixed number of operations (4 possible increments) regardless of the input size.

1Always prioritize larger increments to minimize the number of operations.
2Converting time to a single unit (minutes) simplifies the problem.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.