2224. Minimum Number of Operations to Convert Time
Problem Description
The problem presents a scenario where you are given two strings, current
and correct
, which represent two 24-hour formatted times, using the format "HH:MM". The HH
part corresponds to the hour portion, ranging from 00 to 23, and MM
is the minutes portion, ranging from 00 to 59. The goal is to find the minimum number of operations required to change the current
time to the correct
time. An operation is defined as adding 1, 5, 15, or 60 minutes to the current
time, and you are allowed to perform as many operations as necessary to achieve the correct
time.
Intuition
To solve this problem, we need to minimize the number of operations required to convert current
to correct
. The intuition is to use the largest possible increment (60 minutes) as many times as we can without exceeding the correct
time, then proceed to the next largest (15 minutes), and so on down to the smallest increment (1 minute). This greedy approach ensures that we are always making the biggest leap towards the correct
time at each step, thereby minimizing the total number of steps.
We convert both current
and correct
times to minutes since midnight, which makes the calculation easier since we are now dealing with integers. This is done by multiplying the hours by 60 and adding the minutes. Then, we calculate the difference d
between correct
and current
.
Next, we use a loop to iterate over the list of increments [60, 15, 5, 1]
. In each iteration, we add to ans
the integer division of d
by the current increment, which is the maximum number of times we can perform the current operation. We then update d
to the remainder of this division to process the remaining time with the next smaller increment. This continues until we have used all permissible increments, at which point d
should be 0, and ans
should contain the minimum number of operations.
Learn more about Greedy patterns.
Solution Approach
The implementation of the solution closely mirrors the intuition explained earlier. The algorithm is straightforward and can be broken down into the following steps:
-
Convert both
current
andcorrect
into minutes: This is the first operation in the solution which involves parsing the hour and minute components of both time strings separately, converting them to integers, multiplying the hours by 60 to get the total minutes for the hours part, and finally adding the minutes to this result. In Python, this looks likeint(current[:2]) * 60 + int(current[3:])
, where[:2]
takes the first two characters of the string (hours) and[3:]
takes the characters from the fourth to the end (minutes). -
Calculate the difference: We then subtract the value of
current
in minutes from the value ofcorrect
in minutes to get the total number of minutes that need to be added tocurrent
to reachcorrect
. This difference is represented byd
. -
Perform operations to minimize the time difference: Using a for loop, the solution iterates over a list
[60, 15, 5, 1]
which contains the allowed minute increments for the operations, in descending order.- Within the loop, the solution uses integer division
d // i
to find out how many times a particular increment can be used. It adds this number to theans
variable, which accumulates the total number of operations needed. - Then,
d
is updated to the remainder of this division withd %= i
. This remainder represents the minutes that are yet to be matched and is efficiently reduced with each larger increment before moving to smaller ones.
- Within the loop, the solution uses integer division
The choice of data structures in this solution is minimal. An integer ans
is used to store the cumulative count of operations, and a simple list of increments is iterated upon.
This approach is a common algorithmic pattern called the greedy algorithm, which tries to solve a problem by making the best possible decision at the current step without looking ahead to the future steps. It's called "greedy" because the decision is based on what seems to be the best immediate choice.
Lastly, the algorithm executes in O(1) time complexity since the operations are bound by a fixed number of possible increment values and the loop runs at most 4 times (once for each increment value: 60, 15, 5, and 1).
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Start EvaluatorExample Walkthrough
Let's walk through a simple example to illustrate the solution approach:
Suppose we have current
time as "02:30"
and correct
time as "03:10"
.
-
Convert both times to minutes:
- The
current
time is converted to 150 minutes (2*60 + 30
). - The
correct
time is converted to 190 minutes (3*60 + 10
).
- The
-
Calculate the difference:
- The difference
d
betweencorrect
andcurrent
is 40 minutes (190 - 150
).
- The difference
-
Perform operations:
- We start by iterating over the increments
[60, 15, 5, 1]
. - We cannot use the 60-minute operation since
d
is less than 60. - For the 15-minute operation, we get
40 // 15 = 2
, so we can add 15 minutes two times.ans
becomes2
, andd
becomes10
(the remainder of dividing 40 by 15). - The 5-minute operation can be used twice on the remaining 10 minutes. Now,
ans
is2 + 2 = 4
, andd
is0
. - Since
d
is now 0, we do not need to perform any 1-minute operations.
- We start by iterating over the increments
In this example, the minimum number of operations required to change the current
time to the correct
time is 4. This includes performing the 15-minute increment twice and the 5-minute increment twice.
Solution Implementation
1class Solution:
2 def convertTime(self, current: str, correct: str) -> int:
3 # Convert the 'current' time to minutes
4 current_minutes = int(current[:2]) * 60 + int(current[3:])
5
6 # Convert the 'correct' time to minutes
7 correct_minutes = int(correct[:2]) * 60 + int(correct[3:])
8
9 # Calculate the difference in minutes
10 delta_minutes = correct_minutes - current_minutes
11
12 # Initialize the number of operations used
13 operations_used = 0
14
15 # List of available operation increments in minutes
16 operation_increments = [60, 15, 5, 1]
17
18 # For each increment option, calculate the maximum number of operations possible
19 # And then, reduce the remaining delta_minutes accordingly
20 for increment in operation_increments:
21 operations_used += delta_minutes // increment # Perform the largest operations possible
22 delta_minutes %= increment # Reduce delta_minutes by the amount operated on
23
24 # Return the total number of operations used to adjust the time
25 return operations_used
26
1class Solution {
2 public int convertTime(String current, String correct) {
3 // Convert the 'current' time into minutes
4 int currentMinutes = Integer.parseInt(current.substring(0, 2)) * 60
5 + Integer.parseInt(current.substring(3));
6
7 // Convert the 'correct' time into minutes
8 int correctMinutes = Integer.parseInt(correct.substring(0, 2)) * 60
9 + Integer.parseInt(correct.substring(3));
10
11 // Initialize answer to store the number of operations required
12 int operations = 0;
13
14 // Calculate the difference in minutes
15 int difference = correctMinutes - currentMinutes;
16
17 // Array containing possible operations in descending order by their time value
18 Integer[] operationsArray = new Integer[] {60, 15, 5, 1};
19
20 // Iterate over the operations array to find the minimum number
21 // of operations needed to reach the correct time
22 for (int minutesPerOperation : operationsArray) {
23 // Divide the time difference by the time value of operation
24 operations += difference / minutesPerOperation;
25
26 // Update the difference to reflect the remaining time after performing this operation
27 difference %= minutesPerOperation;
28 }
29
30 // Return the minimum number of operations required
31 return operations;
32 }
33}
34
1#include <string>
2#include <vector>
3
4class Solution {
5public:
6 // Function to convert time from 'current' to 'correct' using minimum operations
7 int convertTime(std::string current, std::string correct) {
8 // Convert 'current' time from hours:minutes format to minutes
9 int currentMinutes = std::stoi(current.substr(0, 2)) * 60 + std::stoi(current.substr(3, 2));
10 // Convert 'correct' time from hours:minutes format to minutes
11 int correctMinutes = std::stoi(correct.substr(0, 2)) * 60 + std::stoi(correct.substr(3, 2));
12
13 // Calculate the difference in minutes
14 int difference = correctMinutes - currentMinutes;
15 // Variable to store the number of operations needed
16 int operationsCount = 0;
17 // Define increments in minutes that can be used to adjust the time
18 std::vector<int> increments = {60, 15, 5, 1};
19
20 // Iterate over possible increments
21 for (int increment : increments) {
22 // Use the largest increment to reduce the difference as much as possible
23 operationsCount += difference / increment;
24 // Update the remaining difference
25 difference %= increment;
26 }
27 // Return the total number of operations required
28 return operationsCount;
29 }
30};
31
1// Import statements are not typically used in TypeScript as they are in C++
2// Instead, TypeScript uses modules and export/import syntax
3
4// Function to convert time from 'current' to 'correct' using minimum operations
5function convertTime(current: string, correct: string): number {
6 // Convert 'current' time from hours:minutes format to minutes
7 const currentMinutes: number = parseInt(current.substring(0, 2)) * 60 + parseInt(current.substring(3));
8
9 // Convert 'correct' time from hours:minutes format to minutes
10 const correctMinutes: number = parseInt(correct.substring(0, 2)) * 60 + parseInt(correct.substring(3));
11
12 // Calculate the difference in minutes
13 let difference: number = correctMinutes - currentMinutes;
14
15 // Variable to store the number of operations needed
16 let operationsCount: number = 0;
17
18 // Define increments in minutes that can be used to adjust the time
19 const increments: number[] = [60, 15, 5, 1];
20
21 // Iterate over possible increments
22 for (const increment of increments) {
23 // Use the largest increment to reduce the difference as much as possible
24 operationsCount += Math.floor(difference / increment);
25
26 // Update the remaining difference
27 difference %= increment;
28 }
29
30 // Return the total number of operations required
31 return operationsCount;
32}
33
34// The code as per TypeScript conventions does not define global variables or methods.
35// Instead, they are typically encapsulated in modules or classes.
36// However, as per the request, this function is defined globally.
37
Time and Space Complexity
Time Complexity
The time complexity of this function is O(1)
. This is because the function performs a constant number of operations irrespective of the size of the input. The calculations performed to convert the times from string format to minutes, the loop to calculate the number of operations needed to reach the correct time, and the arithmetic operations inside the loop all execute in constant time. There are no iterative statements that depend on the input size, and the loop runs a maximum of four times (once for each value in the [60, 15, 5, 1]
list).
Space Complexity
The space complexity of this function is also O(1)
. The function only uses a fixed amount of additional memory: the variables a
, b
, ans
, and d
hold single integer values, and the list [60, 15, 5, 1]
has a fixed size. Therefore, the amount of memory used does not scale with the input size, guaranteeing constant space complexity.
Learn more about how to find time and space complexity quickly using problem constraints.
What's the output of running the following function using the following tree as input?
1def serialize(root):
2 res = []
3 def dfs(root):
4 if not root:
5 res.append('x')
6 return
7 res.append(root.val)
8 dfs(root.left)
9 dfs(root.right)
10 dfs(root)
11 return ' '.join(res)
12
1import java.util.StringJoiner;
2
3public static String serialize(Node root) {
4 StringJoiner res = new StringJoiner(" ");
5 serializeDFS(root, res);
6 return res.toString();
7}
8
9private static void serializeDFS(Node root, StringJoiner result) {
10 if (root == null) {
11 result.add("x");
12 return;
13 }
14 result.add(Integer.toString(root.val));
15 serializeDFS(root.left, result);
16 serializeDFS(root.right, result);
17}
18
1function serialize(root) {
2 let res = [];
3 serialize_dfs(root, res);
4 return res.join(" ");
5}
6
7function serialize_dfs(root, res) {
8 if (!root) {
9 res.push("x");
10 return;
11 }
12 res.push(root.val);
13 serialize_dfs(root.left, res);
14 serialize_dfs(root.right, res);
15}
16
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