What is the time complexity of Minimum Time Difference?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Minimum Time Difference?

This problem uses Array, Math, String, Sorting. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Minimum Time Difference?

Insight 1: The core idea is to convert all the time points to minutes from the beginning of the day (00:00) to simplify the comparison. Insight 2: Sorting the time points (in minutes) is not strictly required for O(n) solution but converting to minutes array can make the code simpler to implement. Instead using the approach of checking for the presence of minute in 1440 element boolean array to avoid sorting complexity

Which companies ask LeetCode 539. Minimum Time Difference in interviews?

This problem is frequently asked at Palantir Technologies, Visa, Zoho, carwale.

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Minimum Time Difference - LeetCode 539 Solution

Q: How to solve Minimum Time Difference?

The provided solution first checks if there are more than 1440 time points. If so, it returns 0 because there must be at least one duplicate. Otherwise, it creates a boolean array of size 1440, representing each minute of the day. It iterates through the time points, converts each to minutes, and marks the corresponding index in the boolean array as true. It also checks for duplicate time entries while marking. Then, it iterates through the boolean array to find the first and last occupied minutes, as well as the minimum difference between adjacent occupied minutes. Finally, it calculates the difference between the last and first minutes (wrap-around case) and returns the overall minimum difference.

Minimum Time Difference - Complete Solution Guide

Minimum Time Difference is LeetCode problem 539, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Given a list of 24-hour clock time points in "HH:MM" format, return the minimum minutes difference between any two time-points in the list . Example 1: Input: timePoints = ["23:59","00:00"] Output: 1 Example 2: Input: timePoints = ["00:00","23:59","00:00"] Output: 0 Constraints: 2 <= timePoints.length <= 2 * 10 4 timePoints[i] is in the format "HH:MM" .

Detailed Explanation

The problem asks us to find the minimum time difference between any two time points given in a list of 24-hour clock time points in "HH:MM" format. The output should be the minimum difference in minutes.

Solution Approach

The provided solution first checks if there are more than 1440 time points. If so, it returns 0 because there must be at least one duplicate. Otherwise, it creates a boolean array of size 1440, representing each minute of the day. It iterates through the time points, converts each to minutes, and marks the corresponding index in the boolean array as true. It also checks for duplicate time entries while marking. Then, it iterates through the boolean array to find the first and last occupied minutes, as well as the minimum difference between adjacent occupied minutes. Finally, it calculates the difference between the last and first minutes (wrap-around case) and returns the overall minimum difference.

Step-by-Step Algorithm

Step 1: Check if the number of time points exceeds 1440. If so, return 0.
Step 2: Create a boolean array `minutes_present` of size 1440, initialized to false.
Step 3: Iterate through the time points. For each time point:
Step 4: Convert the time point from "HH:MM" format to total minutes (hours * 60 + minutes).
Step 5: If `minutes_present[total_minutes]` is already true, return 0 (duplicate time point found).
Step 6: Set `minutes_present[total_minutes]` to true.
Step 7: Initialize `min_diff` to infinity, `first_minute` to -1, and `prev_minute` to -1.
Step 8: Iterate through the `minutes_present` array from index 0 to 1439:
Step 9: If `minutes_present[m]` is true:
Step 10: If `first_minute` is -1, set `first_minute` to `m`.
Step 11: If `prev_minute` is not -1, update `min_diff` to the minimum of `min_diff` and `m - prev_minute`.
Step 12: Set `prev_minute` to `m`.
Step 13: Calculate the wrap-around difference: `wrap_around_diff = (first_minute + 1440) - prev_minute`.
Step 14: Update `min_diff` to the minimum of `min_diff` and `wrap_around_diff`.
Step 15: Return `min_diff`.

Key Insights

Insight 1: The core idea is to convert all the time points to minutes from the beginning of the day (00:00) to simplify the comparison.
Insight 2: Sorting the time points (in minutes) is not strictly required for O(n) solution but converting to minutes array can make the code simpler to implement. Instead using the approach of checking for the presence of minute in 1440 element boolean array to avoid sorting complexity
Insight 3: Consider the wrap-around case, where the minimum difference may be between the largest and smallest time points (e.g., 23:59 and 00:00).
Insight 4: If the number of time points exceeds 1440 (the number of minutes in a day), there must be duplicate time points, resulting in a minimum difference of 0.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, Math, String, Sorting.

Companies

Asked at: Palantir Technologies, Visa, Zoho, carwale.

Minimum Time Difference | LeetCode 539

Medium

Array Math String Sorting

Time: O(n)Space: O(1)

Solution Code

Available in:

import java.util.List;

class Solution {
    public int findMinDifference(List<String> timePoints) {
        if (timePoints.size() > 1440) {
            return 0;
        }

        boolean[] minutes_present = new boolean[1440];
        for (String time_point : timePoints) {
            int hours = Integer.parseInt(time_point.substring(0, 2));
            int minutes = Integer.parseInt(time_point.substring(3));
            int total_minutes = hours * 60 + minutes;
            if (minutes_present[total_minutes]) {
                return 0;
            }
            minutes_present[total_minutes] = true;
        }

        int min_diff = Integer.MAX_VALUE;
        int first_minute = -1;
        int prev_minute = -1;

        for (int m = 0; m < 1440; m++) {
            if (minutes_present[m]) {
                if (first_minute == -1) {
                    first_minute = m;
                }
                if (prev_minute != -1) {
                    min_diff = Math.min(min_diff, m - prev_minute);
                }
                prev_minute = m;
            }
        }

        int wrap_around_diff = (first_minute + 1440) - prev_minute;
        min_diff = Math.min(min_diff, wrap_around_diff);

        return min_diff;
    }
}

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Problem Info

Description

Problem 539: Minimum Time Difference - Detailed explanation and solution approaches.

Topics

Array Math String Sorting

Companies

Palantir TechnologiesVisaZohocarwale

Stats

Difficulty:Medium

Languages:4

Companies:4

Problem #:539

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

15 steps

1
Step 1: Check if the number of time points exceeds 1440. If so, return 0.
2
Step 2: Create a boolean array `minutes_present` of size 1440, initialized to false.
3
Step 3: Iterate through the time points. For each time point:
4
Step 4: Convert the time point from "HH:MM" format to total minutes (hours * 60 + minutes).
5
Step 5: If `minutes_present[total_minutes]` is already true, return 0 (duplicate time point found).
6
Step 6: Set `minutes_present[total_minutes]` to true.
7
Step 7: Initialize `min_diff` to infinity, `first_minute` to -1, and `prev_minute` to -1.
8
Step 8: Iterate through the `minutes_present` array from index 0 to 1439:
9
Step 9: If `minutes_present[m]` is true:
10
Step 10: If `first_minute` is -1, set `first_minute` to `m`.
11
Step 11: If `prev_minute` is not -1, update `min_diff` to the minimum of `min_diff` and `m - prev_minute`.
12
Step 12: Set `prev_minute` to `m`.
13
Step 13: Calculate the wrap-around difference: `wrap_around_diff = (first_minute + 1440) - prev_minute`.
14
Step 14: Update `min_diff` to the minimum of `min_diff` and `wrap_around_diff`.
15
Step 15: Return `min_diff`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is determined by the loop that iterates through the timePoints to convert them into minutes. The boolean array takes constant space.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Sorting Approach

Convert the time points to minutes, sort the minutes array, and then iterate through the sorted array to find the minimum difference between adjacent elements. Also consider the difference between the last and first elements to handle the wrap-around case.

Time: O(n log n) due to the sorting step.Space: O(n) in many sorting algorithm implementations (e.g., merge sort), as we're creating a copy of the list of times in minutes for sorting.

Trade-offs: This approach is more intuitive for some but has a higher time complexity than the boolean array approach. Useful if n is small, or if a sorting algorithm is already available.

✨

Key Insights

1Insight 1: The core idea is to convert all the time points to minutes from the beginning of the day (00:00) to simplify the comparison.
2Insight 2: Sorting the time points (in minutes) is not strictly required for O(n) solution but converting to minutes array can make the code simpler to implement. Instead using the approach of checking for the presence of minute in 1440 element boolean array to avoid sorting complexity
3Insight 3: Consider the wrap-around case, where the minimum difference may be between the largest and smallest time points (e.g., 23:59 and 00:00).
4Insight 4: If the number of time points exceeds 1440 (the number of minutes in a day), there must be duplicate time points, resulting in a minimum difference of 0.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

String

A sequence of characters. Like a word or sentence in programming!

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: Duplicate time points. The solution correctly handles this by returning 0 if any two time points are identical.
Edge case 2: Wrap-around case (e.g., 23:59 and 00:00). The solution explicitly calculates the difference between the last and first time points and considers it in the minimum difference calculation.
Edge case 3: Number of time points exceeding 1440. If there are more than 1440 time points, there must be duplicate entries and the minimum difference is 0. The code handles this at the beginning.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the wrap-around case correctly. For example, missing the comparison between '23:59' and '00:00'.

✗Common mistake 2: Using string comparison directly instead of converting to minutes. String comparison would not accurately reflect the time differences.

✗Common mistake 3: Forgetting the case where the time points are the same (return 0 if duplicates exist).

✗Common mistake 4: Assuming timePoints are already sorted and not accounting for the need to sort or process the data.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Scheduling problems. Finding the minimum time gap between scheduled appointments or tasks.

🏦Real-world scenario 2: Network latency analysis. Determining the smallest difference in time between network packets arriving at a destination.

🎮Real-world scenario 3: Meeting scheduling. Finding the smallest time difference between meeting requests to determine the most efficient scheduling.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: November 27, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 13

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 6

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Minimum Time Difference"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether math techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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