What is the time complexity of Video Stitching?

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

What data structures are used to solve Video Stitching?

This problem uses Array, Dynamic Programming, Greedy. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Video Stitching?

Insight 1: We need to sort the clips based on their starting times to efficiently find the optimal clip to extend our coverage. Insight 2: A greedy approach works here. At each point, we select the clip that extends our coverage the furthest.

Which companies ask LeetCode 1024. Video Stitching in interviews?

This problem is frequently asked at Anduril, Verily.

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Video Stitching - LeetCode 1024 Solution

Video Stitching - Complete Solution Guide

Video Stitching is LeetCode problem 1024, 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

You are given a series of video clips from a sporting event that lasted time seconds. These video clips can be overlapping with each other and have varying lengths. Each video clip is described by an array clips where clips[i] = [start i , end i ] indicates that the ith clip started at start i and ended at end i . We can cut these clips into segments freely. For example, a clip [0, 7] can be cut into segments [0, 1] + [1, 3] + [3, 7] . Return the minimum number of clips needed so that we can cut

Detailed Explanation

The problem asks us to find the minimum number of video clips needed to cover a given time interval [0, time]. We are provided with a list of video clips, each represented as a [start, end] interval. The clips can overlap, and we can cut them into smaller segments. If it's impossible to cover the entire interval [0, time], we should return -1.

Solution Approach

The solution uses a greedy approach. First, the clips are sorted based on their start times. Then, we iterate through the sorted clips, maintaining the current end time of our coverage. At each step, we select the clip that starts before or at our current end time and has the maximum end time. This ensures that we extend our coverage as much as possible with the minimum number of clips. If we ever reach a point where we can't extend our coverage, it means the entire interval cannot be covered, and we return -1.

Step-by-Step Algorithm

Step 1: Sort the `clips` array based on the start time of each clip.
Step 2: Initialize `count` to 0, `current_end` to 0, and `clip_idx` to 0. `count` tracks the number of clips used, `current_end` tracks the current maximum coverage achieved, and `clip_idx` iterates through the sorted clips.
Step 3: While `current_end` is less than `time`, continue the loop.
Step 4: Inside the loop, initialize `max_reach` to `current_end`. This variable will store the maximum end time reachable from `current_end` using available clips.
Step 5: Iterate through the `clips` array (starting from `clip_idx`) as long as the start time of the current clip is less than or equal to `current_end`. In each iteration, update `max_reach` to be the maximum of its current value and the end time of the current clip.
Step 6: After the inner loop, if `max_reach` is still equal to `current_end`, it means no clip could extend the coverage, so return -1.
Step 7: Increment the `count` and update `current_end` to `max_reach`. Move the `clip_idx` to the next set of possible clips.
Step 8: If the loop completes without returning -1, return the `count`.

Key Insights

Insight 1: We need to sort the clips based on their starting times to efficiently find the optimal clip to extend our coverage.
Insight 2: A greedy approach works here. At each point, we select the clip that extends our coverage the furthest.
Insight 3: If we can't extend our coverage from a certain point, it means that there's a gap and we cannot cover the entire interval, so we should return -1.

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(1)

Topics

This problem involves: Array, Dynamic Programming, Greedy.

Companies

Asked at: Anduril, Verily.

Video Stitching | LeetCode 1024

Medium

Array Dynamic Programming Greedy

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

Solution Code

Available in:

import java.util.Arrays;
import java.util.Comparator;

class Solution {
    public int videoStitching(int[][] clips, int time) {
        Arrays.sort(clips, Comparator.comparingInt(a -> a[0]));
        
        int count = 0;
        int current_end = 0;
        int clip_idx = 0;
        int n = clips.length;
        
        while (current_end < time) {
            int max_reach = current_end;
            
            while (clip_idx < n && clips[clip_idx][0] <= current_end) {
                max_reach = Math.max(max_reach, clips[clip_idx][1]);
                clip_idx++;
            }
            
            if (max_reach == current_end) {
                return -1;
            }
            
            count++;
            current_end = max_reach;
            
        }
        
        return count;
    }
}

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Problem Info

Description

Problem 1024: Video Stitching - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming Greedy

Companies

AndurilVerily

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:1024

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

8 steps

1
Step 1: Sort the `clips` array based on the start time of each clip.
2
Step 2: Initialize `count` to 0, `current_end` to 0, and `clip_idx` to 0. `count` tracks the number of clips used, `current_end` tracks the current maximum coverage achieved, and `clip_idx` iterates through the sorted clips.
3
Step 3: While `current_end` is less than `time`, continue the loop.
4
Step 4: Inside the loop, initialize `max_reach` to `current_end`. This variable will store the maximum end time reachable from `current_end` using available clips.
5
Step 5: Iterate through the `clips` array (starting from `clip_idx`) as long as the start time of the current clip is less than or equal to `current_end`. In each iteration, update `max_reach` to be the maximum of its current value and the end time of the current clip.
6
Step 6: After the inner loop, if `max_reach` is still equal to `current_end`, it means no clip could extend the coverage, so return -1.
7
Step 7: Increment the `count` and update `current_end` to `max_reach`. Move the `clip_idx` to the next set of possible clips.
8
Step 8: If the loop completes without returning -1, return the `count`.

💡

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 log n)

✨ Log-linear - Great for sorting algorithms!

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is determined by sorting the input array which requires O(n log n) time. The while loops iterate through the clips at most once, taking O(n) time. The space complexity is O(1) as we are only using a few variables to store the count, current end, and clip index, irrespective of the input size.

📖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

Dynamic Programming

We can use dynamic programming to solve this problem. dp[i] stores the minimum number of clips needed to cover the interval [0, i]. We iterate through all possible intervals [start, end] and update dp[end] based on dp[start].

Time: O(time * n), where n is the number of clips and time is the target time.Space: O(time), where time is the target time.

Trade-offs: The dynamic programming approach has a higher time complexity O(time * n) than the greedy approach O(n log n). It also has a space complexity of O(time), while the greedy approach uses constant space. The dynamic programming approach could be preferred if there are overlapping subproblems which the current greedy algorithm is not utilizing and it could potentially lead to a more optimized runtime for a specific set of edge cases with extreme overlap.

✨

Key Insights

1Insight 1: We need to sort the clips based on their starting times to efficiently find the optimal clip to extend our coverage.
2Insight 2: A greedy approach works here. At each point, we select the clip that extends our coverage the furthest.
3Insight 3: If we can't extend our coverage from a certain point, it means that there's a gap and we cannot cover the entire interval, so we should return -1.

📚

Key Terms Explained

For Students

Array

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

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Greedy Algorithm

Making the best choice at each step without looking ahead.

Edge Cases

Edge case 1: `clips` is empty. The solution correctly handles this because the while loop condition `current_end < time` will not be satisfied at the beginning, so it will return -1 if `time` is greater than 0 and 0 otherwise.
Edge case 2: No clip starts at 0. After sorting, if the first clip's start time is greater than 0, the `max_reach` will remain equal to `current_end` (which is 0), causing the solution to correctly return -1.
Edge case 3: Clips do not fully cover the range [0, time]. The inner while loop searches for the best clip. If no clip extends the range from the current `current_end`, `max_reach` remains the same. The check `if (max_reach == current_end)` catches this scenario and returns -1.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not sorting the clips before applying the greedy algorithm. Sorting is crucial for the greedy approach to work correctly.

✗Common mistake 2: Failing to handle the case where the entire interval cannot be covered. Forgetting to check if `max_reach` is equal to `current_end` inside the loop leads to incorrect results when the clips have gaps.

✗Common mistake 3: Assuming that the clips are non-overlapping. The clips can overlap, and the algorithm must handle this correctly to select the optimal clips to minimize the number of clips required.

✅

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: Video editing software. The algorithm can be used to find the minimum number of video segments needed to create a trailer or highlight reel from a longer video.

🏦Real-world scenario 2: Resource allocation. Suppose we want to allocate resources over a period. Each resource has a start and end time for its availability. The algorithm can be used to find the minimum number of resources required to cover the entire period.

💡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

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ 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: October 3, 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

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I identify if a problem needs dynamic programming?

DP problems typically have: 1) Overlapping subproblems (same calculation repeated), 2) Optimal substructure (optimal solution uses optimal solutions to subproblems). Look for keywords like "minimum/maximum," "count ways," "is it possible," or "longest/shortest."

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Video Stitching"?

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 dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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