Leetcode 1024: Video Stitching

Input: The input consists of an array of clips, where each clip is represented by a pair of integers [start, end] indicating the start and end times of the clip. Additionally, a single integer, time, specifies the duration of the event to be covered.

Example: clips = [[0, 3], [4, 7], [2, 5], [6, 8], [5, 9]], time = 9

Constraints:

• 1 <= clips.length <= 100

• 0 <= starti <= endi <= 100

• 1 <= time <= 100

Output: The output should be a single integer representing the minimum number of clips needed to cover the entire event. If it's impossible to cover the event, return -1.

Example: Output: 3

Constraints:

• If the clips cannot cover the entire event, return -1.

Goal: The task is to determine the smallest number of clips that can be used to cover the entire duration of the sporting event. This is a greedy problem where we need to select the clips that maximize the coverage of the event, ensuring we cover the entire time range from 0 to time.

Steps:

• 1. Sort the clips based on their start times.

• 2. Use a greedy approach to select clips, always choosing the one that extends the coverage the most without skipping any part of the event.

• 3. If at any point the selected clips cannot extend the coverage, return -1.

• 4. Otherwise, return the number of clips selected.

Goal: Ensure the solution works for the given input size constraints.

Steps:

• The solution should handle up to 100 clips, with start and end times between 0 and 100 seconds.

• The event duration is between 1 and 100 seconds.

Assumptions:

• The clips array will be non-empty and the event duration will be positive.

• Input: Input: clips = [[0, 3], [4, 7], [2, 5], [6, 8], [5, 9]], time = 9

• Explanation: In this case, we can take the clips [0, 3], [2, 5], and [5, 9], which cover the entire event from 0 to 9. So, the output is 3.

• Input: Input: clips = [[0, 1], [1, 2]], time = 5

• Explanation: In this case, it is impossible to cover the event duration from 0 to 5 using just the clips [0, 1] and [1, 2], so the output is -1.

Approach: The approach uses a greedy strategy where we attempt to cover the event by iteratively selecting the clip that extends the current coverage the furthest.

Observations:

• Sorting the clips by start time allows us to systematically check which clips to select for covering the event.

• By always picking the clip that extends the coverage the most, we can minimize the number of clips needed.

Steps:

• 1. Sort the clips by their start times.

• 2. Initialize variables to track the current end time of the covered event and the number of clips selected.

• 3. For each clip, check if it extends the current coverage. If it does, update the coverage and increment the count of clips.

• 4. If at any point we cannot extend the coverage, return -1.

• 5. Return the number of clips selected once the entire event is covered.

Empty Inputs:

• If there are no clips, return -1 unless the event duration is 0.

Large Inputs:

• The solution must efficiently handle inputs with up to 100 clips.

Special Values:

• If a clip exactly matches the time, it should be counted as one clip.

Constraints:

• Ensure that the solution handles cases where no set of clips can cover the entire event.

int videoStitching(vector<vector<int>>& clips, int time) {
    sort(clips.begin(), clips.end());
    int res = 0;
    for(auto i = 0, st = 0, end = 0; st < time; st=end, ++res){
        for(; i < clips.size() && clips[i][0] <= st; ++i)
            end = max(end, clips[i][1]);
        if(st == end) return -1;
    }
    return res;
}

1 : Function Definition

int videoStitching(vector<vector<int>>& clips, int time) {

Define the function `videoStitching`, which takes a list of clips and a target time interval to cover.

2 : Sorting

    sort(clips.begin(), clips.end());

Sort the clips by their start times to make it easier to find overlapping intervals and select the optimal clips.

3 : Variable Initialization

    int res = 0;

Initialize the result variable `res` to 0, which will store the number of clips used to cover the time interval.

4 : Looping

    for(auto i = 0, st = 0, end = 0; st < time; st=end, ++res){

Start a loop where `i` is the index for iterating through the clips, `st` is the start of the current covered interval, and `end` is the end of the last selected clip.

5 : Looping

        for(; i < clips.size() && clips[i][0] <= st; ++i)

Loop through the clips and find the clip whose start time is less than or equal to the current start time (`st`).

6 : Max Calculation

            end = max(end, clips[i][1]);

Update the `end` time by taking the maximum of the current `end` and the end time of the current clip, which extends the covered interval.

7 : Condition Check

        if(st == end) return -1;

If the current start time `st` is the same as the end time `end`, it means no clip can extend the coverage. Return -1 to indicate it's impossible to cover the entire time interval.

8 : Return Statement

    return res;

Return the number of clips used to cover the time interval.

Best Case: O(n log n)

Average Case: O(n log n)

Worst Case: O(n log n)

Description: The time complexity is dominated by the sorting step, which takes O(n log n), where n is the number of clips.

Best Case: O(1)

Worst Case: O(n)

Description: The space complexity is O(n) due to the space required for sorting the clips.

Leetcode 1024: Video Stitching

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