LeetCode 1024: Video Stitching Solution

Master LeetCode problem 1024 (Video Stitching), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1024. Video Stitching

Medium

Array Dynamic Programming Greedy

Problem Explanation

Explanation

To solve this problem, we can use a greedy approach. We need to find the minimum number of clips needed to cover the entire sporting event. We can start by sorting the clips based on their start time. Then, we iterate over the clips to find the clip that can cover the farthest end within the current coverage. If we cannot extend the coverage further, we need to select a new clip that can cover the current end.

Algorithm

Sort the clips array based on the start time.
Initialize end to 0 to store the current coverage end.
Initialize count to 0 to store the number of clips used.
Iterate over the sorted clips array:
- If the current clip's start time is greater than end, return -1 as it means there is a gap in coverage.
- Update end to the maximum of current end and current clip's end time.
- If end is greater than or equal to the required time, return count.
- Increment count by 1.
If coverage hasn't reached time, return -1 as it is not possible.

Time Complexity

The time complexity of this algorithm is O(nlogn) due to sorting the clips array, where n is the number of clips.

Space Complexity

The space complexity is O(1) as we are using only a constant amount of extra space.

Solution Code

class Solution {
    public int videoStitching(int[][] clips, int time) {
        Arrays.sort(clips, (a, b) -> a[0] - b[0]);
        int end = 0, count = 0;
        
        for (int i = 0; i < clips.length; ) {
            if (clips[i][0] > end) {
                return -1;
            }
            
            int newEnd = end;
            while (i < clips.length && clips[i][0] <= end) {
                newEnd = Math.max(newEnd, clips[i][1]);
                i++;
            }
            
            count++;
            end = newEnd;
            
            if (end >= time) {
                return count;
            }
        }
        
        return -1;
    }
}

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