Problem Description

You have a collection of video clips from a sporting event that lasted time seconds. Each clip is represented as [start, end], where start is when the clip begins and end is when it ends. These clips can overlap with each other and have different lengths.

Your task is to select the minimum number of clips needed to cover the entire event from second 0 to second time. You can use any portion of a clip - for example, a clip [0, 7] can be cut and used as segments like [0, 1], [1, 3], and [3, 7].

Given:

• An array clips where clips[i] = [start_i, end_i] represents the i-th clip
• An integer time representing the duration of the event

Return:

• The minimum number of clips needed to cover [0, time]
• Return -1 if it's impossible to cover the entire duration

For example, if you have clips [[0,2], [1,5], [3,7]] and time = 6, you could use clips [0,2] and [1,5] to cover [0,5], then use [3,7] to cover the remaining time, requiring 3 clips total. However, a better solution would be to use [0,2] and [1,5] (which covers up to 5) and [3,7] (which covers from 3 to 7), but since [1,5] already reaches position 5, we only need 2 clips: [0,2] extended by [1,5] to cover [0,5], and then [3,7] to complete the coverage.

The key insight is that you want to greedily select clips that extend your coverage as far as possible while ensuring no gaps exist in the coverage from 0 to time.

Intuition

Think of this problem like trying to build a bridge across a river using planks of different lengths. You want to use the minimum number of planks to get from one side (position 0) to the other (position time).

The key observation is that at each position, we want to choose the clip that can take us the farthest. If multiple clips start at the same position, we should always pick the one that extends the furthest - there's no benefit to choosing a shorter clip when a longer one is available from the same starting point.

This leads us to a preprocessing step: for each position i, we can calculate the farthest position reachable from clips starting at i. We store this in an array last[i].

Now imagine walking along the timeline from position 0. At each step, we keep track of:

• mx: the farthest position we can reach using all clips we've seen so far
• pre: the right boundary of the last clip we actually selected
• ans: the count of clips we've used

As we walk forward, we continuously update mx with the farthest reachable position. When we reach the boundary of our current clip (when i == pre), we must select a new clip. The best choice is to jump to the farthest position we can reach (mx), incrementing our clip count.

If at any point mx <= i, it means we can't move forward anymore - there's a gap in coverage that makes it impossible to reach time.

This greedy strategy works because once we've decided to use a clip to cover up to position pre, the optimal choice for the next clip is always the one that extends the furthest from any position we've already covered. There's never a benefit to choosing a shorter extension when a longer one is available.

The approach is similar to the "Jump Game" problem where you want to reach the end with minimum jumps - at each forced jump point, you jump to the farthest position possible based on all the positions you've explored.

Learn more about Greedy and Dynamic Programming patterns.

Solution Approach

The implementation follows a greedy approach with preprocessing to optimize clip selection.

Step 1: Preprocessing - Build the last array

We create an array last of size time where last[i] stores the farthest position reachable from any clip starting at position i.

last = [0] * time
for a, b in clips:
    if a < time:
        last[a] = max(last[a], b)

For each clip [a, b], if the start position a is within our range [0, time), we update last[a] to be the maximum of its current value and b. This ensures that for clips with the same starting point, we only keep track of the one with the farthest endpoint.

Step 2: Greedy Traversal

We initialize three variables:

• mx = 0: tracks the farthest position reachable from all positions we've seen
• pre = 0: marks the right boundary of the last selected clip
• ans = 0: counts the number of clips used

ans = mx = pre = 0
for i, v in enumerate(last):
    mx = max(mx, v)
    if mx <= i:
        return -1
    if pre == i:
        ans += 1
        pre = mx

We iterate through each position i from 0 to time-1:

1. Update maximum reach: mx = max(mx, last[i]) - we update the farthest reachable position considering the clip starting at position i.

2. Check for gaps: If mx <= i, it means we cannot go beyond the current position i. This creates a gap in coverage, making it impossible to reach time, so we return -1.

3. Select new clip: When pre == i, we've reached the boundary of our currently selected clip and must choose a new one. We:

   • Increment ans to count this new clip
   • Update pre = mx to jump to the farthest position we can reach

The algorithm essentially creates "jump points" where we must select a new clip, and at each jump point, we greedily choose to extend as far as possible.

Time Complexity: O(n + m) where n is the number of clips and m is the value of time

• O(n) for preprocessing the clips
• O(m) for the traversal

Space Complexity: O(m) for the last array

This approach guarantees the minimum number of clips because at each forced selection point, we're making the optimal choice by extending as far as possible, eliminating the need to backtrack or consider alternative paths.

Example Walkthrough

Let's walk through the algorithm with clips = [[0,2], [1,6], [3,5], [5,7]] and time = 6.

Step 1: Build the last array

We create last = [0, 0, 0, 0, 0, 0] (size 6 for positions 0-5).

Processing each clip:

• [0,2]: Set last[0] = 2 (from position 0, we can reach position 2)
• [1,6]: Set last[1] = 6 (from position 1, we can reach position 6)
• [3,5]: Set last[3] = 5 (from position 3, we can reach position 5)
• [5,7]: Set last[5] = 7 (from position 5, we can reach position 7)

Result: last = [2, 6, 0, 5, 0, 7]

Step 2: Greedy Traversal

Initialize: ans = 0, mx = 0, pre = 0

After the loop, pre = 6 ≥ time = 6, so we've successfully covered the entire duration.

Result: Return ans = 2

The algorithm selected 2 clips:

1. First selection at position 0: Used a clip to cover [0,2]
2. Second selection at position 2: Jumped to position 6 (using the clip starting at position 1)

This gives us coverage from 0 to 6 with just 2 clips, which is optimal.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n + T) where n is the number of clips and T is the time value.

• The first loop iterates through all n clips to populate the last array, taking O(n) time
• The second loop iterates through the last array of size time, taking O(T) time
• Overall time complexity is O(n + T)

Space Complexity: O(T) where T is the time value.

• The algorithm creates a last array of size time to store the maximum reachable position from each starting point
• Only constant extra variables (ans, mx, pre) are used besides the last array
• Overall space complexity is O(T)

Learn more about how to find time and space complexity quickly.

Common Pitfalls

Pitfall 1: Not Handling Clips That Start Beyond the Target Time

The Problem: The current implementation only checks if start < time when building the furthest_reach array, but this creates an issue when we have clips that start at exactly time or beyond. While these clips shouldn't be included in our solution, the array initialization with size time means we're not tracking the furthest reach from position time-1 correctly if there's a clip starting there.

Example:

clips = [[0, 1], [1, 2]]
time = 2

The furthest_reach array has size 2 (indices 0 and 1), but we never check if we can actually reach position 2 from position 1.

Solution: Ensure we properly track whether we can reach the target time:

def videoStitching(self, clips: List[List[int]], time: int) -> int:
    if time == 0:
        return 0
  
    # Store the furthest end point reachable from each starting position
    furthest_reach = [0] * (time + 1)  # Include one extra position
  
    for start, end in clips:
        if start <= time:  # Changed from < to <=
            furthest_reach[start] = max(furthest_reach[start], end)
  
    num_clips = 0
    max_reach = 0
    prev_end = 0
  
    for current_pos in range(time):  # Explicit range instead of enumerate
        max_reach = max(max_reach, furthest_reach[current_pos])
      
        if max_reach <= current_pos:
            return -1
      
        if prev_end == current_pos:
            num_clips += 1
            prev_end = max_reach
          
            # Early termination if we can already reach the target
            if prev_end >= time:
                return num_clips
  
    # Check if we can reach the target time
    return num_clips if prev_end >= time else -1

Pitfall 2: Incorrect Handling of Edge Case When time = 0

The Problem: When time = 0, we don't need any clips to cover the interval [0, 0]. However, the current implementation creates an empty array for furthest_reach which causes the loop to not execute at all, returning 0 by default. While this gives the correct answer, it's by accident rather than design.

Solution: Add explicit handling for this edge case at the beginning:

def videoStitching(self, clips: List[List[int]], time: int) -> int:
    if time == 0:
        return 0  # No clips needed to cover empty interval
  
    # Rest of the implementation...

Pitfall 3: Not Verifying Final Coverage

The Problem: The current implementation assumes that if we can traverse through all positions from 0 to time-1 without gaps, we've successfully covered the entire interval. However, it doesn't explicitly verify that the last selected clip actually reaches or exceeds time.

Example:

clips = [[0, 2], [1, 3], [2, 4]]
time = 5

The algorithm might select clips that cover up to position 4, but never actually reach position 5.

Solution: Add a final check after the loop:

def videoStitching(self, clips: List[List[int]], time: int) -> int:
    # ... previous code ...
  
    for current_pos in range(time):
        max_reach = max(max_reach, furthest_reach[current_pos])
      
        if max_reach <= current_pos:
            return -1
      
        if prev_end == current_pos:
            num_clips += 1
            prev_end = max_reach
  
    # Verify that we can actually reach the target time
    return num_clips if max_reach >= time else -1

Complete Corrected Solution:

def videoStitching(self, clips: List[List[int]], time: int) -> int:
    if time == 0:
        return 0
  
    # Store the furthest end point reachable from each starting position
    furthest_reach = [0] * time
  
    for start, end in clips:
        if 0 <= start < time:
            furthest_reach[start] = max(furthest_reach[start], end)
  
    num_clips = 0
    max_reach = 0
    prev_end = 0
  
    for current_pos in range(time):
        max_reach = max(max_reach, furthest_reach[current_pos])
      
        # Can't reach beyond current position - gap in coverage
        if max_reach <= current_pos:
            return -1
      
        # Need to select a new clip
        if prev_end == current_pos:
            num_clips += 1
            prev_end = max_reach
          
            # Early termination if we've covered the target
            if prev_end >= time:
                return num_clips
  
    # Final check if we've covered the entire duration
    return num_clips if prev_end >= time else -1