LeetCode #1024 — MEDIUM

Video Stitching

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

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 the clips into segments that cover the entire sporting event [0, time]. If the task is impossible, return -1.

Example 1:

Input: clips = [[0,2],[4,6],[8,10],[1,9],[1,5],[5,9]], time = 10
Output: 3
Explanation: We take the clips [0,2], [8,10], [1,9]; a total of 3 clips.
Then, we can reconstruct the sporting event as follows:
We cut [1,9] into segments [1,2] + [2,8] + [8,9].
Now we have segments [0,2] + [2,8] + [8,10] which cover the sporting event [0, 10].

Example 2:

Input: clips = [[0,1],[1,2]], time = 5
Output: -1
Explanation: We cannot cover [0,5] with only [0,1] and [1,2].

Example 3:

Input: clips = [[0,1],[6,8],[0,2],[5,6],[0,4],[0,3],[6,7],[1,3],[4,7],[1,4],[2,5],[2,6],[3,4],[4,5],[5,7],[6,9]], time = 9
Output: 3
Explanation: We can take clips [0,4], [4,7], and [6,9].

Constraints:

1 <= clips.length <= 100
0 <= start_i <= end_i <= 100
1 <= time <= 100

Step 01

Brute Force Baseline

Problem summary: 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] = [starti, endi] indicates that the ith clip started at starti and ended at endi. 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 the clips into segments that cover the entire sporting event [0, time]. If the task is impossible, return -1.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Dynamic Programming · Greedy

Example 1

[[0,2],[4,6],[8,10],[1,9],[1,5],[5,9]]
10

Example 2

[[0,1],[1,2]]
5

Example 3

[[0,1],[6,8],[0,2],[5,6],[0,4],[0,3],[6,7],[1,3],[4,7],[1,4],[2,5],[2,6],[3,4],[4,5],[5,7],[6,9]]
9

Step 02

Core Insight

What unlocks the optimal approach

What if we sort the intervals? Considering the sorted intervals, how can we solve the problem with dynamic programming?
Let's consider a DP(pos, limit) where pos represents the position of the current interval we are gonna take the decision and limit is the current covered area from [0 - limit]. This DP returns the minimum number of taken intervals or infinite if it's not possible to cover the [0 - T] section.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1024: Video Stitching
class Solution {
    public int videoStitching(int[][] clips, int time) {
        int[] last = new int[time];
        for (var e : clips) {
            int a = e[0], b = e[1];
            if (a < time) {
                last[a] = Math.max(last[a], b);
            }
        }
        int ans = 0, mx = 0, pre = 0;
        for (int i = 0; i < time; ++i) {
            mx = Math.max(mx, last[i]);
            if (mx <= i) {
                return -1;
            }
            if (pre == i) {
                ++ans;
                pre = mx;
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1024: Video Stitching
func videoStitching(clips [][]int, time int) int {
	last := make([]int, time)
	for _, v := range clips {
		a, b := v[0], v[1]
		if a < time {
			last[a] = max(last[a], b)
		}
	}
	ans, mx, pre := 0, 0, 0
	for i, v := range last {
		mx = max(mx, v)
		if mx <= i {
			return -1
		}
		if pre == i {
			ans++
			pre = mx
		}
	}
	return ans
}

# Accepted solution for LeetCode #1024: Video Stitching
class Solution:
    def videoStitching(self, clips: List[List[int]], time: int) -> int:
        last = [0] * time
        for a, b in clips:
            if a < time:
                last[a] = max(last[a], b)
        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
        return ans

// Accepted solution for LeetCode #1024: Video Stitching
struct Solution;

use std::cmp::Reverse;

impl Solution {
    fn video_stitching(mut clips: Vec<Vec<i32>>, t: i32) -> i32 {
        clips.sort_by_key(|v| (v[0], Reverse(v[1])));
        let n = clips.len();
        let mut res = 0;
        let mut left = 0;
        let mut right = 0;
        for i in 0..n {
            if clips[i][0] >= t {
                break;
            }
            if clips[i][0] < left {
                right = right.max(clips[i][1]);
            } else if clips[i][0] <= right {
                right = right.max(clips[i][1]);
                left = right;
                res += 1;
            } else {
                return -1;
            }
        }
        if right < t {
            -1
        } else {
            if left < t {
                res + 1
            } else {
                res
            }
        }
    }
}

#[test]
fn test() {
    let clips = vec_vec_i32![[0, 2], [4, 6], [8, 10], [1, 9], [1, 5], [5, 9]];
    let t = 10;
    let res = 3;
    assert_eq!(Solution::video_stitching(clips, t), res);
    let clips = vec_vec_i32![[0, 1], [1, 2]];
    let t = 5;
    let res = -1;
    assert_eq!(Solution::video_stitching(clips, t), res);
    let clips = vec_vec_i32![
        [0, 1],
        [6, 8],
        [0, 2],
        [5, 6],
        [0, 4],
        [0, 3],
        [6, 7],
        [1, 3],
        [4, 7],
        [1, 4],
        [2, 5],
        [2, 6],
        [3, 4],
        [4, 5],
        [5, 7],
        [6, 9]
    ];
    let t = 9;
    let res = 3;
    assert_eq!(Solution::video_stitching(clips, t), res);
    let clips = vec_vec_i32![[0, 4], [2, 8]];
    let t = 5;
    let res = 2;
    assert_eq!(Solution::video_stitching(clips, t), res);
    let clips = vec_vec_i32![
        [5, 7],
        [1, 8],
        [0, 0],
        [2, 3],
        [4, 5],
        [0, 6],
        [5, 10],
        [7, 10]
    ];
    let t = 5;
    let res = 1;
    assert_eq!(Solution::video_stitching(clips, t), res);
}

// Accepted solution for LeetCode #1024: Video Stitching
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #1024: Video Stitching
// class Solution {
//     public int videoStitching(int[][] clips, int time) {
//         int[] last = new int[time];
//         for (var e : clips) {
//             int a = e[0], b = e[1];
//             if (a < time) {
//                 last[a] = Math.max(last[a], b);
//             }
//         }
//         int ans = 0, mx = 0, pre = 0;
//         for (int i = 0; i < time; ++i) {
//             mx = Math.max(mx, last[i]);
//             if (mx <= i) {
//                 return -1;
//             }
//             if (pre == i) {
//                 ++ans;
//                 pre = mx;
//             }
//         }
//         return ans;
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n+m)

Space

O(m)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.