How do you solve Number of Sets of K Non-Overlapping Line Segments on LeetCode?

Number of Sets of K Non-Overlapping Line Segments (LeetCode #1621) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * k) time complexity and O(n * k) space complexity. Key insight: Understanding how to break down the problem into smaller subproblems is crucial for dynamic programming.

What is the brute force solution for Number of Sets of K Non-Overlapping Line Segments?

The brute force approach for Number of Sets of K Non-Overlapping Line Segments is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible combinations of line segments and counting those that meet the criteria. This is straightforward but inefficient, as it checks every possible configuration. The optimal Optimal Solution approach improves this to O(n * k).

What are the interview tips for Number of Sets of K Non-Overlapping Line Segments?

Always explain your thought process and approach before jumping into coding. Consider edge cases and constraints while designing your solution. Practice writing clean and efficient code, as clarity is important in interviews.

What are common mistakes when solving Number of Sets of K Non-Overlapping Line Segments?

Failing to account for overlapping segments when checking combinations. Not initializing the DP table correctly or misunderstanding the dimensions.

#1621

Number of Sets of K Non-Overlapping Line Segments

Medium

MathDynamic ProgrammingCombinatoricsPrefix SumDynamic ProgrammingCombinatorics

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n * k)
Space	O(1)	O(n * k)

💡

Intuition

Time O(n * k)Space O(n * k)

The optimal solution uses dynamic programming to efficiently compute the number of ways to form k segments by building on previously computed results. This reduces the complexity significantly by avoiding redundant calculations.

⚙️

Algorithm

3 steps

1Step 1: Initialize a DP table where dp[i][j] represents the number of ways to draw j segments using the first i points.
2Step 2: Iterate through each point and each possible number of segments, updating the DP table based on valid segment placements.
3Step 3: Use prefix sums to quickly calculate the number of valid segments that can be formed.

solution.py12 lines

1# Full working Python code
2MOD = 10**9 + 7
3
4def count_segments_optimal(n, k):
5    dp = [[0] * (k + 1) for _ in range(n + 1)]
6    dp[0][0] = 1
7    for i in range(1, n + 1):
8        for j in range(1, k + 1):
9            for l in range(1, i):
10                dp[i][j] = (dp[i][j] + dp[l][j - 1]) % MOD
11    return dp[n][k]
12

ℹ

Complexity note: The time complexity is O(n * k) because we iterate through all points and segments, updating our DP table based on previous results, which is much more efficient than the brute force method.

1Understanding how to break down the problem into smaller subproblems is crucial for dynamic programming.
2Recognizing overlapping subproblems can help optimize the solution significantly.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.