LeetCode #1621 — MEDIUM

Number of Sets of K Non-Overlapping Line Segments

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

The Problem

Problem Statement

Given n points on a 1-D plane, where the i^th point (from 0 to n-1) is at x = i, find the number of ways we can draw exactly k non-overlapping line segments such that each segment covers two or more points. The endpoints of each segment must have integral coordinates. The k line segments do not have to cover all n points, and they are allowed to share endpoints.

Return the number of ways we can draw k non-overlapping line segments. Since this number can be huge, return it modulo 10⁹ + 7.

Example 1:

Input: n = 4, k = 2
Output: 5
Explanation: The two line segments are shown in red and blue.
The image above shows the 5 different ways {(0,2),(2,3)}, {(0,1),(1,3)}, {(0,1),(2,3)}, {(1,2),(2,3)}, {(0,1),(1,2)}.

Example 2:

Input: n = 3, k = 1
Output: 3
Explanation: The 3 ways are {(0,1)}, {(0,2)}, {(1,2)}.

Example 3:

Input: n = 30, k = 7
Output: 796297179
Explanation: The total number of possible ways to draw 7 line segments is 3796297200. Taking this number modulo 10⁹ + 7 gives us 796297179.

Constraints:

2 <= n <= 1000
1 <= k <= n-1

Step 01

Brute Force Baseline

Problem summary: Given n points on a 1-D plane, where the ith point (from 0 to n-1) is at x = i, find the number of ways we can draw exactly k non-overlapping line segments such that each segment covers two or more points. The endpoints of each segment must have integral coordinates. The k line segments do not have to cover all n points, and they are allowed to share endpoints. Return the number of ways we can draw k non-overlapping line segments. Since this number can be huge, return it modulo 109 + 7.

Baseline thinking

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

Pattern signal: Math · Dynamic Programming

Example 1

4
2

Example 2

3
1

Example 3

30
7

Step 02

Core Insight

What unlocks the optimal approach

Try to use dynamic programming where the current index and remaining number of line segments to form can describe any intermediate state.
To make the computation of each state in constant time, we could add another flag to the state that indicates whether or not we are in the middle of placing a line (placed start point but no endpoint).

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 #1621: Number of Sets of K Non-Overlapping Line Segments
class Solution {
    private static final int MOD = (int) 1e9 + 7;

    public int numberOfSets(int n, int k) {
        int[][] f = new int[n + 1][k + 1];
        int[][] g = new int[n + 1][k + 1];
        f[1][0] = 1;
        for (int i = 2; i <= n; ++i) {
            for (int j = 0; j <= k; ++j) {
                f[i][j] = (f[i - 1][j] + g[i - 1][j]) % MOD;
                g[i][j] = g[i - 1][j];
                if (j > 0) {
                    g[i][j] += f[i - 1][j - 1];
                    g[i][j] %= MOD;
                    g[i][j] += g[i - 1][j - 1];
                    g[i][j] %= MOD;
                }
            }
        }
        return (f[n][k] + g[n][k]) % MOD;
    }
}

// Accepted solution for LeetCode #1621: Number of Sets of K Non-Overlapping Line Segments
func numberOfSets(n int, k int) int {
	f := make([][]int, n+1)
	g := make([][]int, n+1)
	for i := range f {
		f[i] = make([]int, k+1)
		g[i] = make([]int, k+1)
	}
	f[1][0] = 1
	var mod int = 1e9 + 7
	for i := 2; i <= n; i++ {
		for j := 0; j <= k; j++ {
			f[i][j] = (f[i-1][j] + g[i-1][j]) % mod
			g[i][j] = g[i-1][j]
			if j > 0 {
				g[i][j] += f[i-1][j-1]
				g[i][j] %= mod
				g[i][j] += g[i-1][j-1]
				g[i][j] %= mod
			}
		}
	}
	return (f[n][k] + g[n][k]) % mod
}

# Accepted solution for LeetCode #1621: Number of Sets of K Non-Overlapping Line Segments
class Solution:
    def numberOfSets(self, n: int, k: int) -> int:
        mod = 10**9 + 7
        f = [[0] * (k + 1) for _ in range(n + 1)]
        g = [[0] * (k + 1) for _ in range(n + 1)]
        f[1][0] = 1
        for i in range(2, n + 1):
            for j in range(k + 1):
                f[i][j] = (f[i - 1][j] + g[i - 1][j]) % mod
                g[i][j] = g[i - 1][j]
                if j:
                    g[i][j] += f[i - 1][j - 1]
                    g[i][j] %= mod
                    g[i][j] += g[i - 1][j - 1]
                    g[i][j] %= mod
        return (f[-1][-1] + g[-1][-1]) % mod

// Accepted solution for LeetCode #1621: Number of Sets of K Non-Overlapping Line Segments
struct Solution;

use std::collections::HashMap;

const MOD: i64 = 1_000_000_007;

impl Solution {
    fn number_of_sets(n: i32, k: i32) -> i32 {
        let mut memo: HashMap<(i32, i32, bool), i64> = HashMap::new();
        Self::dp(n, k, false, &mut memo) as i32
    }

    fn dp(n: i32, k: i32, started: bool, memo: &mut HashMap<(i32, i32, bool), i64>) -> i64 {
        if k == 0 {
            return 1;
        }
        if n == 0 {
            return 0;
        }
        if let Some(&res) = memo.get(&(n, k, started)) {
            return res;
        }
        let mut res = Self::dp(n - 1, k, started, memo);
        if started {
            res += Self::dp(n, k - 1, false, memo);
        } else {
            res += Self::dp(n - 1, k, true, memo);
        }
        res %= MOD;
        memo.insert((n, k, started), res);
        res
    }
}

#[test]
fn test() {
    let n = 4;
    let k = 2;
    let res = 5;
    assert_eq!(Solution::number_of_sets(n, k), res);
    let n = 3;
    let k = 1;
    let res = 3;
    assert_eq!(Solution::number_of_sets(n, k), res);
    let n = 30;
    let k = 7;
    let res = 796297179;
    assert_eq!(Solution::number_of_sets(n, k), res);
}

// Accepted solution for LeetCode #1621: Number of Sets of K Non-Overlapping Line Segments
function numberOfSets(n: number, k: number): number {
    const f = Array.from({ length: n + 1 }, _ => new Array(k + 1).fill(0));
    const g = Array.from({ length: n + 1 }, _ => new Array(k + 1).fill(0));
    f[1][0] = 1;
    const mod = 10 ** 9 + 7;
    for (let i = 2; i <= n; ++i) {
        for (let j = 0; j <= k; ++j) {
            f[i][j] = (f[i - 1][j] + g[i - 1][j]) % mod;
            g[i][j] = g[i - 1][j];
            if (j) {
                g[i][j] += f[i - 1][j - 1];
                g[i][j] %= mod;
                g[i][j] += g[i - 1][j - 1];
                g[i][j] %= mod;
            }
        }
    }
    return (f[n][k] + g[n][k]) % mod;
}

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(n × 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.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

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.