Leetcode 1621: Number of Sets of K Non-Overlapping Line Segments
You are given n points on a 1-D plane, where each point i is at x = i. Your task is to find the number of ways to draw exactly k non-overlapping line segments that cover two or more points, such that the endpoints are integral. Return the result modulo 10^9 + 7.
Problem
Approach
Steps
Complexity
Input: You are given two integers n (2 <= n <= 1000) and k (1 <= k <= n - 1), where n represents the number of points on the 1-D plane and k represents the number of non-overlapping line segments to draw.
Example: n = 4, k = 2
Constraints:
• 2 <= n <= 1000
• 1 <= k <= n - 1
Output: Return the number of ways to draw k non-overlapping line segments, modulo 10^9 + 7.
Example: Output: 5
Constraints:
• The result should be returned modulo 10^9 + 7.
Goal: You need to find the number of distinct ways to draw k non-overlapping line segments.
Steps:
• Use dynamic programming to calculate the number of ways to draw the segments.
• Memoize intermediate results to optimize the calculation.
Goal: The constraints ensure that the problem size is manageable with dynamic programming.
Steps:
• 2 <= n <= 1000
• 1 <= k <= n - 1
• The answer should be modulo 10^9 + 7.
Assumptions:
• The input will always satisfy the given constraints.
• You can use dynamic programming with memoization to solve the problem efficiently.
• Input: n = 4, k = 2
• Explanation: The 5 different ways to draw 2 non-overlapping line segments are: (0, 2) and (2, 3), (0, 1) and (1, 3), (0, 1) and (2, 3), (1, 2) and (2, 3), (0, 1) and (1, 2).
• Input: n = 3, k = 1
• Explanation: The 3 ways to draw 1 non-overlapping line segment are: (0, 1), (0, 2), (1, 2).
Approach: We use dynamic programming to compute the number of ways to draw k non-overlapping line segments by considering each point and whether we can start or end a line segment at that point.
Observations:
• The problem can be solved using dynamic programming to track the number of ways to draw the segments.
• The segments must be non-overlapping, and we have to take care of memoization to avoid recomputation.
• We need a recursive function that calculates the number of ways to draw segments from the current position and memoizes the results for efficiency.
Steps:
• Initialize a memoization table to store the results of subproblems.
• Define a recursive function that calculates the number of ways to draw segments from a given index.
• Use the base case where no more segments are needed, and return the result modulo 10^9 + 7.
Empty Inputs:
• n cannot be less than 2, so there is no need to handle empty inputs.
Large Inputs:
• Ensure that the solution handles the upper limit where n is up to 1000.
Special Values:
• The number of ways may become large, so remember to return the result modulo 10^9 + 7.
Constraints:
• The value of k must always be between 1 and n-1, inclusive.
int n;
int mod = (int) 1e9 + 7;
int memo[1001][1001][2];
int dp(int idx, int seg, bool startHere) {
if(seg == 0) return 1;
if(idx == n) return 0;
if(memo[idx][seg][startHere] != -1) return memo[idx][seg][startHere];
int ans = dp(idx + 1, seg, startHere); // will start on next or continue the seg
if(startHere) {
ans = (ans + dp(idx + 1, seg, false)) % mod; // stared new line
} else {
ans = (ans + dp(idx, seg - 1, true)) % mod; // end the line here
}
return memo[idx][seg][startHere] = ans;
}
int numberOfSets(int n, int k) {
this->n = n;
memset(memo, -1, sizeof(memo));
return dp(0, k, true);
}
1 : Variable Initialization
int n;
Declare a variable `n` to store the length of the sequence to be divided.
2 : Variable Initialization
int mod = (int) 1e9 + 7;
Initialize a variable `mod` with a large prime number, used for modulo operations to prevent overflow in the final result.
3 : Array Initialization
int memo[1001][1001][2];
Declare a 3D array `memo` for memoization, to store results of subproblems and avoid redundant calculations. It stores results for different states of the recursion.
4 : Function Definition
int dp(int idx, int seg, bool startHere) {
Define the recursive function `dp`, which takes the current index `idx`, the number of segments left `seg`, and a boolean `startHere` indicating whether we are starting a new segment.
5 : Base Case
if(seg == 0) return 1;
Base case: if there are no more segments to divide, return 1, indicating one valid way to split the sequence.
6 : Base Case
if(idx == n) return 0;
Base case: if the index has reached the end of the sequence (`n`), return 0, as there are no ways to further split the sequence.
7 : Memoization Check
if(memo[idx][seg][startHere] != -1) return memo[idx][seg][startHere];
Check if the result for the current subproblem is already computed and stored in the `memo` array. If so, return the cached result.
8 : Recursive Call
int ans = dp(idx + 1, seg, startHere); // will start on next or continue the seg
Recursively call `dp` to explore the option of either starting a new segment or continuing the current one, incrementing the index `idx`.
9 : Conditional Branching
if(startHere) {
Check if we are starting a new segment. If `startHere` is true, we can split the sequence at the current index and start a new segment.
10 : Recursive Call
ans = (ans + dp(idx + 1, seg, false)) % mod; // started new line
If we start a new segment, make a recursive call to `dp`, marking `startHere` as false and continue dividing the sequence.
11 : Else Branch
} else {
If we are not starting a new segment, then we end the current segment and decrease the segment count.
12 : Recursive Call
ans = (ans + dp(idx, seg - 1, true)) % mod; // end the line here
If not starting a new segment, recursively call `dp` with a reduced segment count `seg - 1` and mark `startHere` as true to start the next segment.
13 : Memoization Update
return memo[idx][seg][startHere] = ans;
Store the result of the current subproblem in the `memo` array to avoid redundant calculations in future calls.
14 : Function Definition
int numberOfSets(int n, int k) {
Define the main function `numberOfSets`, which initializes the necessary variables and calls the `dp` function to calculate the number of ways to divide the sequence into `k` segments.
15 : Variable Assignment
this->n = n;
Assign the value of `n` (number of elements in the sequence) to the instance variable `n`.
16 : Array Initialization
memset(memo, -1, sizeof(memo));
Initialize the `memo` array to -1 to indicate that no subproblem results have been calculated yet.
17 : Return Statement
return dp(0, k, true);
Call the `dp` function starting from the first index (`0`), with `k` segments to be formed, and return the result.
Best Case: O(n * k)
Average Case: O(n * k)
Worst Case: O(n * k)
Description: The dynamic programming solution evaluates each subproblem once, and there are at most n * k subproblems to solve.
Best Case: O(n * k)
Worst Case: O(n * k)
Description: The space complexity is O(n * k) due to the memoization table storing results for each subproblem.
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