What is the time complexity of Distinct Points Reachable After Substring Removal?

The optimal solution for Distinct Points Reachable After Substring Removal runs in O(n) time and uses O(n) space. Prefix sums are computed in O(n) and each coordinate calculation is O(1), leading to O(n) overall.

What is the brute force solution for Distinct Points Reachable After Substring Removal?

The brute force approach for Distinct Points Reachable After Substring Removal is Brute Force, with O(n²) time complexity and O(1) space complexity. Remove each possible substring of length k and calculate the reachable coordinates for each case. This method is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Distinct Points Reachable After Substring Removal?

Distinct Points Reachable After Substring Removal uses the following concepts: Hash Table, String, Sliding Window, Prefix Sum. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Distinct Points Reachable After Substring Removal?

Explain your thought process clearly. Discuss edge cases like very small strings. Be prepared to optimize your initial solution.

What are common mistakes when solving Distinct Points Reachable After Substring Removal?

Not accounting for all possible substrings to remove. Incorrectly calculating coordinates after substring removal.

#3694

Distinct Points Reachable After Substring Removal

Medium

Hash TableStringSliding WindowPrefix SumHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

Utilize prefix sums to track movements, allowing efficient calculation of reachable coordinates after removing any substring of length k.

⚙️

Algorithm

3 steps

1Step 1: Create prefix sum arrays for x and y coordinates based on the moves in the string.
2Step 2: For each possible removal index, calculate the new final coordinates using the prefix sums.
3Step 3: Store the unique coordinates in a set and return its size.

solution.py12 lines

1def distinct_points_optimal(s, k):
2    n = len(s)
3    x_prefix, y_prefix = [0] * (n + 1), [0] * (n + 1)
4    for i in range(n):
5        x_prefix[i + 1] = x_prefix[i] + (1 if s[i] == 'R' else -1 if s[i] == 'L' else 0)
6        y_prefix[i + 1] = y_prefix[i] + (1 if s[i] == 'U' else -1 if s[i] == 'D' else 0)
7    unique_coords = set()
8    for i in range(n - k + 1):
9        x = x_prefix[n] - (x_prefix[i + k] - x_prefix[i])
10        y = y_prefix[n] - (y_prefix[i + k] - y_prefix[i])
11        unique_coords.add((x, y))
12    return len(unique_coords)

ℹ

Complexity note: Prefix sums are computed in O(n) and each coordinate calculation is O(1), leading to O(n) overall.

1Removing a substring affects the final coordinates based on the prefix sums.
2Distinct final coordinates can be efficiently tracked using a set.

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