How do you solve Minimum Deletions to Make Alternating Substring on LeetCode?

Minimum Deletions to Make Alternating Substring (LeetCode #3777) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Use data structures like Fenwick trees for efficient range queries.

What is the time complexity of Minimum Deletions to Make Alternating Substring?

The optimal solution for Minimum Deletions to Make Alternating Substring runs in O(n log n) time and uses O(n) space. Fenwick tree allows efficient updates and queries, making it optimal for this problem.

What is the brute force solution for Minimum Deletions to Make Alternating Substring?

The brute force approach for Minimum Deletions to Make Alternating Substring is Brute Force, with O(n²) time complexity and O(1) space complexity. Check every substring for alternating characters by counting adjacent duplicates. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Minimum Deletions to Make Alternating Substring?

Minimum Deletions to Make Alternating Substring uses the following concepts: String, Segment Tree. The recommended patterns to study are: Fenwick Tree, Segment Tree.

What are the interview tips for Minimum Deletions to Make Alternating Substring?

Explain your thought process clearly. Discuss trade-offs between brute force and optimal solutions. Practice coding on a whiteboard or online editor.

What are common mistakes when solving Minimum Deletions to Make Alternating Substring?

Not updating the auxiliary structure after a flip. Misunderstanding the range of queries.

#3777

Minimum Deletions to Make Alternating Substring

Hard

StringSegment TreeFenwick TreeSegment Tree

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n log n)Space O(n)

Use a Fenwick tree to efficiently track and update adjacent duplicates, allowing quick calculations for deletions.

⚙️

Algorithm

3 steps

1Step 1: Initialize an array eq where eq[i] = 1 if s[i] == s[i-1], else 0.
2Step 2: Use a Fenwick tree to maintain prefix sums of eq for quick range queries.
3Step 3: For a query of type [2, l, r], return the sum of eq from l to r.

solution.py36 lines

1class FenwickTree:
2    def __init__(self, size):
3        self.size = size
4        self.tree = [0] * (size + 1)
5    def update(self, index, delta):
6        while index <= self.size:
7            self.tree[index] += delta
8            index += index & -index
9    def query(self, index):
10        total = 0
11        while index > 0:
12            total += self.tree[index]
13            index -= index & -index
14        return total
15
16def min_deletions_optimal(s, queries):
17    n = len(s)
18    eq = [0] * n
19    for i in range(1, n):
20        eq[i] = 1 if s[i] == s[i - 1] else 0
21    fenwick = FenwickTree(n)
22    for i in range(1, n):
23        fenwick.update(i, eq[i])
24    results = []
25    for query in queries:
26        if query[0] == 1:
27            idx = query[1]
28            s = s[:idx] + ('B' if s[idx] == 'A' else 'A') + s[idx + 1:]
29            if idx > 0:
30                fenwick.update(idx, 1 if s[idx] == s[idx - 1] else -1)
31            if idx < n - 1:
32                fenwick.update(idx + 1, 1 if s[idx + 1] == s[idx] else -1)
33        else:
34            l, r = query[1], query[2]
35            results.append(fenwick.query(r) - fenwick.query(l))
36    return results

ℹ

Complexity note: Fenwick tree allows efficient updates and queries, making it optimal for this problem.

1Use data structures like Fenwick trees for efficient range queries.
2Flipping characters can be managed with careful updates to auxiliary structures.

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