How do you solve Minimum Time to Activate String on LeetCode?

Minimum Time to Activate String (LeetCode #3639) 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: The number of valid substrings increases as more '*' are added.

What is the time complexity of Minimum Time to Activate String?

The optimal solution for Minimum Time to Activate String runs in O(n log n) time and uses O(n) space. Binary search reduces the number of checks to O(log n), while counting valid substrings takes O(n).

What is the brute force solution for Minimum Time to Activate String?

The brute force approach for Minimum Time to Activate String is Brute Force, with O(n²) time complexity and O(1) space complexity. This approach checks every possible time step to see if the string becomes active. It counts valid substrings after each character is replaced by '*', which can be inefficient. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Minimum Time to Activate String?

Minimum Time to Activate String uses the following concepts: Array, Binary Search. The recommended patterns to study are: Binary Search, Dynamic Programming.

What are the interview tips for Minimum Time to Activate String?

Explain your thought process clearly. Discuss edge cases and constraints. Practice similar problems to build confidence.

What are common mistakes when solving Minimum Time to Activate String?

Not accounting for all valid substrings correctly. Misunderstanding the binary search bounds.

#3639

Minimum Time to Activate String

Medium

ArrayBinary SearchBinary SearchDynamic Programming

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)

This approach uses binary search on time t and calculates valid substrings efficiently by counting non-'*' segments. It reduces unnecessary computations.

⚙️

Algorithm

3 steps

1Step 1: Use binary search to find the minimum time t where valid substrings >= k.
2Step 2: For each mid in binary search, mark the first mid+1 positions in order as '*'.
3Step 3: Count valid substrings by subtracting the count of non-'*' segments from total substrings.

solution.py24 lines

1def minTimeToActivate(s, order, k):
2    n = len(s)
3    def count_valid(mid):
4        modified = [''] * n
5        for i in range(mid + 1):
6            modified[order[i]] = '*'
7        count = 0
8        length = 0
9        for char in modified:
10            if char == '*':
11                count += (length * (length + 1)) // 2
12                length = 0
13            else:
14                length += 1
15        count += (length * (length + 1)) // 2
16        return (n * (n + 1)) // 2 - count
17    left, right = 0, n - 1
18    while left <= right:
19        mid = (left + right) // 2
20        if count_valid(mid) >= k:
21            right = mid - 1
22        else:
23            left = mid + 1
24    return left if left < n else -1

ℹ

Complexity note: Binary search reduces the number of checks to O(log n), while counting valid substrings takes O(n).

1The number of valid substrings increases as more '*' are added.
2Binary search efficiently narrows down the time needed to activate the string.

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