Asked in: AMAZON
Image of the Question
All Testcases Passed ✔
Solution
def findMinimumTime(password, attackOrder, m):
# Write your code here
ans = 1
n = len(password)
// ... rest of solution available after purchaseExplanation
```
To solve this problem, you need to understand how the password is transformed over time and how the number of substrings containing at least one malicious character '*' evolves as the virus attacks.
---
### Step 1: Understand the problem setting and key definitions
- You have a password string of length n.
- The virus attacks the password one character at a time, according to a given order attackOrder (a permutation of [1..n]).
- At the ith second, the character at position attackOrder[i] (1-based indexing) is replaced by '*'.
- After each replacement, count how many substrings of the updated password contain at least one '*'.
- The password becomes irrecoverable once the count of such substrings is at least m.
- Your goal is to find the minimum time (i.e., smallest i) such that the password becomes irrecoverable.
---
### Step 2: Analyzing the effect of malicious characters on substrings
First, consider what it means for a substring to contain at least one '*'. Such substrings are any contiguous segments of the password that include at least one attacked character.
At the start (before any attack), there are zero '*' characters, so no substrings with '*'.
As characters get replaced by '*', more substrings become "infected."
---
### Step 3: Total number of substrings and complementary counting
The total number of substrings in a string of length n is:
total_substrings = n * (n + 1) / 2
The problem involves counting substrings that contain '*'. It might be easier to think about the complementary set: substrings that contain no '*' at all (i.e., only original characters).
- If you know the number of substrings with no '*', then:
substrings_with_star = total_substrings - substrings_without_star
This complementary approach can be more efficient than enumerating all substrings explicitly.
---
### Step 4: Representing the password after attacks as segments of normal characters
After some characters are replaced by '*', the password splits into several segments of contiguous normal characters (i.e., characters not replaced by '*') separated by '*'.
For example, if the password length is 10 and '*' appear at positions 3 and 7, the segments of normal characters might be:
- positions 1 to 2 (length 2)
- positions 4 to 6 (length 3)
- positions 8 to 10 (length 3)
---
### Step 5: Counting substrings without '*'
For each segment of length L, the number of substrings without any '*' inside that segment is:
substrings_in_segment = L * (L + 1) / 2
Since these segments are disjoint and substrings cannot cross a '*', the total number of substrings without '*' is the sum over all segments of:
sum(L_i * (L_i + 1) / 2)
---
### Step 6: Update segments efficiently as '*' characters are added
Initially, when there are no '*', the entire password is one segment of length n, and substrings_without_star = total_substrings.
As '*' are introduced (one per second, at position attackOrder[i]):
- The segment that contained the attacked position breaks into two smaller segments (to the left and right of the attacked position).
- For example, if a segment covers positions [start, end] and the attack hits position p (start ≤ p ≤ end):
- The segment is split into at most two segments:
- [start, p-1] (if p > start)
- [p+1, end] (if p < end)
- Remove the old segment's substring count and add counts for the new segments formed.
This process helps maintain the count of substrings_without_star after each attack without recalculating everything from scratch.
---
### Step 7: Checking if the number of substrings with '*' meets or exceeds m
After each attack (each second):
- Calculate:
substrings_with_star = total_substrings - substrings_without_star
- If substrings_with_star ≥ m, then the password is irrecoverable at this time.
Return this time as the minimum time when the password becomes irrecoverable.
---
### Step 8: Algorithm efficiency considerations
- The number of attacks n can be large (up to 8 * 10^5), so the approach must be efficient.
- Use a data structure to quickly find and update segments containing the attacked position.
- One option is to keep track of segments in a balanced tree or interval tree, enabling efficient segment splitting and sum updates.
- This data structure supports:
- Finding which segment contains a given position in O(log n) time.
- Removing a segment and adding new smaller segments after splitting in O(log n) time.
- Update the running sum of substrings_without_star accordingly.
---
### Step 9: Edge cases and initial conditions
- If m = 0, then the password is irrecoverable at time 1 because zero substrings with '*' is already ≥ m (which is 0).
- If the password is initially irrecoverable (e.g., m = 1 and password has no '*', so no substrings with '*'), the problem states to return 1 immediately.
- Handle the case when all characters get replaced, resulting in substrings_without_star = 0, so substrings_with_star = total_substrings.
---
### Summary of the approach
1. Compute total_substrings = n * (n + 1) / 2.
2. Initialize segments with one segment: [1, n], substrings_without_star = total_substrings.
3. Iterate over attackOrder from i = 1 to n:
- Find the segment containing attackOrder[i].
- Remove substrings contributed by that segment from substrings_without_star.
- Split the segment into up to two new segments excluding the attacked position.
- Add substrings contributed by new segments back to substrings_without_star.
- Compute substrings_with_star = total_substrings - substrings_without_star.
- If substrings_with_star ≥ m, return i.
4. If no such time found, return n (all replaced).
This strategy efficiently tracks how substrings without '*' decrease over time, enabling quick calculation of when substrings containing '*' reach or exceed m.
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