Asked in: AMAZON
Image of the Question
All Testcases Passed ✔
Solution
def getMinOperations(s, m, k):
n = len(s)
arr = [0 for i in range(n+2*k)]
current = 0
// ... rest of solution available after purchaseExplanation
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To solve this problem, you are given a binary string `s` composed of '0's and '1's. Your goal is to ensure that no segment of **exactly** `m` consecutive '0's exists in the final string. You are allowed to perform operations where, in each one, you flip an entire contiguous substring of length `k` — turning all the bits in that segment into '1's.
Your task is to determine the **minimum number of such operations** needed to satisfy the condition that no substring of exactly `m` zeros exists.
Let’s break down the logical flow to understand how to approach this problem.
### Understanding the Problem
You need to eliminate **all** occurrences of exactly `m` consecutive '0's. These are critical regions where an operation may be needed.
However, there's a subtlety: your goal isn't to remove **all** zeros or all long chains of zeros. It's specifically to ensure that no segment of zeros of length **m** exists. Segments of length **less than m** are acceptable, and so are those **longer than m**, as long as the longer ones are also broken down such that the resulting segments don't leave any exact m-length chain of zeros.
### Step-by-Step Strategy
#### 1. Traverse the String
Begin by scanning the string linearly to find contiguous blocks of zeros. You want to identify their start and end positions and compute their lengths. These blocks are the regions where potential violations of the rule might exist.
#### 2. Identify Target Segments
Once you find a block of zeros, you need to decide if it contains any **exactly m-length** segments of zeros. For example, if you find a block of zeros with length `L`:
- If `L < m`, it's fine — skip it.
- If `L == m`, it needs **at least one** operation — you must break it.
- If `L > m`, you must ensure that **no part** of it ends up with an exact m-length segment after your operations. That means you'll need to break it strategically.
#### 3. Break the Segments Strategically
You are allowed to flip any **k-length segment**, and all flipped bits will become '1'. The challenge is to position these operations smartly to minimize how many times you apply them.
For example, consider a long block of zeros (length L >= m). The number of **bad** m-length segments (i.e., exact-length violations) depends on how the zeros are distributed.
The key idea is to **eliminate all possible exact m-length subsegments** in such a block. If L is a long zero block, then you must ensure that for every m-length window in that block, **at least one zero inside it** is flipped (i.e., turned to '1').
You can think of this as a greedy interval covering problem: For a sequence of zeros, cover it with as few length-k windows as possible such that every m-length subsegment is touched by at least one operation.
The important observation is that if you start at the beginning of a zero block and jump ahead by (m) units each time, you're skipping potential bad segments. Instead, you must ensure that **every window of size m** inside a zero block is interrupted.
The optimal greedy approach involves:
- Starting at the first position of a zero block.
- For every point in that block where a length-m segment could start, ensure there's an operation covering that segment.
- Strategically apply operations to cover as many of those m-length segments as possible using a single operation of length k.
#### 4. Greedy Application of Operations
Within each zero block, you want to **greedily** place operations as far to the left as possible to cover the most upcoming violations. For a given position, if you find a potential violation, apply an operation such that its range covers this position and the maximum number of next m-length windows.
Once you've applied an operation, skip ahead accordingly — you can jump beyond the range you've just covered. Keep track of the end of your last operation to avoid redundant operations.
Also, be cautious not to re-apply operations on bits that have already been flipped by a previous one — maintain a variable that keeps track of the current "end" of the last flipped segment.
#### 5. Repeat Until Entire String is Processed
Do this for every zero block in the string. Sum up the total number of operations you needed, and that gives your answer.
### Example
Take `s = "000000"` with `m = 3`, `k = 2`:
- There's a block of length 6.
- Possible m-length segments: positions 0-2, 1-3, 2-4, 3-5.
- You must ensure each of these 4 segments has at least one bit flipped.
- A greedy solution is to flip positions [2,3] (which interrupts all four windows), so only **1 operation** is needed.
### Summary of Key Ideas
- Traverse the string to find contiguous zero blocks.
- For each block of zeros:
- Ignore if length < m.
- If length >= m, cover all m-length segments inside the block.
- Use a greedy method to place as few operations as possible by leveraging the fixed k-length operation window.
- Track positions already covered to avoid unnecessary operations.
- Count and return the total operations used.
This approach ensures correctness and efficiency, even for the maximum constraints of the problem.
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