Track contiguous ones by updating group lengths dynamically as you scan the array. Use a hash map to record lengths of groups ending at each position, allowing constant-time updates when groups merge or extend. Start from the end when possible to quickly identify the latest step matching the target size m.
Problem Statement
You are given a permutation array arr of length n containing numbers from 1 to n. Start with a binary string of size n initialized to all zeros. At each step i, set the bit at position arr[i] to 1, creating contiguous groups of ones dynamically.
Given an integer m, determine the latest step at which there exists at least one contiguous group of ones of length exactly m. A group of ones is defined as consecutive 1's that cannot be extended left or right. If no such group ever exists, return -1.
Examples
Example 1
Input: arr = [3,5,1,2,4], m = 1
Output: 4
Step 1: "00100", groups: ["1"] Step 2: "00101", groups: ["1", "1"] Step 3: "10101", groups: ["1", "1", "1"] Step 4: "11101", groups: ["111", "1"] Step 5: "11111", groups: ["11111"] The latest step at which there exists a group of size 1 is step 4.
Example 2
Input: arr = [3,1,5,4,2], m = 2
Output: -1
Step 1: "00100", groups: ["1"] Step 2: "10100", groups: ["1", "1"] Step 3: "10101", groups: ["1", "1", "1"] Step 4: "10111", groups: ["1", "111"] Step 5: "11111", groups: ["11111"] No group of size 2 exists during any step.
Constraints
- n == arr.length
- 1 <= m <= n <= 105
- 1 <= arr[i] <= n
- All integers in arr are distinct.
Solution Approach
Use a Hash Map to Track Group Boundaries
Maintain a hash map where keys are positions and values are the lengths of the group ending or starting at that position. When a new one is placed, check left and right neighbors to merge groups, update lengths, and check if a group of size m appears. This ensures O(1) updates per step.
Scan the Array Once
Iterate through arr sequentially. At each step, update the group lengths in the hash map and track counts of groups of length m. This approach directly aligns with the array scanning plus hash lookup pattern, efficiently identifying when target-sized groups appear or disappear.
Consider Reverse Search for Latest Step
Since the problem asks for the latest step, optionally scan from the end or maintain a running latest-step variable whenever a group of size m exists. This prevents unnecessary full iteration and aligns with the failure mode where early detection might be misleading.
Complexity Analysis
| Metric | Value |
|---|---|
| Time | Depends on the final approach |
| Space | Depends on the final approach |
Time complexity can be O(n) since each position is updated at most once, and hash lookups are O(1). Space complexity is O(n) to store group boundaries and counts of groups of specific lengths.
What Interviewers Usually Probe
- Asks if groups can merge or split dynamically and how to track that efficiently.
- Questions whether scanning from the end helps in quickly finding the latest step.
- Checks understanding of hash-based bookkeeping for constant-time group length updates.
Common Pitfalls or Variants
Common pitfalls
- Failing to update both ends of a merged group, leading to incorrect group length tracking.
- Counting overlapping groups incorrectly, especially after merges or splits.
- Assuming the first occurrence is enough instead of tracking the latest step.
Follow-up variants
- Find the earliest step a group of size m exists instead of the latest.
- Return the number of steps where at least one group of size m exists.
- Change the problem to track groups of at least size m instead of exact size.
How GhostInterview Helps
- Automatically identifies when and how to merge contiguous groups efficiently with hash maps.
- Highlights the importance of tracking both ends of a group to prevent miscounting.
- Suggests scanning strategies that optimize finding the latest step rather than iterating naively.
Topic Pages
Related GhostInterview Pages
- LeetCode Interview Copilot - Use GhostInterview as a live solver when you want direct help with LeetCode-style coding questions.
- Coding Interview Assistant - See how GhostInterview supports array, string, linked list, graph, and tree interview workflows.
- How GhostInterview Works - Review the screenshot, reasoning, and answer flow before using the solver in a live interview.
FAQ
What is the core pattern in Find Latest Group of Size M?
It is array scanning plus hash lookup to maintain group boundaries and quickly detect groups of exact length m.
Can we solve this problem without a hash map?
Yes, using a union-find or array-based boundary tracking is possible, but hash maps simplify dynamic length updates and merges.
Why is scanning from the end recommended?
Scanning from the end lets you detect the latest step quickly, aligning with the problem's request to find the last occurrence of a group of size m.
How do we handle overlapping or merging groups?
Update both start and end positions in the hash map when a new one connects neighboring groups to prevent miscounting.
What if no group of size m ever exists?
Return -1, indicating that during the entire process no contiguous group of ones matched the required length.
Need direct help with Find Latest Group of Size M instead of spending more time grinding it?
Download GhostInterview when you want a LeetCode solver, not another long practice loop. Capture Find Latest Group of Size M from a screenshot, get the answer path and complexity, and use supported stealth workflows that stay outside captured layers.
Capture the prompt fast instead of rewriting the problem by hand.
Get the solution path, trade-offs, and complexity summary in one pass.
Stay outside captured layers on supported screen-share workflows.
Stay in the same pattern family
This problem asks you to avoid flooding by deciding when to dry lakes between rain events.
Open problem page#1477 Find Two Non-overlapping Sub-arrays Each With Target SumFind two non-overlapping sub-arrays with a given target sum and return the minimal total length efficiently using array scanning and hash lookup.
Open problem page#1658 Minimum Operations to Reduce X to ZeroFind the minimum number of operations to reduce a value x to zero by removing elements from an array.
Open problem page