Find Latest Group of Size M: Reverse Simulation with Group Tracking

You are given an array arr that is a permutation of [1, 2, ..., n]. At step i, you set the bit at position arr[i] to 1. After each step, you check all groups of consecutive 1s. Return the latest step at which there exists a group of exactly m consecutive 1s. If no such group ever exists, return -1.

This is LeetCode 1562: Find Latest Group of Size M, a medium problem that looks like it needs brute-force simulation but can be solved in O(n) with a clever group-tracking trick.

Why this problem matters

The naive approach is to simulate every step: flip the bit, then scan the entire binary string to check group sizes. That is O(n^2), and for n up to 100,000, it will time out.

The key insight is that you do not need to scan the whole string after every step. When you set a bit, only the groups adjacent to that position change. You can track group sizes at the boundaries and update them in O(1) per step.

This boundary-tracking technique shows up in several interval merging and union-find problems. Once you see how it works here, you will recognize it immediately in other contexts.

The key insight: boundary length tracking

Maintain an array length[] where length[left_boundary] and length[right_boundary] both store the size of the group that spans from left_boundary to right_boundary. When you set a new bit at position pos, check the neighbors:

• length[pos - 1] tells you the size of any group to the left
• length[pos + 1] tells you the size of any group to the right
• The new merged group has size left + right + 1
• Update only the new boundary endpoints

Also maintain a count[] array where count[k] is the number of groups of size k. When merging, decrement count[left] and count[right], and increment count[new_len]. Then check if count[m] > 0.

The solution

def findLatestStep(arr: list[int], m: int) -> int:
    n = len(arr)
    length = [0] * (n + 2)
    count = [0] * (n + 1)
    result = -1

    for step, pos in enumerate(arr, 1):
        left = length[pos - 1]
        right = length[pos + 1]
        new_len = left + right + 1

        length[pos - left] = new_len
        length[pos + right] = new_len

        count[new_len] += 1
        if left:
            count[left] -= 1
        if right:
            count[right] -= 1

        if count[m] > 0:
            result = step

    return result

Let's walk through the logic line by line.

Setup. length has size n + 2 so we can safely check length[pos - 1] and length[pos + 1] without bounds issues. count[k] tracks how many groups currently have size k.

Each step. When we set the bit at pos:

1. left = length[pos - 1] gives the size of the group immediately to the left (0 if none).
2. right = length[pos + 1] gives the size of the group immediately to the right (0 if none).
3. The new merged group has size left + right + 1.
4. We update the boundary endpoints: length[pos - left] is the new left boundary, length[pos + right] is the new right boundary. Both store new_len.
5. We increment count[new_len] and decrement count[left] and count[right] (those old groups no longer exist as separate groups).
6. If count[m] > 0, at least one group of size m exists, so we update result to the current step.

After processing all steps, result holds the latest step number, or -1 if no group of size m ever appeared.

Visual walkthrough

Let's trace through arr = [3, 5, 1, 2, 4] with m = 1. Watch how groups form, merge, and how the count array tracks everything.

Notice how at step 4, position 2 merges with its left neighbor (position 1) and right neighbor (position 3). The two groups of size 1 at positions 1 and 3 disappear, but the group of size 1 at position 5 still exists. So count[1] drops from 3 to 1, and we still have a group of size m = 1.

At step 5, position 4 merges everything into one big group of size 5. Now count[1] = 0, so no group of size 1 remains. The answer is step 4.

Complexity analysis

Time: O(n). Each step does O(1) work: two lookups, two boundary updates, three count updates, and one comparison. There are n steps total.

Space: O(n). The length array is size n + 2 and the count array is size n + 1. Both are linear.

The building blocks

This problem combines two reusable patterns:

1. Boundary-based group tracking

The pattern of storing group metadata at boundary positions so you can query and update in O(1):

left_size = length[pos - 1]
right_size = length[pos + 1]
new_size = left_size + right_size + 1
length[pos - left_size] = new_size
length[pos + right_size] = new_size

This is a lightweight alternative to union-find for 1D merging problems. When you insert an element and need to merge with adjacent groups, you only need to know the size of the groups to the left and right. The boundary endpoints give you that in O(1). You will see the same idea in problems like "longest consecutive sequence" and interval merge problems.

2. Count-based existence tracking

Instead of scanning all groups to check if any has size m, maintain a frequency count:

count[new_len] += 1
if left:
    count[left] -= 1
if right:
    count[right] -= 1
if count[m] > 0:
    result = step

This turns an O(n) scan into an O(1) check. Any time you need to repeatedly ask "does a group/value of size X exist?", maintaining a count array (or hash map) lets you answer instantly. The same pattern shows up in sliding window frequency problems and bucket sort approaches.

Edge cases

Before submitting, make sure your solution handles these:

• m equals n: the only time all bits form a single group is after the very last step. Return n.
• m equals 1: every individual bit that is not adjacent to another set bit counts. The answer is usually late in the process.
• Single element arr = [1], m = 1: one step, one group of size 1. Return 1.
• No group of size m ever exists: return -1. This can happen when m is large relative to n and the insertion order never creates a group that big before it gets merged into something bigger. Actually, this is rare since at step m you always have at most m bits set, but the arrangement might not form a consecutive block of size m.
• Large n (100,000): the O(n) solution handles this easily. The brute force O(n^2) would time out.

From understanding to recall

You have read through the boundary-tracking solution and it makes sense. Two arrays, O(1) merges, a count check at each step. Clean and efficient. But can you write it from scratch in an interview?

The tricky parts are remembering the boundary update formula (pos - left and pos + right), getting the count decrements right (only decrement if left > 0 or right > 0), and sizing the length array to n + 2 for boundary safety. These are small details that are easy to mess up under pressure if you have not practiced them.

Spaced repetition closes that gap. You practice writing the boundary-tracking loop from memory at increasing intervals. After a few rounds, the index math and count bookkeeping are automatic. You see a merging problem and the skeleton flows out without hesitation.

The boundary-based group tracking pattern is one of roughly 60 reusable building blocks that cover hundreds of LeetCode problems. Learning them individually and drilling them with spaced repetition is far more effective than grinding random problems and hoping they stick.

Related posts

• Jump Game - Another array problem where tracking a single running value avoids brute force simulation
• Number of Islands - Connected component counting, the 2D analogue of group tracking
• Longest Consecutive Sequence - Uses a similar idea of tracking group boundaries to find the longest run

CodeBricks breaks Find Latest Group of Size M into its boundary-tracking and count-existence building blocks, then drills them independently with spaced repetition. You type each piece from scratch until the pattern is automatic. When a group merging problem shows up in your interview, you do not think about it. You just write it.