Difficulty: 🟡 Medium
Topics: Array, Hash Table, Binary Search, Simulation

Solution

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LeetCode 1562 - Find Latest Group of Size M

Problem Understanding

This problem gives us a permutation array arr containing integers from 1 to n, where n is the size of the binary string we are building. Initially, the binary string contains only zeroes. At each step, we flip exactly one position from 0 to 1, according to the order specified in arr.

The key requirement is to determine the latest step at which there exists at least one contiguous group of 1s whose length is exactly m.

A group of ones means a maximal contiguous segment of 1s. The group cannot be extended to the left or right because either the boundary of the string is reached or a 0 appears.

For example, if the binary string is:

The valid groups are:

["111", "11"]

The middle 0 separates the groups, so we treat them independently.

The challenge comes from the fact that the binary string evolves over time. At each step, flipping one bit can merge adjacent groups, split the effective count of valid groups, or create entirely new ones. We are not simply checking the final string, we must identify the latest moment during construction when a group of size exactly m exists.

The input constraints are important:

n == arr.length
1 <= m <= n <= 10^5
arr is a permutation of numbers from 1 to n

Since n can be as large as 100,000, an algorithm worse than roughly O(n log n) risks timing out. A naive simulation that repeatedly scans the entire string after every flip would be too slow.

Several edge cases deserve attention upfront. If m == n, then the only possible group of size n appears at the very end when all bits become 1. If no group of size m ever forms, we must return -1. Another subtle case occurs when a group of size m exists temporarily but later disappears due to merging with neighboring groups. Since we want the latest step, we must continuously update the answer rather than stopping at the first occurrence.

Approaches

Brute Force Approach

A straightforward solution is to simulate the binary string step by step.

At each step, we flip the specified position to 1, then scan the entire binary string to identify all contiguous groups of 1s. If any group has size exactly m, we update the answer to the current step.

This method is correct because it directly models the problem statement. After every modification, we inspect the full state and check whether a valid group exists.

However, the performance is poor. Since there are n steps and each step may require scanning the entire binary string of length n, the total complexity becomes O(n²).

With n = 100,000, this would require roughly 10^10 operations in the worst case, which is far too slow.

Key Insight for an Optimal Solution

The crucial observation is that flipping a bit affects only the groups immediately adjacent to that position.

Suppose we flip position x from 0 to 1.

Only three segments matter:

The group ending at x - 1
The group starting at x + 1
The new merged group formed after flipping x

Everything else in the binary string remains unchanged.

Instead of rescanning the whole array, we can efficiently maintain the lengths of contiguous groups using a boundary length array.

The idea is to store the size of each group only at its boundaries. If a group spans from left to right, then:

length[left] = group_size
length[right] = group_size

Intermediate positions do not matter.

When we flip a new position:

Check the size of the adjacent left group.
Check the size of the adjacent right group.
Merge them into one larger segment.
Update only the new boundaries.
Keep track of how many groups of size m currently exist.

This avoids rescanning the binary string and allows us to process each step in constant time.

Approach	Time Complexity	Space Complexity	Notes
Brute Force	`O(n²)`	`O(n)`	Simulates the string and scans all groups after every step
Optimal	`O(n)`	`O(n)`	Maintains segment lengths at boundaries and tracks groups of size `m`

Algorithm Walkthrough

We use an array called length of size n + 2.

The extra padding prevents out-of-bounds checks when accessing neighbors.

At any moment:

length[i]

stores the length of the contiguous group if i is a group boundary.

We also maintain:

count_m

which tracks how many groups of size m currently exist.

Finally:

result

stores the latest step where such a group exists.

Step-by-step Algorithm

Initialize a length array of size n + 2 filled with zeroes.

Zero means the position has not yet been flipped to 1. 2. Initialize:

result = -1
count_m = 0

Iterate through arr.

At step i, let:

pos = arr[i]

Find neighboring group sizes.

The left neighboring group length is:

left_size = length[pos - 1]

The right neighboring group length is:

right_size = length[pos + 1]

These values tell us whether adjacent groups already exist. 5. Before merging, remove old groups of size m.

If:

left_size == m

then that valid group disappears after merging.

Similarly for:

right_size == m

Compute the merged group size.

The new segment size becomes:

merged_size = left_size + right_size + 1

Update boundaries.

The new segment spans:

left_boundary = pos - left_size
right_boundary = pos + right_size

Store:

length[left_boundary] = merged_size
length[right_boundary] = merged_size

If the new merged segment has size m, increment count_m.
If count_m > 0, update:

result = current_step

After processing all steps, return result.

Why it works

The algorithm maintains the invariant that each contiguous group of 1s stores its correct size at both boundaries.

Whenever a new bit is flipped, only the immediately adjacent groups can change. By updating only those affected boundaries, we preserve correctness without rescanning the entire string.

The count_m variable always reflects how many valid groups of size m currently exist. Therefore, whenever count_m > 0, we know that the current step contains a valid group, and storing the latest such step guarantees the correct answer.

Python Solution

from typing import List

class Solution:
    def findLatestStep(self, arr: List[int], m: int) -> int:
        n = len(arr)

        if m == n:
            return n

        length = [0] * (n + 2)
        count_m = 0
        result = -1

        for step, position in enumerate(arr, 1):
            left_size = length[position - 1]
            right_size = length[position + 1]

            merged_size = left_size + right_size + 1

            left_boundary = position - left_size
            right_boundary = position + right_size

            length[left_boundary] = merged_size
            length[right_boundary] = merged_size

            if left_size == m:
                count_m -= 1

            if right_size == m:
                count_m -= 1

            if merged_size == m:
                count_m += 1

            if count_m > 0:
                result = step

        return result

The implementation begins by handling a useful edge case:

if m == n:
    return n

If the desired group size equals the total array length, the only possible valid group appears after every position has been flipped.

The length array stores segment sizes only at boundaries. At every step, we inspect neighboring group lengths around the newly flipped position.

The left and right segment sizes are retrieved in constant time:

left_size = length[position - 1]
right_size = length[position + 1]

These represent any adjacent groups that will merge with the current position.

We then compute the merged group size and determine its boundaries.

The crucial detail is updating count_m. Any neighboring group of size m disappears once merged, so we decrement the count before creating the new merged group. If the merged group itself has size m, we increment the count.

Finally, whenever count_m > 0, we update result because the current step contains at least one valid group of size m.

Go Solution

func findLatestStep(arr []int, m int) int {
	n := len(arr)

	if m == n {
		return n
	}

	length := make([]int, n+2)
	countM := 0
	result := -1

	for step, position := range arr {
		leftSize := length[position-1]
		rightSize := length[position+1]

		mergedSize := leftSize + rightSize + 1

		leftBoundary := position - leftSize
		rightBoundary := position + rightSize

		length[leftBoundary] = mergedSize
		length[rightBoundary] = mergedSize

		if leftSize == m {
			countM--
		}

		if rightSize == m {
			countM--
		}

		if mergedSize == m {
			countM++
		}

		if countM > 0 {
			result = step + 1
		}
	}

	return result
}

The Go implementation follows the same logic as the Python version.

One difference is indexing during iteration. Since Go's range loop starts at 0, we store the answer using:

step + 1

to match the problem's one-indexed step numbering.

The length slice is initialized with size n + 2, providing safe boundary padding exactly like the Python version.

No integer overflow concerns exist here because values never exceed n, which is at most 100,000.

Worked Examples

Example 1

arr = [3,5,1,2,4]
m = 1

Step	Position	Left Size	Right Size	Merged Size	count_m	Result
1	3	0	0	1	1	1
2	5	0	0	1	2	2
3	1	0	0	1	3	3
4	2	1	1	3	1	4
5	4	3	1	5	0	4

Step 1:

One group of size 1 exists.

Step 2:

Two groups of size 1 exist.

Step 3:

Three groups of size 1 exist.

Step 4:

The groups around position 2 merge into "111", removing two size-1 groups. One size-1 group remains.

Step 5:

Everything merges into a single size-5 group. No size-1 group remains.

Final answer:

Example 2

arr = [3,1,5,4,2]
m = 2

Step	Position	Left Size	Right Size	Merged Size	Result
1	3	0	0	1	-1
2	1	0	0	1	-1
3	5	0	0	1	-1
4	4	1	1	3	-1
5	2	1	3	5	-1

At no step does a segment of size exactly 2 appear.

Final answer:

-1

Complexity Analysis

Measure	Complexity	Explanation
Time	`O(n)`	Each position is processed once, with constant work per step
Space	`O(n)`	The boundary length array requires linear memory

The time complexity is linear because every flip operation performs only a fixed number of array accesses and updates. No rescanning of the binary string occurs.

The space complexity is also linear because the length array stores one integer per position.

Test Cases

sol = Solution()

assert sol.findLatestStep([3, 5, 1, 2, 4], 1) == 4  # Example 1
assert sol.findLatestStep([3, 1, 5, 4, 2], 2) == -1  # Example 2

assert sol.findLatestStep([1], 1) == 1  # Single element array
assert sol.findLatestStep([1, 2], 2) == 2  # Entire array becomes one group
assert sol.findLatestStep([2, 1], 1) == 1  # Group of size 1 disappears later

assert sol.findLatestStep([1, 2, 3, 4], 4) == 4  # m equals n
assert sol.findLatestStep([1, 2, 3, 4], 3) == 3  # Group appears before final step

assert sol.findLatestStep([2, 1, 4, 3], 2) == 4  # Multiple merges
assert sol.findLatestStep([1, 3, 2], 2) == 2  # Temporary size-2 group
assert sol.findLatestStep([2, 3, 1], 1) == 1  # Latest valid step not final

assert sol.findLatestStep([5, 4, 3, 2, 1], 1) == 1  # Repeated merging removes singles
assert sol.findLatestStep([1, 5, 2, 4, 3], 1) == 4  # Multiple isolated groups

Test	Why
`[3,5,1,2,4], m=1`	Validates first provided example
`[3,1,5,4,2], m=2`	Validates no valid group case
`[1], m=1`	Smallest possible input
`[1,2], m=2`	Entire array becomes valid group
`[2,1], m=1`	Group exists temporarily
`[1,2,3,4], m=4`	Edge case where `m == n`
`[1,2,3,4], m=3`	Group appears before completion
`[2,1,4,3], m=2`	Multiple segment merges
`[1,3,2], m=2`	Temporary valid interval
`[2,3,1], m=1`	Latest valid step occurs early
`[5,4,3,2,1], m=1`	Continuous merging behavior
`[1,5,2,4,3], m=1`	Several independent groups

Edge Cases

Case 1: `m == n`

When the target group size equals the entire array length, there is only one possible valid moment, the final step when all positions have become 1.

This case can easily be mishandled if the algorithm only relies on local merging logic. The implementation explicitly checks:

if m == n:
    return n

which guarantees correctness and avoids unnecessary processing.

Case 2: A Valid Group Exists Temporarily

A group of size m may appear and later disappear due to merging.

For example:

arr = [2,1]
m = 1

Step 1 creates:

which contains a size-1 group.

Step 2 creates:

which removes it.

A buggy implementation might return the first occurrence or only inspect the final configuration. Our implementation updates result whenever count_m > 0, ensuring we capture the latest valid step.

Case 3: Merging Destroys Multiple Valid Groups

Sometimes a newly flipped bit merges two neighboring groups that are both size m.

For example:

left group = size m
right group = size m

Both valid groups disappear simultaneously after merging.

A naive implementation may forget to decrement the count twice. Our algorithm correctly handles this:

if left_size == m:
    count_m -= 1

if right_size == m:
    count_m -= 1

This guarantees that count_m always accurately reflects the number of currently valid groups.

LeetCode 1562 - Find Latest Group of Size M

Solution

LeetCode 1562 - Find Latest Group of Size M

Problem Understanding

Approaches

Brute Force Approach

Key Insight for an Optimal Solution

Algorithm Walkthrough

Step-by-step Algorithm

Why it works

Python Solution

Go Solution

Worked Examples

Example 1

Example 2

Complexity Analysis

Test Cases

Edge Cases

Case 1: m == n

Case 2: A Valid Group Exists Temporarily

Case 3: Merging Destroys Multiple Valid Groups

Case 1: `m == n`