How do you solve Find Latest Group of Size M on LeetCode?

Find Latest Group of Size M (LeetCode #1562) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: The problem requires tracking contiguous groups of ones efficiently.

What is the brute force solution for Find Latest Group of Size M?

The brute force approach for Find Latest Group of Size M is Brute Force, with O(n²) time complexity and O(1) space complexity. We can simulate the process of setting bits in the binary string for each step and check for groups of ones of size m after each step. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Find Latest Group of Size M?

Find Latest Group of Size M uses the following concepts: Array, Hash Table, Binary Search, Simulation. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Find Latest Group of Size M?

Always clarify the requirements and constraints of the problem. Think about how you can optimize your brute force solution. Practice explaining your thought process clearly as you code.

What are common mistakes when solving Find Latest Group of Size M?

Not resetting the count of contiguous ones correctly. Overlooking edge cases where groups of size m might exist at the beginning or end.

#1562

Find Latest Group of Size M

Medium

ArrayHash TableBinary SearchSimulationHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

Instead of checking the entire binary string after each step, we can maintain a count of contiguous ones and track the latest step where a group of size m exists using a HashMap.

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Algorithm

5 steps

1Step 1: Initialize a binary string of size n with all bits set to 0 and a HashMap to track the count of contiguous ones.
2Step 2: For each index i from 1 to n, set the bit at position arr[i] to 1.
3Step 3: Maintain a count of contiguous ones and update the HashMap with the latest step for each count of ones.
4Step 4: If the count of ones equals m, update the latest step.
5Step 5: Return the latest step where a group of size m was found, or -1 if none was found.

solution.py20 lines

1# Full working Python code
2
3def latestStep(arr, m):
4    n = len(arr)
5    binary = [0] * n
6    latest_step = -1
7    count = 0
8    for step in range(n):
9        binary[arr[step] - 1] = 1
10        count = 0
11        for i in range(n):
12            if binary[i] == 1:
13                count += 1
14            else:
15                if count == m:
16                    latest_step = step + 1
17                count = 0
18        if count == m:
19            latest_step = step + 1
20    return latest_step

ℹ

Complexity note: The time complexity is O(n) because we only need to iterate through the array once, and the space complexity is O(n) for storing the binary representation.

1The problem requires tracking contiguous groups of ones efficiently.
2Using a HashMap can help in tracking the latest step for each group size.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.