How do you solve Find the Maximum Number of Marked Indices on LeetCode?

Find the Maximum Number of Marked Indices (LeetCode #2576) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(1) space complexity. Key insight: The problem can be efficiently solved by sorting the array and using a two-pointer technique.

What is the brute force solution for Find the Maximum Number of Marked Indices?

The brute force approach for Find the Maximum Number of Marked Indices is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible pair of indices to see if they can be marked. This is straightforward but inefficient as it requires checking all combinations. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Find the Maximum Number of Marked Indices?

Find the Maximum Number of Marked Indices uses the following concepts: Array, Two Pointers, Binary Search, Greedy, Sorting. The recommended patterns to study are: Two Pointers, Greedy Algorithms, Sorting.

What are the interview tips for Find the Maximum Number of Marked Indices?

Always consider sorting the input when dealing with pairing problems. Use clear variable names and maintain a clean state to avoid confusion during pointer movements. Think about the constraints and edge cases before finalizing your approach.

What are common mistakes when solving Find the Maximum Number of Marked Indices?

Not considering the sorted order of the array, leading to missed pairs. Failing to handle edge cases where no pairs can be formed.

#2576

Find the Maximum Number of Marked Indices

Medium

ArrayTwo PointersBinary SearchGreedySortingTwo PointersGreedy AlgorithmsSorting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n log n)Space O(1)

The optimal approach sorts the array and uses a two-pointer technique to efficiently find pairs that can be marked. This reduces the number of checks needed by leveraging the sorted order.

⚙️

Algorithm

5 steps

1Step 1: Sort the array nums.
2Step 2: Initialize two pointers, one at the start (left) and one at the middle (right) of the sorted array.
3Step 3: While both pointers are within bounds, check if 2 * nums[left] <= nums[right]. If true, mark both and move both pointers forward.
4Step 4: If the condition is not met, move the left pointer forward to find a larger value.
5Step 5: Continue until either pointer goes out of bounds, and return the total marked count.

solution.py18 lines

1# Full working Python code
2
3def maxMarkedIndices(nums):
4    nums.sort()
5    n = len(nums)
6    left, right = 0, n // 2
7    count = 0
8    while left < n // 2 and right < n:
9        if 2 * nums[left] <= nums[right]:
10            count += 2
11            left += 1
12            right += 1
13        else:
14            left += 1
15    return count
16
17# Example usage
18print(maxMarkedIndices([3, 5, 2, 4]))  # Output: 2

ℹ

Complexity note: The time complexity is O(n log n) due to the sorting step, while the two-pointer traversal is O(n). The space complexity is O(1) since we are not using any additional data structures that grow with input size.

1The problem can be efficiently solved by sorting the array and using a two-pointer technique.
2The relationship 2 * nums[i] <= nums[j] suggests a pairing strategy where smaller values can be matched with larger values.

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