How do you solve Find the Number of Ways to Place People II on LeetCode?

Find the Number of Ways to Place People II (LeetCode #3027) 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: Sorting the points helps in efficiently determining valid pairs.

What is the time complexity of Find the Number of Ways to Place People II?

The optimal solution for Find the Number of Ways to Place People II runs in O(n log n) time and uses O(1) space. The sorting step takes O(n log n) time, and the nested loop runs in O(n²) in the worst case, but the sorting helps us reduce unnecessary checks.

What is the brute force solution for Find the Number of Ways to Place People II?

The brute force approach for Find the Number of Ways to Place People II is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute force approach, we check every possible pair of points to see if they can be Alice's and Bob's positions. This is simple but inefficient, as we have to compare each pair directly. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Find the Number of Ways to Place People II?

Find the Number of Ways to Place People II uses the following concepts: Array, Math, Geometry, Sorting, Enumeration. The recommended patterns to study are: Sorting, Two Pointers.

What are the interview tips for Find the Number of Ways to Place People II?

Always clarify the problem constraints and requirements before jumping into coding. Think about edge cases, such as when all points are in a straight line. Practice explaining your thought process as you code, as this helps in interviews.

What are common mistakes when solving Find the Number of Ways to Place People II?

Not sorting the points correctly before checking pairs. Confusing the conditions for the upper-left and lower-right corners.

#3027

Find the Number of Ways to Place People II

Hard

ArrayMathGeometrySortingEnumerationSortingTwo Pointers

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)

By sorting the points and using a single loop, we can efficiently count valid pairs without needing to check every combination. This reduces the time complexity significantly.

⚙️

Algorithm

6 steps

1Step 1: Sort the points by x-coordinate in non-decreasing order and by y-coordinate in non-increasing order.
2Step 2: Initialize a count variable to 0.
3Step 3: Use a nested loop to iterate through pairs of points (i, j) where i < j.
4Step 4: For each pair, check if points[i][0] <= points[j][0] and points[i][1] >= points[j][1].
5Step 5: If valid, increment the count.
6Step 6: Return the count.

solution.py12 lines

1# Full working Python code
2
3def count_valid_pairs(points):
4    points.sort(key=lambda p: (p[0], -p[1]))
5    count = 0
6    n = len(points)
7    for i in range(n):
8        for j in range(i + 1, n):
9            if points[i][0] <= points[j][0] and points[i][1] >= points[j][1]:
10                count += 1
11    return count
12

ℹ

Complexity note: The sorting step takes O(n log n) time, and the nested loop runs in O(n²) in the worst case, but the sorting helps us reduce unnecessary checks.

1Sorting the points helps in efficiently determining valid pairs.
2The conditions for valid pairs are based on the relative positions of the points in the sorted order.

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