How do you solve Minimum Number of Moves to Seat Everyone on LeetCode?

Minimum Number of Moves to Seat Everyone (LeetCode #2037) 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 helps in minimizing the distance moved by aligning students and seats.

What is the time complexity of Minimum Number of Moves to Seat Everyone?

The optimal solution for Minimum Number of Moves to Seat Everyone runs in O(n log n) time and uses O(1) space. The time complexity is O(n log n) due to the sorting step, while the space complexity is O(1) since we are using a constant amount of extra space.

What is the brute force solution for Minimum Number of Moves to Seat Everyone?

The brute force approach for Minimum Number of Moves to Seat Everyone is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute force approach, we will try to match each student to each seat and calculate the total number of moves required for each combination. This method is straightforward but inefficient for larger inputs. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Minimum Number of Moves to Seat Everyone?

Minimum Number of Moves to Seat Everyone uses the following concepts: Array, Greedy, Sorting, Counting Sort. The recommended patterns to study are: Greedy, Sorting, Array.

What are the interview tips for Minimum Number of Moves to Seat Everyone?

Always consider sorting when dealing with matching problems. Explain your thought process clearly; interviewers appreciate understanding your reasoning. Practice similar problems to recognize patterns and improve your problem-solving speed.

What are common mistakes when solving Minimum Number of Moves to Seat Everyone?

Not sorting the arrays before matching students to seats. Overcomplicating the problem by trying to find permutations instead of direct matches.

#2037

Minimum Number of Moves to Seat Everyone

Easy

ArrayGreedySortingCounting SortGreedySortingArray

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 both the seats and students arrays, we can match each student to the closest available seat in a greedy manner. This minimizes the total distance moved by each student.

⚙️

Algorithm

3 steps

1Step 1: Sort both the seats and students arrays.
2Step 2: Initialize a variable to keep track of total moves.
3Step 3: Iterate through both arrays and calculate the total moves required to seat each student in the corresponding seat.

solution.py9 lines

1# Full working Python code
2
3def minMovesToSeat(seats, students):
4    seats.sort()
5    students.sort()
6    moves = 0
7    for i in range(len(seats)):
8        moves += abs(seats[i] - students[i])
9    return moves

ℹ

Complexity note: The time complexity is O(n log n) due to the sorting step, while the space complexity is O(1) since we are using a constant amount of extra space.

1Sorting helps in minimizing the distance moved by aligning students and seats.
2Greedy approach works effectively when the problem requires minimizing total costs.

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