2037. Minimum Number of Moves to Seat Everyone LeetCode Solution
In this guide, you will get 2037. Minimum Number of Moves to Seat Everyone LeetCode Solution with the best time and space complexity. The solution to Minimum Number of Moves to Seat Everyone problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.
Problem Statement of Minimum Number of Moves to Seat Everyone
There are n availabe seats and n students standing in a room. You are given an array seats of length n, where seats[i] is the position of the ith seat. You are also given the array students of length n, where students[j] is the position of the jth student.
You may perform the following move any number of times:
Increase or decrease the position of the ith student by 1 (i.e., moving the ith student from position x to x + 1 or x – 1)
Return the minimum number of moves required to move each student to a seat such that no two students are in the same seat.
Note that there may be multiple seats or students in the same position at the beginning.
Example 1:
Input: seats = [3,1,5], students = [2,7,4]
Output: 4
Explanation: The students are moved as follows:
– The first student is moved from position 2 to position 1 using 1 move.
– The second student is moved from position 7 to position 5 using 2 moves.
– The third student is moved from position 4 to position 3 using 1 move.
In total, 1 + 2 + 1 = 4 moves were used.
Input: seats = [4,1,5,9], students = [1,3,2,6]
Output: 7
Explanation: The students are moved as follows:
– The first student is not moved.
– The second student is moved from position 3 to position 4 using 1 move.
– The third student is moved from position 2 to position 5 using 3 moves.
– The fourth student is moved from position 6 to position 9 using 3 moves.
In total, 0 + 1 + 3 + 3 = 7 moves were used.
Example 3:
Input: seats = [2,2,6,6], students = [1,3,2,6]
Output: 4
Explanation: Note that there are two seats at position 2 and two seats at position 6.
The students are moved as follows:
– The first student is moved from position 1 to position 2 using 1 move.
– The second student is moved from position 3 to position 6 using 3 moves.
– The third student is not moved.
– The fourth student is not moved.
In total, 1 + 3 + 0 + 0 = 4 moves were used.
Constraints:
n == seats.length == students.length
1 <= n <= 100
1 <= seats[i], students[j] <= 100
Complexity Analysis
Time Complexity: O(\texttt{sort})
Space Complexity: O(n)
2037. Minimum Number of Moves to Seat Everyone LeetCode Solution in C++
class Solution {
public:
int minMovesToSeat(vector<int>& seats, vector<int>& students) {
int ans = 0;
ranges::sort(seats);
ranges::sort(students);
for (int i = 0; i < seats.size(); ++i)
ans += abs(seats[i] - students[i]);
return ans;
}
}; /* code provided by PROGIEZ */
2037. Minimum Number of Moves to Seat Everyone LeetCode Solution in Java
class Solution {
public int minMovesToSeat(int[] seats, int[] students) {
int res = 0;
Arrays.sort(seats);
Arrays.sort(students);
for (int i = 0; i < seats.length; ++i)
res += Math.abs(seats[i] - students[i]);
return res;
}
} // code provided by PROGIEZ
2037. Minimum Number of Moves to Seat Everyone LeetCode Solution in Python
class Solution:
def minMovesToSeat(self, seats: list[int], students: list[int]) -> int:
return sum(
abs(seat - student) for seat,
student in zip(sorted(seats),
sorted(students))) # code by PROGIEZ
