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:
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.
Example 2:
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.length1 <= n <= 1001 <= seats[i], students[j] <= 100This approach relies on sorting both the seats and students arrays. After sorting them, we pair the i-th seat with the i-th student. The sum of absolute differences between corresponding elements gives us the minimum number of moves required.
We define a comparison function cmpfunc for sorting. Using qsort, both arrays are sorted. The absolute differences between paired elements from sorted arrays are summed to get the minimum moves.
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Time Complexity: O(n log n) due to sorting.
Space Complexity: O(1) since sorting is in-place.
This approach involves creating frequency arrays or maps for both seats and students positions. Align students to seats by finding the closest available seat iteratively, simulating a greedy approach.
Initialize frequency arrays for both seats and students. Then, use two pointers to traverse and match the closest available seat to each student, updating movements accordingly.
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Time Complexity: O(n + m), where m is the range of possible positions (maximum 100).
Space Complexity: O(m) due to frequency arrays.
| Approach | Complexity |
|---|---|
| Sorting and Pairing | Time Complexity: O(n log n) due to sorting. |
| Greedy with Frequency Arrays | Time Complexity: O(n + m), where m is the range of possible positions (maximum 100). |
Minimum Number of Moves to Seat Everyone - Leetcode 2037 - Python • NeetCodeIO • 8,845 views views
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