Minimum Number of Moves to Seat Everyone

class Solution {
public:
    int seatStudents(vector<int>& seats, vector<int>& students) {
        // Calculate the greatest position among seats and students
        int maxPosition = max(getMaxPosition(seats), getMaxPosition(students));

        // Differences array keeps track of seat and student disparity at positions
        vector<int> positionDifferences(maxPosition, 0);

        // Increment for each seat at a specific position
        for (int seat : seats) {
            positionDifferences[seat - 1]++;
        }

        // Decrement for each student at a specific position
        for (int student : students) {
            positionDifferences[student - 1]--;
        }

        // Accumulate the total moves required to seat all students
        int totalMoves = 0, balance = 0;
        for (int difference : positionDifferences) {
            totalMoves += abs(balance);
            balance += difference;
        }

        return totalMoves;
    }

private:
    int getMaxPosition(const vector<int>& data) {
        int maxVal = 0;
        for (int value : data) {
            if (value > maxVal) {
                maxVal = value;
            }
        }
        return maxVal;
    }
};

In the given C++ solution, the problem of seating students in the minimum number of moves is solved efficiently. Here's a summary of how the solution operates:

• The solution defines a Solution class with two primary methods: seatStudents for solving the problem and a helper method getMaxPosition to find the maximum seat or student position.

• Inside seatStudents, it starts by determining the maximum position value among the seats and students using the helper method. This helps in creating an adequately sized vector positionDifferences to track the difference in seat availability at each position.

• The function then iterates through the seats array, increasing the corresponding index in the positionDifferences array for each seat found. Following that, it decreases the count at each index in the same positionDifferences array for each student position.

• A running balance is maintained while iterating through positionDifferences to calculate how many moves are required to seat all students properly. The balance tracks the net position discrepancies between available seats and students, accumulating the absolute value of this running balance to totalMoves.

This approach ensures that the total number of moves made to seat all students is minimized by directly resolving seat and student count imbalances across possible positions.

class Solution {

    public int calculateMinimumMoves(int[] seatsArray, int[] studentsArray) {
        int highestPosition = Math.max(getMaxValue(seatsArray), getMaxValue(studentsArray));

        int[] countDifferences = new int[highestPosition];

        for (int seat : seatsArray) {
            countDifferences[seat - 1]++;
        }

        for (int student : studentsArray) {
            countDifferences[student - 1]--;
        }

        int totalMoves = 0;
        int unpaired = 0;
        for (int diff : countDifferences) {
            totalMoves += Math.abs(unpaired);
            unpaired += diff;
        }

        return totalMoves;
    }

    private int getMaxValue(int[] values) {
        int max = 0;
        for (int value : values) {
            if (value > max) {
                max = value;
            }
        }
        return max;
    }
}

This Java solution addresses the problem of finding the minimum number of moves required to seat everyone by comparing two integer arrays: seatsArray, representing seat positions, and studentsArray, representing student positions.

The key steps involved in the solution are as follows:

1. Calculate the highest seat or student position to define the range of position differences.
2. Initialize an array countDifferences with a size equal to the highest position, used to record the imbalance between seats and students at each position.
3. Iterate through seatsArray and increment the corresponding index in countDifferences for populated seats.
4. Iterate through studentsArray and decrement the corresponding index in countDifferences for each student.
5. Calculate the total moves required to balance the differences:
   • Initialize a counter totalMoves to accumulate the total moves necessary.
   • Use an intermediate variable unpaired to track the cumulative surplus or deficit of seats to students as you iterate through countDifferences.
   • For each iteration, movements required are updated by adding the absolute value of unpaired to totalMoves.
   • Update unpaired with the current difference from countDifferences.

The function getMaxValue helps in determining the maximum value in an array, which is essential for properly sizing the countDifferences array and avoiding indexing errors.

Following these steps effectively helps ensure that seat assignments require the minimal amount of movement, optimizing the process of seating every student.

class Solution:
    def minMoves(self, seats: List[int], students: List[int]) -> int:
        maximum_value = max(max(seats), max(students))
        difference_counts = [0] * maximum_value

        for seat in seats:
            difference_counts[seat - 1] += 1

        for student in students:
            difference_counts[student - 1] -= 1

        total_moves = 0
        excess = 0
        for diff in difference_counts:
            total_moves += abs(excess)
            excess += diff

        return total_moves

The provided Python code solves the problem of finding the minimum number of moves required to seat everyone by matching seats and students. The core idea is to minimize the absolute distance seats have to move to accommodate students.

Here's how the solution works:

• Determine the maximum_value from the union of the maximum values in both seats and students lists.
• Create a list, difference_counts, initialized to zeroes with a length based on maximum_value. This list tracks the net availability of seats at each position.
• Iterate over each seat and increment the corresponding index in difference_counts, indicating that a seat is available.
• Iterate over each student and decrement the corresponding index in difference_counts, indicating a student needs a seat.
• Initialize total_moves to zero, which will accumulate the total moves needed. excess keeps track of the current imbalance of seats vs students as the algorithm assesses each position in difference_counts.
• Loop through each value in difference_counts, using excess to adjust and track the net offset of students versus available seats and updating total_moves with the absolute value of excess.

Finally, the function returns total_moves, representing the minimum moves necessary to seat all students.