Minimize Maximum Value in a Grid

class Solution {
public:
    vector<vector<int>> minimumScore(vector<vector<int>>& matrix) {
        int rowCount = matrix.size(), colCount = matrix[0].size();

        priority_queue<pair<int, pair<int, int>>,
                       vector<pair<int, pair<int, int>>>,
                       greater<pair<int, pair<int, int>>>>
            pq;

        vector<int> rowMax(rowCount, 1);
        vector<int> colMax(colCount, 1);

        for (int r = 0; r < rowCount; r++) {
            for (int c = 0; c < colCount; c++) {
                pq.push({matrix[r][c], {r, c}});
            }
        }

        while (!pq.empty()) {
            auto current = pq.top();
            pq.pop();

            int value = current.first, r = current.second.first, c = current.second.second;
            int assignValue = max(rowMax[r], colMax[c]);
            matrix[r][c] = assignValue;

            rowMax[r] = assignValue + 1;
            colMax[c] = assignValue + 1;
        }

        return matrix;
    }
};

This C++ solution employs a method to minimize the maximum value in a matrix grid. Implement this solution by following the structured process outlined below:

1. Begin with defining a class Solution that includes the public member function minimumScore.
2. Declare and initialize variables for storing the number of rows (rowCount) and columns (colCount) derived from the matrix dimensions.
3. Utilize a priority queue pq that contains elements arranged in ascending order based on their first pair element (matrix value). Each queue element consists of a matrix value paired with its row and column indices.
4. Set up two vectors rowMax and colMax to track the maximum adjusted value for each row and column, respectively, initialized to 1.
5. Loop through the matrix using nested for-loops, and for each matrix element, push a tuple containing the element value and its position (row and column indices) into the priority queue.
6. Process the priority queue with a while loop until it's empty:
   • Extract the top element (smallest based on value), noting its original row r, column c, and value.
   • Determine the new value to assign to matrix[r][c] using the maximum existing values from rowMax[r] and colMax[c].
   • Update the matrix element at the position with the new maximum value.
   • Increase the values in rowMax and colMax for the respective row and column by 1.
7. After processing all elements, the function returns the adjusted matrix where the maximum values have been minimized subject to the constraints of their respective rows and columns.

By following these steps, the function efficiently adjusts the matrix to minimize the maximum value possible within the constraints provided by the existing matrix structure.

class Solution {

    public int[][] assignMinScores(int[][] matrix) {
        int numRows = matrix.length, numCols = matrix[0].length;

        PriorityQueue<int[]> priorityQueue = new PriorityQueue<>(
            (entry1, entry2) -> entry1[0] - entry2[0]
        );

        int[] maxRowValues = new int[numRows];
        int[] maxColValues = new int[numCols];
        for (int r = 0; r < numRows; r++) maxRowValues[r] = 1;
        for (int c = 0; c < numCols; c++) maxColValues[c] = 1;

        for (int r = 0; r < numRows; r++) {
            for (int c = 0; c < numCols; c++) {
                priorityQueue.offer(new int[] { matrix[r][c], r, c });
            }
        }

        while (!priorityQueue.isEmpty()) {
            int[] current = priorityQueue.poll();
            int value = current[0];
            int row = current[1];
            int col = current[2];

            int assignedValue = Math.max(maxRowValues[row], maxColValues[col]);
            matrix[row][col] = assignedValue;

            maxRowValues[row] = assignedValue + 1;
            maxColValues[col] = assignedValue + 1;
        }

        return matrix;
    }
}

The given Java solution focuses on minimizing the maximum value in a grid, using a priority queue to efficiently manage and update elements based on their values and positions. Here’s a concise breakdown of how the solution works:

• Initialization:

  • Extract the number of rows (numRows) and columns (numCols) from the matrix.
  • Create a PriorityQueue to store matrix elements along with their row and column indices, sorting these elements by their values in ascending order.
  • Initialize two arrays, maxRowValues and maxColValues, to track the maximum values assigned to each row and column, respectively. Initially, every entry in these arrays is set to 1.

• Matrix Traversal:

  • Populate the priority queue with all matrix elements. Each entry in the queue contains the value of the element and its position (row and column indices).

• Processing Each Element:

  • Dequeue elements one by one. For each element:
    • Determine the maximum value that has been assigned to the respective row or column thus far.
    • Assign the maximum of these values to the current matrix cell.
    • Update the maxRowValues for the row and the maxColValues for the column to the newly assigned value plus one.

• Result:

  • The process continues until all elements have been processed and updated.
  • The modified matrix is returned, now having minimized the maximum values in alignment with the constraints defined by their respective row and column.

By strategically updating the matrix using the priority queue, this solution ensures that the values are adjusted in a minimal way while respecting the maximum values observed in their respective rows and columns throughout the process. The use of priority queue helps in handling the elements in the order of their original values, facilitating a systematic and efficient update mechanism.

class Solution:
    def minimumScore(self, matrix: List[List[int]]) -> List[List[int]]:
        rowCount, colCount = len(matrix), len(matrix[0])

        priorityQueue = []
        rowTrack = [1] * rowCount
        colTrack = [1] * colCount

        for x in range(rowCount):
            for y in range(colCount):
                heapq.heappush(priorityQueue, (matrix[x][y], x, y))

        while priorityQueue:
            _, x, y = heapq.heappop(priorityQueue)
            minValue = max(rowTrack[x], colTrack[y])
            matrix[x][y] = minValue
            rowTrack[x] = minValue + 1
            colTrack[y] = minValue + 1

        return matrix

In the provided Python code, you find a method aimed at minimizing the maximum value in a grid represented by a matrix. Here’s a breakdown of how the code operates:

• Initialization

  • The dimensions of the matrix are defined using rowCount and colCount.
  • Tracks rowTrack and colTrack are initialized to keep count of occurrences in each row and column, respectively.
  • A priority queue is utilized to store the matrix elements and their respective positions.

• Processing Each Element in Priority Queue

  • Elements of the matrix are added to the priority queue, ordered by their numerical value.
  • The priority queue is then processed using a while loop. For each element:
    • The method calculates the minimum value between the current values in rowTrack and colTrack for the specific row and column.
    • The matrix element at position (x, y) is updated to this minimum value.
    • The counter in both rowTrack and colTrack at indices x and y respectively are incremented by 1.

• Output

  • After processing all elements, the method returns the updated matrix where each entry is minimized against its maximum possible value considering its row and column history.

This method efficiently minimizes matrix values and manages the spread of larger numbers using a combination of min-heap for orderly processing and tracking arrays for dynamic updating based on previously processed values. This approach ensures that each value only increases minimally from its row or column's preceding values.