LeetCode-in-Java

1632. Rank Transform of a Matrix

Hard

Given an m x n matrix, return a new matrix answer where answer[row][col] is the rank of matrix[row][col].

The rank is an integer that represents how large an element is compared to other elements. It is calculated using the following rules:

• The rank is an integer starting from 1.
• If two elements p and q are in the same row or column, then:
  • If p < q then rank(p) < rank(q)
  • If p == q then rank(p) == rank(q)
  • If p > q then rank(p) > rank(q)
• The rank should be as small as possible.

The test cases are generated so that answer is unique under the given rules.

Example 1:

Input: matrix = [[1,2],[3,4]]

Output: [[1,2],[2,3]]

Explanation:

The rank of matrix[0][0] is 1 because it is the smallest integer in its row and column.

The rank of matrix[0][1] is 2 because matrix[0][1] > matrix[0][0] and matrix[0][0] is rank 1.

The rank of matrix[1][0] is 2 because matrix[1][0] > matrix[0][0] and matrix[0][0] is rank 1.

The rank of matrix[1][1] is 3 because matrix[1][1] > matrix[0][1], matrix[1][1] > matrix[1][0], and both matrix[0][1] and matrix[1][0] are rank 2.

Example 2:

Input: matrix = [[7,7],[7,7]]

Output: [[1,1],[1,1]]

Example 3:

Input: matrix = [[20,-21,14],[-19,4,19],[22,-47,24],[-19,4,19]]

Output: [[4,2,3],[1,3,4],[5,1,6],[1,3,4]]

Constraints:

• m == matrix.length
• n == matrix[i].length
• 1 <= m, n <= 500
• -109 <= matrix[row][col] <= 109

Solution

import java.util.Arrays;

public class Solution {
    public int[][] matrixRankTransform(int[][] matrix) {
        int rowCount = matrix.length;
        int colCount = matrix[0].length;
        long[] nums = new long[rowCount * colCount];
        int numsIdx = 0;
        int[] rows = new int[rowCount];
        int[] cols = new int[colCount];

        for (int r = rowCount - 1; r >= 0; r--) {
            for (int c = colCount - 1; c >= 0; c--) {
                nums[numsIdx++] = ((long) matrix[r][c] << 32) | ((long) r << 16) | c;
            }
        }
        Arrays.sort(nums);
        int nIdx = 0;
        while (nIdx < numsIdx) {
            long num = nums[nIdx] & 0xFFFFFFFF00000000L;
            int endIdx = nIdx + 1;
            while (endIdx < numsIdx && ((nums[endIdx] & 0xFFFFFFFF00000000L) == num)) {
                endIdx++;
            }
            doGroup(matrix, nums, nIdx, endIdx, rows, cols);
            nIdx = endIdx;
        }
        return matrix;
    }

    private void doGroup(
            int[][] matrix, long[] nums, int startIdx, int endIdx, int[] rows, int[] cols) {
        if (startIdx + 1 == endIdx) {
            int r = ((int) nums[startIdx] >> 16) & 0xFFFF;
            int c = (int) nums[startIdx] & 0xFFFF;
            matrix[r][c] = rows[r] = cols[c] = Math.max(rows[r], cols[c]) + 1;
        } else {
            int rowCount = matrix.length;
            int[] ufind = new int[rowCount + matrix[0].length];
            Arrays.fill(ufind, -1);
            for (int nIdx = startIdx; nIdx < endIdx; nIdx++) {
                int r = ((int) nums[nIdx] >> 16) & 0xFFFF;
                int c = (int) nums[nIdx] & 0xFFFF;
                int pr = getIdx(ufind, r);
                int pc = getIdx(ufind, rowCount + c);
                if (pr != pc) {
                    ufind[pr] =
                            Math.min(
                                    Math.min(ufind[pr], ufind[pc]),
                                    -Math.max(rows[r], cols[c]) - 1);
                    ufind[pc] = pr;
                }
            }
            for (int nIdx = startIdx; nIdx < endIdx; nIdx++) {
                int r = ((int) nums[nIdx] >> 16) & 0xFFFF;
                int c = (int) nums[nIdx] & 0xFFFF;
                matrix[r][c] = rows[r] = cols[c] = -ufind[getIdx(ufind, r)];
            }
        }
    }

    private int getIdx(int[] ufind, int idx) {
        if (ufind[idx] < 0) {
            return idx;
        } else {
            ufind[idx] = getIdx(ufind, ufind[idx]);
            return ufind[idx];
        }
    }
}

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