1632. Rank Transform of a Matrix

Description

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

 

Solutions

Solution: Union Find

• Time complexity: O(mnlogmn)
• Space complexity: O(mn)

 

JavaScript

/**
 * @param {number[][]} matrix
 * @return {number[][]}
 */
const matrixRankTransform = function (matrix) {
  const m = matrix.length;
  const n = matrix[0].length;
  const valueMap = new Map();

  for (let row = 0; row < m; row++) {
    for (let col = 0; col < n; col++) {
      const value = matrix[row][col];

      if (!valueMap.has(value)) {
        valueMap.set(value, []);
      }

      valueMap.get(value).push({ row, col });
    }
  }

  const sortedValues = [...valueMap.keys()].sort((a, b) => a - b);
  const ranks = Array.from({ length: m + n }, () => 0);
  const result = Array.from({ length: m }, () => new Array(n).fill(0));

  for (const value of sortedValues) {
    const uf = new UnionFind();

    for (const { row, col } of valueMap.get(value)) {
      uf.union(row, col + m);
    }
    const groups = uf.getGroups();

    for (const ids of groups.values()) {
      let maxRank = 0;

      for (const id of ids) {
        maxRank = Math.max(ranks[id], maxRank);
      }

      for (const id of ids) {
        ranks[id] = maxRank + 1;
      }
    }

    for (const { row, col } of valueMap.get(value)) {
      result[row][col] = ranks[row];
    }
  }

  return result;
};

class UnionFind {
  constructor() {
    this.groupMap = new Map();
  }

  find(x) {
    if (!this.groupMap.has(x)) {
      this.groupMap.set(x, x);
    }

    if (this.groupMap.get(x) !== x) {
      const group = this.groupMap.get(x);

      this.groupMap.set(x, this.find(group));
    }

    return this.groupMap.get(x);
  }

  union(x, y) {
    const groupX = this.find(x);
    const groupY = this.find(y);

    if (groupX === groupY) return false;

    this.groupMap.set(groupX, groupY);

    return true;
  }

  getGroups() {
    const groups = new Map();

    for (const x of this.groupMap.keys()) {
      const group = this.find(x);

      if (!groups.has(group)) {
        groups.set(group, []);
      }

      groups.get(group).push(x);
    }

    return groups;
  }
}