README_EN.md

1632. Rank Transform of a Matrix

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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

Python3

class Solution:
    def matrixRankTransform(self, matrix: List[List[int]]) -> List[List[int]]:
        def find(x):
            if p.setdefault(x, x) != x:
                p[x] = find(p[x])
            return p[x]

        def union(a, b):
            p[find(a)] = find(b)

        m, n = len(matrix), len(matrix[0])
        d = defaultdict(list)
        for i, row in enumerate(matrix):
            for j, v in enumerate(row):
                d[v].append((i, j))
        row_max = [0] * m
        col_max = [0] * n
        ans = [[0] * n for _ in range(m)]
        for v in sorted(d):
            p, rank = {}, defaultdict(int)
            for i, j in d[v]:
                union(i, j + 500)
            for i, j in d[v]:
                rank[find(i)] = max(rank[find(i)], row_max[i], col_max[j])
            for i, j in d[v]:
                ans[i][j] = row_max[i] = col_max[j] = 1 + rank[find(i)]
        return ans

Java

class Solution {
    private Map<Integer, Integer> p = new HashMap<>();
    private Map<Integer, Integer> rank = new HashMap<>();

    public int[][] matrixRankTransform(int[][] matrix) {
        int m = matrix.length, n = matrix[0].length;
        TreeMap<Integer, List<int[]>> d = new TreeMap<>();
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                d.computeIfAbsent(matrix[i][j], k -> new ArrayList<>()).add(new int[] {i, j});
            }
        }
        int[] rowMax = new int[m];
        int[] colMax = new int[n];
        int[][] ans = new int[m][n];
        for (var e : d.entrySet()) {
            int v = e.getKey();
            var g = e.getValue();
            p.clear();
            rank.clear();
            for (int[] x : g) {
                union(x[0], x[1] + 500);
            }
            for (int[] x : g) {
                int i = x[0], j = x[1];
                rank.put(find(i),
                    Math.max(rank.getOrDefault(find(i), 0), Math.max(rowMax[i], colMax[j])));
            }
            for (int[] x : g) {
                int i = x[0], j = x[1];
                ans[i][j] = 1 + rank.getOrDefault(find(i), 0);
                rowMax[i] = ans[i][j];
                colMax[j] = ans[i][j];
            }
        }
        return ans;
    }

    private void union(int a, int b) {
        p.put(find(a), find(b));
    }

    private int find(int x) {
        p.putIfAbsent(x, x);
        if (p.get(x) != x) {
            p.put(x, find(p.get(x)));
        }
        return p.get(x);
    }
}

C++

class Solution {
public:
    vector<vector<int>> matrixRankTransform(vector<vector<int>>& matrix) {
        int m = matrix.size(), n = matrix[0].size();
        map<int, vector<pair<int, int>>> d;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                d[matrix[i][j]].push_back({i, j});
            }
        }
        vector<int> rowMax(m);
        vector<int> colMax(n);
        vector<vector<int>> ans(m, vector<int>(n));
        for (auto& [v, g] : d) {
            unordered_map<int, int> p;
            unordered_map<int, int> rank;
            for (auto [i, j] : g) {
                unite(i, j + 500, p);
            }
            for (auto [i, j] : g) {
                rank[find(i, p)] = max(rank[find(i, p)], max(rowMax[i], colMax[j]));
            }
            for (auto [i, j] : g) {
                ans[i][j] = rowMax[i] = colMax[j] = 1 + rank[find(i, p)];
            }
        }
        return ans;
    }

    void unite(int a, int b, unordered_map<int, int>& p) {
        p[find(a, p)] = find(b, p);
    }

    int find(int x, unordered_map<int, int>& p) {
        if (!p.count(x)) p[x] = x;
        if (p[x] != x) p[x] = find(p[x], p);
        return p[x];
    }
};

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