# 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

• class UnionFind {
private int[] p;
private int[] size;

public UnionFind(int n) {
p = new int[n];
size = new int[n];
for (int i = 0; i < n; ++i) {
p[i] = i;
size[i] = 1;
}
}

public int find(int x) {
if (p[x] != x) {
p[x] = find(p[x]);
}
return p[x];
}

public void union(int a, int b) {
int pa = find(a), pb = find(b);
if (pa != pb) {
if (size[pa] > size[pb]) {
p[pb] = pa;
size[pa] += size[pb];
} else {
p[pa] = pb;
size[pb] += size[pa];
}
}
}

public void reset(int x) {
p[x] = x;
size[x] = 1;
}
}

class Solution {
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];
UnionFind uf = new UnionFind(m + n);
int[] rank = new int[m + n];
for (var ps : d.values()) {
for (var p : ps) {
uf.union(p[0], p[1] + m);
}
for (var p : ps) {
int i = p[0], j = p[1];
rank[uf.find(i)] = Math.max(rank[uf.find(i)], Math.max(rowMax[i], colMax[j]));
}
for (var p : ps) {
int i = p[0], j = p[1];
ans[i][j] = 1 + rank[uf.find(i)];
rowMax[i] = ans[i][j];
colMax[j] = ans[i][j];
}
for (var p : ps) {
uf.reset(p[0]);
uf.reset(p[1] + m);
}
}
return ans;
}
}

• class UnionFind {
public:
UnionFind(int n) {
p = vector<int>(n);
size = vector<int>(n, 1);
iota(p.begin(), p.end(), 0);
}

void unite(int a, int b) {
int pa = find(a), pb = find(b);
if (pa != pb) {
if (size[pa] > size[pb]) {
p[pb] = pa;
size[pa] += size[pb];
} else {
p[pa] = pb;
size[pb] += size[pa];
}
}
}

int find(int x) {
if (p[x] != x) {
p[x] = find(p[x]);
}
return p[x];
}

void reset(int x) {
p[x] = x;
size[x] = 1;
}

private:
vector<int> p, size;
};

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));
UnionFind uf(m + n);
vector<int> rank(m + n);
for (auto& [_, ps] : d) {
for (auto& [i, j] : ps) {
uf.unite(i, j + m);
}
for (auto& [i, j] : ps) {
rank[uf.find(i)] = max({rank[uf.find(i)], rowMax[i], colMax[j]});
}
for (auto& [i, j] : ps) {
ans[i][j] = rowMax[i] = colMax[j] = 1 + rank[uf.find(i)];
}
for (auto& [i, j] : ps) {
uf.reset(i);
uf.reset(j + m);
}
}
return ans;
}
};

• class UnionFind:
def __init__(self, n):
self.p = list(range(n))
self.size = [1] * n

def find(self, x):
if self.p[x] != x:
self.p[x] = self.find(self.p[x])
return self.p[x]

def union(self, a, b):
pa, pb = self.find(a), self.find(b)
if pa != pb:
if self.size[pa] > self.size[pb]:
self.p[pb] = pa
self.size[pa] += self.size[pb]
else:
self.p[pa] = pb
self.size[pb] += self.size[pa]

def reset(self, x):
self.p[x] = x
self.size[x] = 1

class Solution:
def matrixRankTransform(self, matrix: List[List[int]]) -> List[List[int]]:
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)]
uf = UnionFind(m + n)
for v in sorted(d):
rank = defaultdict(int)
for i, j in d[v]:
uf.union(i, j + m)
for i, j in d[v]:
rank[uf.find(i)] = max(rank[uf.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[uf.find(i)]
for i, j in d[v]:
uf.reset(i)
uf.reset(j + m)
return ans


• type unionFind struct {
p, size []int
}

func newUnionFind(n int) *unionFind {
p := make([]int, n)
size := make([]int, n)
for i := range p {
p[i] = i
size[i] = 1
}
return &unionFind{p, size}
}

func (uf *unionFind) find(x int) int {
if uf.p[x] != x {
uf.p[x] = uf.find(uf.p[x])
}
return uf.p[x]
}

func (uf *unionFind) union(a, b int) {
pa, pb := uf.find(a), uf.find(b)
if pa != pb {
if uf.size[pa] > uf.size[pb] {
uf.p[pb] = pa
uf.size[pa] += uf.size[pb]
} else {
uf.p[pa] = pb
uf.size[pb] += uf.size[pa]
}
}
}

func (uf *unionFind) reset(x int) {
uf.p[x] = x
uf.size[x] = 1
}

func matrixRankTransform(matrix [][]int) [][]int {
m, n := len(matrix), len(matrix[0])
type pair struct{ i, j int }
d := map[int][]pair{}
for i, row := range matrix {
for j, v := range row {
d[v] = append(d[v], pair{i, j})
}
}
rowMax := make([]int, m)
colMax := make([]int, n)
ans := make([][]int, m)
for i := range ans {
ans[i] = make([]int, n)
}
vs := []int{}
for v := range d {
vs = append(vs, v)
}
sort.Ints(vs)
uf := newUnionFind(m + n)
rank := make([]int, m+n)
for _, v := range vs {
ps := d[v]
for _, p := range ps {
uf.union(p.i, p.j+m)
}
for _, p := range ps {
i, j := p.i, p.j
rank[uf.find(i)] = max(rank[uf.find(i)], max(rowMax[i], colMax[j]))
}
for _, p := range ps {
i, j := p.i, p.j
ans[i][j] = 1 + rank[uf.find(i)]
rowMax[i], colMax[j] = ans[i][j], ans[i][j]
}
for _, p := range ps {
uf.reset(p.i)
uf.reset(p.j + m)
}
}
return ans
}