# Rank of a Matrix - Google Top Interview Questions

### Problem Statement :

```You are given a two-dimensional list of integers matrix.

Return a new matrix A with the same dimensions such that each A[r][c] is the rank of matrix[r][c].

The ranks of any two elements a and b that are either on the same row or column are calculated by:

rank(a) < rank(b) if a < b

rank(a) > rank(b) if a > b

rank(a) = rank(b) if a = b

Ranks are greater than or equal to 1 but are as small as possible.

Constraints

0 ≤ n, m ≤ 250 where n and m are the number of rows and columns in matrix

Example 1

Input

matrix = [

[5, 1, 3],

[1, 6, 1],

[2, 3, 1]

]

Output

[

[3, 1, 2],

[1, 4, 1],

[2, 3, 1]

]

Example 2

Input

matrix = [

[3, 3],

[3, 3]

]

Output

[

[1, 1],

[1, 1]

]```

### Solution :

```                        ```Solution in C++ :

pair<int, int> find(vector<vector<pair<int, int>>>& parent, pair<int, int> ele) {
auto [x, y] = ele;
if (parent[x][y] == ele) return ele;
return parent[x][y] = find(parent, parent[x][y]);
}

// this function does the union of two elements who have same value and share either the row or
// column.
void unio(vector<vector<pair<int, int>>>& parent, pair<int, int>& a, pair<int, int> b) {
auto [x1, y1] = find(parent, a);
auto [x2, y2] = find(parent, b);
if (x1 == x2 and y1 == y2) {
return;
}
parent[x1][y1] = {x2, y2};
}
vector<vector<int>> solve(vector<vector<int>>& nums) {
int n = nums.size(), m = nums[0].size();

// here we create a parent array with each element pointing to itself.
vector<vector<pair<int, int>>> parent(n, vector<pair<int, int>>(m));

// for keeping the highest rank's in rows and columns.
vector<int> row(n, 0), col(m, 0);

// initializing the dsu
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
parent[i][j] = {i, j};
}
}

// union all the element which share the common row and have equal values.
for (int i = 0; i < n; i++) {
unordered_map<int, pair<int, int>> rows;
for (int j = 0; j < m; j++) {
if (rows.find(nums[i][j]) != rows.end()) {
unio(parent, rows[nums[i][j]], {i, j});
}
rows[nums[i][j]] = {i, j};
}
}

// union all the element which share the common columns and have equal values
for (int j = 0; j < m; j++) {
unordered_map<int, pair<int, int>> cols;
for (int i = 0; i < n; i++) {
if (cols.find(nums[i][j]) != cols.end()) {
unio(parent, cols[nums[i][j]], {i, j});
}
cols[nums[i][j]] = {i, j};
}
}

// now let's make the actual parent mapping. means all the element point to their real parents.
vector<vector<pair<int, int>>> parentmapping(n, vector<pair<int, int>>(m));
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
parentmapping[i][j] = find(parent, {i, j});
}
}

// now let's make the reverse parent mapping where all the elements which are in a connected
// component can be identified.
map<pair<int, int>, vector<pair<int, int>>> rev_mapping;
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
rev_mapping[parentmapping[i][j]].push_back({i, j});
}
}
// so far all the job done for finding the connected component's

// now we start picking the element from the smaller values so we put them in a temporary array
// and start picking the smaller values first.
vector<vector<int>> temp;
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
temp.push_back({nums[i][j], i, j});
}
}
sort(temp.begin(), temp.end());

// this is the array we return
vector<vector<int>> ret(n, vector<int>(m, 0));

// let's start giving the ranks.
vector<vector<int>> visited(n, vector<int>(m, 0));
for (int i = 0; i < temp.size(); i++) {
int vals = temp[i][0], x = temp[i][1], y = temp[i][2];

// if this element has been already ranked we skip the element.
if (visited[x][y] == 1) continue;

// now we find the max rank that has been given to the elements which share the common rows
// and colums to the elements of the connected componenets.
int max_rank_so_far = 0;
pair<int, int> its_parent = parentmapping[x][y];
for (auto& nbrs : rev_mapping[its_parent]) {
max_rank_so_far = max({max_rank_so_far, row[nbrs.first], col[nbrs.second]});
}
// we give the {final rank} as {max rank + 1}.
int f_rank = max_rank_so_far + 1;
// here we give the ranks to the elements and marked them as visited.
for (auto [p, q] : rev_mapping[its_parent]) {
visited[p][q] = 1;
row[p] = col[q] = ret[p][q] = f_rank;
}
}
return ret;
}```
```

```                        ```Solution in Python :

class UnionFind:
def __init__(self):
self._parent = {}
self._size = {}
self._max_rank = {}

def union(self, a, b, rank):
a, b = self.find(a), self.find(b)
if a == b:
return
if self._size[a] < self._size[b]:
a, b = b, a
self._parent[b] = a
self._size[a] += self._size[b]
self._max_rank[a] = max(self._max_rank[a], self._max_rank[b], rank)

def find(self, x):
if x not in self._parent:
self._parent[x] = x
self._size[x] = 1
self._max_rank[x] = 0
return x

while self._parent[x] != x:
self._parent[x] = self._parent[self._parent[x]]
x = self._parent[x]
return x

def max_rank(self, x):
return self._max_rank[self.find(x)]

class Solution:
def solve(self, matrix):
if not matrix or not matrix[0]:
return matrix

m, n = len(matrix), len(matrix[0])
groups = defaultdict(list)

for y, row in enumerate(matrix):
for x, val in enumerate(row):
groups[val].append((y, x))

A = [[0] * n for _ in range(m)]
row_rank = [0] * m
col_rank = [0] * n

for val in sorted(groups):
uf = UnionFind()
for y, x in groups[val]:
uf.union(y, ~x, max(row_rank[y], col_rank[x]))
for y, x in groups[val]:
A[y][x] = row_rank[y] = col_rank[x] = uf.max_rank(y) + 1

return A```
```

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