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This approach utilizes the Union-Find data structure to associate ranks across the same row and column when we process elements in increasing order. We use coordinate compression to ensure that elements are processed efficiently by their unique values in increasing order. The process maintains a rank per element while ensuring rank relationships across the matrix.
Time Complexity: O(m * n * log(m * n)), where m and n are the dimensions of the matrix. The log factor comes from union-find operations.
Space Complexity: O(m * n), due to the storage demands of the union-find structure and result matrix.
1def matrixRankTransform(matrix):
2 import collections
3
4 def find(v):
5 if parent[v] != v:
6 parent[v] = find(parent[v])
7 return parent[v]
8
9 def union(v1, v2):
10 root1 = find(v1)
11 root2 = find(v2)
12 if root1 != root2:
13 if rank[root1] > rank[root2]:
14 parent[root2] = root1
15 else:
16 parent[root1] = root2
17 if rank[root1] == rank[root2]:
18 rank[root2] += 1
19
20 m, n = len(matrix), len(matrix[0])
21 rank_map = collections.defaultdict(list)
22 for r in range(m):
23 for c in range(n):
24 rank_map[matrix[r][c]].append((r, c))
25
26 result = [[0] * n for _ in range(m)]
27 row_max, col_max = [0] * m, [0] * n
28
29 for value in sorted(rank_map):
30 parent = {}
31 rank = {}
32
33 for r, c in rank_map[value]:
34 key = (r, c)
35 parent[key] = key
36 rank[key] = 0
37
38 for r, c in rank_map[value]:
39 if r > 0 and matrix[r-1][c] == value:
40 union((r-1, c), (r, c))
41 if c > 0 and matrix[r][c-1] == value:
42 union((r, c-1), (r, c))
43
44 rank_set = collections.defaultdict(list)
45 for r, c in rank_map[value]:
46 root = find((r, c))
47 rank_set[root].append((r, c))
48
49 for group in rank_set.values():
50 max_rank = 0
51 for r, c in group:
52 max_rank = max(max_rank, row_max[r], col_max[c])
53 new_rank = max_rank + 1
54 for r, c in group:
55 result[r][c] = new_rank
56 row_max[r] = new_rank
57 col_max[c] = new_rank
58
59 return resultThe solution defines a clustering strategy using a union-find structure with path compression and union by rank. Elements in the matrix are grouped based on their values and are processed by increasing order of the element value. For each value, rows and columns are traversed to determine minimal ranks possible, and these ranks are then assigned to the rows and columns. The algorithm utilizes a union-find to merge similar values found in previous and subsequent cells of the matrix.
We can also solve this problem using a greedy strategy where each value is processed independently by sorting rows and columns. This involves sorting and updating ranks based on maintaining column and row data separately with sorting. This technique ensures an optimal rank assignment at each step by having separate row and column maximum rank tracking.
Time Complexity: O(m * n * log(m * n)) because of sorting and complexity from the union-find operations.
Space Complexity: O(m + n) for union-find operation manipulation and storage for returned results.
1function matrixRankTransform(matrix) {
2 const m = matrix.length,
This approach uses coordinate compression with a union-find data structure to efficiently calculate ranks. By representing elements, their row and column constraints via sets, we can manage ranks dynamically as we process elements from smallest to largest.
Time Complexity: O((m * n) log(m * n)) due to sorting, and space complexity is O(m * n) for storing ranks and unions.
1def matrixRankTransform(matrix):
2 from collections
This approach employs a Breadth-First Search (BFS) strategy to propagate the ranks throughout the matrix. By utilizing a leveling mechanism, where each element will check its neighbors and assign the rank based on the relative position, the implementation aims to be more intuitive.
Time Complexity: O(m * n), since each cell is determined in a linear pass across sorted values in range buckets. Space Complexity: O(1) additional space beyond input and output sizes, due to row and col tracking indices.
1#include <vector>
2#include <queue>
3#include <algorithm>
4#include <unordered_map>
5#include <utility>
using namespace std;
vector<vector<int>> matrixRankTransform(vector<vector<int>>& matrix) {
int m = matrix.size(), n = matrix[0].size();
vector<vector<int>> result(m, vector<int>(n, 0));
vector<unordered_map<int, int>> row(m), col(n);
vector<queue<pair<int, int>>> queues(1001);
for (int i = 0; i < m; ++i)
for (int j = 0; j < n; ++j)
queues[matrix[i][j] + 500].push({i, j});
for (int val = -500; val <= 500; ++val) {
auto& currQueue = queues[val + 500];
while (!currQueue.empty()) {
auto [r, c] = currQueue.front(); currQueue.pop();
int rank = max(row[r][val], col[c][val]) + 1;
result[r][c] = rank;
row[r][val] = max(row[r][val], rank);
col[c][val] = max(col[c][val], rank);
}
}
return result;
}
The JavaScript solution involves sorting all entries by their values and unionizing their row and column indices. With each group of same-value entries, maximum possible ranks are assigned based on the leading rank values of rows and columns. The unique ranks are formed by later tracking the maximum of the unions established and annotated back to the full matrix grid.
This solution sorts elements and processes them using a union-find algorithm to ensure that all elements with the same value and in the same row or column receive the same rank. It keeps track of the maximum rank imposed by previous elements in a separate array and updates these in the union-find process.
The C++ code takes advantage of a range-buckets mechanism (from -500 to 500, offsetting with 500) to store indices of matrix values. A BFS-style level processing is utilized to assign ranks iterated linearly through pre-bucketed values.