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:
1.p and q are in the same row or column, then:
p < q then rank(p) < rank(q)p == q then rank(p) == rank(q)p > q then rank(p) > rank(q)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.lengthn == matrix[i].length1 <= m, n <= 500-109 <= matrix[row][col] <= 109This 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.
The 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.
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.
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.
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.
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.
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.
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.
JavaScript
Time Complexity: O((m * n) log(m * n)) due to sorting, and space complexity is O(m * n) for storing ranks and unions.
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.
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.
Java
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.
| Approach | Complexity |
|---|---|
| Union-Find and Coordinate Compression | 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. |
| Sorting Rows and Columns | Time Complexity: O(m * n * log(m * n)) because of sorting and complexity from the union-find operations. |
| Approach 1: Coordinate Compression with Union-Find | Time Complexity: O((m * n) log(m * n)) due to sorting, and space complexity is O(m * n) for storing ranks and unions. |
| Approach 2: BFS Rank Propagation | 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. |
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