O(N^3) solution for 50x50 matrix
In this problem we need to not only traverse the matrix, but also each diagonal, bringing the total complexity to N^3 or 125K iterations... code is below, cheers, ACC
Difference of Number of Distinct Values on Diagonals - LeetCode
Given a 0-indexed 2D grid
of size m x n
, you should find the matrix answer
of size m x n
.
The value of each cell (r, c)
of the matrix answer
is calculated in the following way:
- Let
topLeft[r][c]
be the number of distinct values in the top-left diagonal of the cell(r, c)
in the matrixgrid
. - Let
bottomRight[r][c]
be the number of distinct values in the bottom-right diagonal of the cell(r, c)
in the matrixgrid
.
Then answer[r][c] = |topLeft[r][c] - bottomRight[r][c]|
.
Return the matrix answer
.
A matrix diagonal is a diagonal line of cells starting from some cell in either the topmost row or leftmost column and going in the bottom-right direction until reaching the matrix's end.
A cell (r1, c1)
belongs to the top-left diagonal of the cell (r, c)
, if both belong to the same diagonal and r1 < r
. Similarly is defined bottom-right diagonal.
Example 1:
Input: grid = [[1,2,3],[3,1,5],[3,2,1]] Output: [[1,1,0],[1,0,1],[0,1,1]] Explanation: The 1st diagram denotes the initial grid. The 2nd diagram denotes a grid for cell (0,0), where blue-colored cells are cells on its bottom-right diagonal. The 3rd diagram denotes a grid for cell (1,2), where red-colored cells are cells on its top-left diagonal. The 4th diagram denotes a grid for cell (1,1), where blue-colored cells are cells on its bottom-right diagonal and red-colored cells are cells on its top-left diagonal. - The cell (0,0) contains [1,1] on its bottom-right diagonal and [] on its top-left diagonal. The answer is |1 - 0| = 1. - The cell (1,2) contains [] on its bottom-right diagonal and [2] on its top-left diagonal. The answer is |0 - 1| = 1. - The cell (1,1) contains [1] on its bottom-right diagonal and [1] on its top-left diagonal. The answer is |1 - 1| = 0. The answers of other cells are similarly calculated.
Example 2:
Input: grid = [[1]] Output: [[0]] Explanation: - The cell (0,0) contains [] on its bottom-right diagonal and [] on its top-left diagonal. The answer is |0 - 0| = 0.
Constraints:
m == grid.length
n == grid[i].length
1 <= m, n, grid[i][j] <= 50
public int[][] DifferenceOfDistinctValues(int[][] grid) { int[][] retVal = new int[grid.Length][]; for (int r = 0; r < grid.Length; r++) retVal[r] = new int[grid[r].Length]; for (int r = 0; r < grid.Length; r++) { for (int c = 0; c < grid[r].Length; c++) { Hashtable htLeft = new Hashtable(); int lr = r - 1; int lc = c - 1; while (lr >= 0 && lc >= 0) { if (!htLeft.ContainsKey(grid[lr][lc])) htLeft.Add(grid[lr][lc], true); lr--; lc--; } Hashtable htRight = new Hashtable(); int rr = r + 1; int rc = c + 1; while (rr < grid.Length && rc < grid[rr].Length) { if (!htRight.ContainsKey(grid[rr][rc])) htRight.Add(grid[rr][rc], true); rr++; rc++; } retVal[r][c] = Math.Abs(htLeft.Count - htRight.Count); } } return retVal; }
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