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Formatted question description: https://leetcode.ca/all/1329.html

1329. Sort the Matrix Diagonally (Medium)

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. For example, the matrix diagonal starting from mat[2][0], where mat is a 6 x 3 matrix, includes cells mat[2][0], mat[3][1], and mat[4][2].

Given an m x n matrix mat of integers, sort each matrix diagonal in ascending order and return the resulting matrix.

 

Example 1:

Input: mat = [[3,3,1,1],[2,2,1,2],[1,1,1,2]]
Output: [[1,1,1,1],[1,2,2,2],[1,2,3,3]]

 

Constraints:

• m == mat.length
• n == mat[i].length
• 1 <= m, n <= 100
• 1 <= mat[i][j] <= 100

Related Topics:
Array, Sort

Solution 1.

• Java
• C++
• Python
• Go

• class Solution {
      public int[][] diagonalSort(int[][] mat) {
          int rows = mat.length, columns = mat[0].length;
          int startRow = rows - 2, startColumn = 0;
          while (startRow > 0) {
              List<Integer> diagonalList = new ArrayList<Integer>();
              for (int i = startRow, j = startColumn; i < rows && j < columns; i++, j++)
                  diagonalList.add(mat[i][j]);
              Collections.sort(diagonalList);
              for (int i = startRow, j = startColumn, index = 0; i < rows && j < columns; i++, j++, index++)
                  mat[i][j] = diagonalList.get(index);
              startRow--;
          }
          if (startRow < 0)
              return mat;
          while (startColumn < columns) {
              List<Integer> diagonalList = new ArrayList<Integer>();
              for (int i = startRow, j = startColumn; i < rows && j < columns; i++, j++)
                  diagonalList.add(mat[i][j]);
              Collections.sort(diagonalList);
              for (int i = startRow, j = startColumn, index = 0; i < rows && j < columns; i++, j++, index++)
                  mat[i][j] = diagonalList.get(index);
              startColumn++;
          }
          return mat;
      }
  }
  
  ############
  
  class Solution {
      public int[][] diagonalSort(int[][] mat) {
          int m = mat.length, n = mat[0].length;
          for (int k = 0; k < Math.min(m, n) - 1; ++k) {
              for (int i = 0; i < m - 1; ++i) {
                  for (int j = 0; j < n - 1; ++j) {
                      if (mat[i][j] > mat[i + 1][j + 1]) {
                          int t = mat[i][j];
                          mat[i][j] = mat[i + 1][j + 1];
                          mat[i + 1][j + 1] = t;
                      }
                  }
              }
          }
          return mat;
      }
  }
  

• // OJ: https://leetcode.com/problems/sort-the-matrix-diagonally/
  // Time: O((M + N) * XlogX) where X is min(M, N) 
  // Space: O(X)
  class Solution {
  public:
      vector<vector<int>> diagonalSort(vector<vector<int>>& A) {
          int M = A.size(), N = A[0].size();
          for (int i = 0; i < M; ++i) {
              vector<int> v;
              for (int x = i, y = 0; x < M && y < N; ++x, ++y) v.push_back(A[x][y]);
              sort(begin(v), end(v));
              for (int x = i, y = 0; x < M && y < N; ++x, ++y) A[x][y] = v[y];
          }
          for (int j = 1; j < N; ++j) {
              vector<int> v;
              for (int x = 0, y = j; x < M && y < N; ++x, ++y) v.push_back(A[x][y]);
              sort(begin(v), end(v));
              for (int x = 0, y = j; x < M && y < N; ++x, ++y) A[x][y] = v[x];
          }
          return A;
      }
  };
  

• class Solution:
      def diagonalSort(self, mat: List[List[int]]) -> List[List[int]]:
          m, n = len(mat), len(mat[0])
          for k in range(min(m, n) - 1):
              for i in range(m - 1):
                  for j in range(n - 1):
                      if mat[i][j] > mat[i + 1][j + 1]:
                          mat[i][j], mat[i + 1][j + 1] = mat[i + 1][j + 1], mat[i][j]
          return mat
  
  
  

• func diagonalSort(mat [][]int) [][]int {
  	m, n := len(mat), len(mat[0])
  	for k := 0; k < m-1 && k < n-1; k++ {
  		for i := 0; i < m-1; i++ {
  			for j := 0; j < n-1; j++ {
  				if mat[i][j] > mat[i+1][j+1] {
  					mat[i][j], mat[i+1][j+1] = mat[i+1][j+1], mat[i][j]
  				}
  			}
  		}
  	}
  	return mat
  }
  

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