Toeplitz Matrix LT766

A matrix is Toeplitz if every diagonal from top-left to bottom-right has the same element.

Now given an M x N matrix, return True if and only if the matrix is Toeplitz.

Example 1:

Input:
matrix = [
  [1,2,3,4],
  [5,1,2,3],
  [9,5,1,2]
]
Output: True
Explanation:
In the above grid, the diagonals are:
"[9]", "[5, 5]", "[1, 1, 1]", "[2, 2, 2]", "[3, 3]", "[4]".
In each diagonal all elements are the same, so the answer is True.

Example 2:

Input:
matrix = [
  [1,2],
  [2,2]
]
Output: False
Explanation:
The diagonal "[1, 2]" has different elements.

Note:

matrix will be a 2D array of integers.
matrix will have a number of rows and columns in range [1, 20].
matrix[i][j] will be integers in range [0, 99].

Follow up:

What if the matrix is stored on disk, and the memory is limited such that you can only load at most one row of the matrix into the memory at once?
What if the matrix is so large that you can only load up a partial row into the memory at once?

Idea 1. Observe the example, found that all diagnals have the same k = rowIndex - columIndex, k in the range [M-1, -(N-1)]

Time complexity: O(M * N)

Space comlexity: O(1)

 1 class Solution {
 2     public boolean isToeplitzMatrix(int[][] matrix) {
 3         int M = matrix.length;
 4         int N = matrix[0].length;
 5         
 6         for(int k = M-1; k >= -(N-1); --k) {
 7             for(int i = M-2; i>= 0 && (i - k) >= 0; --i) {
 8                 if(i+1-k <= N-1 && matrix[i][i-k] != matrix[i+1][i+1-k]) {
 9                     return false;
10                 }
11             }
12         }
13         
14         return true;
15     }
16 }

Idea 1.a loop the matrix row by row, use hashMap to store the diffrence r - c, return false if one element has different r-c, much simpler logic

Time complexity: O(M*N)

Space complexity: O(M+N)

 1 class Solution {
 2     public boolean isToeplitzMatrix(int[][] matrix) {
 3         int M = matrix.length;
 4         int N = matrix[0].length;
 5         
 6         Map<Integer, Integer> diagonals = new HashMap<>();
 7         for(int i = 0; i < M; ++i) {
 8             for(int j = 0; j < N; ++j) {
 9                 diagonals.putIfAbsent(i-j, matrix[i][j]);
10                 if(matrix[i][j] != diagonals.get(i-j)) {
11                     return false;
12                 }
13             }
14         }
15         
16         return true;
17       
18     }
19 }

Idea 1.c another observation would be make the code much easier, A[i][j] == A[i+1][j+1] on the same diagonal

Time complexity: O(M * N)

Space complexity: O(1)

 1 class Solution {
 2     public boolean isToeplitzMatrix(int[][] matrix) {
 3         int M = matrix.length;
 4         int N = matrix[0].length;
 5         
 6         for(int i = 0; i < M; ++i) {
 7             for(int j = 0; j < N; ++j) {
 8                if(i+1 < M && j+1 < N && matrix[i][j] != matrix[i+1][j+1]) {
 9                    return false;
10                }
11             }
12         }
13         
14         return true;
15     }
16 }

Follow up 1:

Idea 1. Store the previous row in a queue from right to left, store the element in the queue if it has neighbour on the next row [i+1][j+1], compare buffer[j] = matrix[i+1][j+1]

Time complexity: O(M*N)

Space complexity: O(N)

 1 class Solution {
 2     public boolean isToeplitzMatrix(int[][] matrix) {
 3         int M = matrix.length;
 4         int N = matrix[0].length;
 5         
 6         Deque<Integer> buffer = new ArrayDeque<>();
 7         for(int j = N-2; j >= 0; --j) {
 8             buffer.addLast(matrix[0][j]);
 9         }
10         
11         
12         for(int i = 1; i < M; ++i) {
13             for(int j = N-1; j >= 1; --j) {
14                 int prev = buffer.removeFirst();
15                 if(prev != matrix[i][j]) {
16                     return false;
17                 }
18                 if(j <= N-2) {
19                     buffer.addLast(matrix[i][j]);
20                 }
21             }
22             buffer.addLast(matrix[i][0]);
23         }
24         return true;
25     }
26 }

Follow up 2: split the row into mulitple partial row, and do row comparison like follow up 1 for a block with partial matrix only, then continue for the next block unitl all the blocks are done.

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