2022-10-31 Daily Challenge

Posted on:
Posted modified:
In: DailyChallenge

Today I have done leetcode's October LeetCoding Challenge with cpp.

October LeetCoding Challenge 31

Description

Toeplitz Matrix

Given an m x n matrix, return true if the matrix is Toeplitz. Otherwise, return false.

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

 

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.

 

Constraints:

  • m == matrix.length
  • n == matrix[i].length
  • 1 <= m, n <= 20
  • 0 <= matrix[i][j] <= 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?

Solution

auto speedup = [](){
  cin.tie(nullptr);
  cout.tie(nullptr);
  ios::sync_with_stdio(false);
  return 0;
}();
class Solution {
public:
  bool isToeplitzMatrix(vector<vector<int>>& matrix) {
    int rows = matrix.size();
    int cols = matrix.front().size();
    for(int i = 0; i < cols - 1; ++i) {
      int element = matrix[0][i];
      for(int j = 1; j + i < cols && j < rows; ++j) {
        if(matrix[j][i + j] != element) return false;
      }
    }
    for(int i = 1; i < rows - 1; ++i) {
      int element = matrix[i][0];
      for(int j = 1; i + j < rows && j < cols; ++j) {
        if(matrix[i + j][j] != element) return false;
      }
    }
    return true;
  }
};

// Accepted
// 483/483 cases passed (17 ms)
// Your runtime beats 88.89 % of cpp submissions
// Your memory usage beats 15.16 % of cpp submissions (17.5 MB)
Algorithm