grid47 Exploring patterns and algorithms
Sep 18, 2024
5 min read
Solution to LeetCode 498: Diagonal Traverse Problem
You are given an m x n matrix. The task is to return an array that contains all the elements of the matrix in diagonal order, starting from the top-left and moving diagonally towards the bottom-right. The diagonal order alternates between upward and downward diagonals.
Problem
Approach
Steps
Complexity
Input: The input consists of an m x n matrix where each element is an integer.
Example: [[1,2,3],[4,5,6],[7,8,9]]
Constraints:
• 1 <= m, n <= 10^4
• 1 <= m * n <= 10^4
• -10^5 <= mat[i][j] <= 10^5
Output: The output is an array of integers, which represents all the elements of the matrix in diagonal order.
Example: [1, 2, 4, 7, 5, 3, 6, 8, 9]
Constraints:
• The output array contains all the elements from the matrix in diagonal order.
Goal: The goal is to traverse the matrix in diagonal order, alternating between upward and downward diagonals.
Steps:
• 1. Iterate over all diagonals, starting from the top-left corner.
• 2. For each diagonal, alternate the direction of traversal (upward or downward).
• 3. Add elements to the result array in the correct diagonal order.
Goal: The problem has the following constraints:
Steps:
• 1 <= m, n <= 10^4
• 1 <= m * n <= 10^4
• -10^5 <= mat[i][j] <= 10^5
Assumptions:
• The matrix will always have valid integer values.
• The matrix will have at least one element.
• Input: [[1,2,3],[4,5,6],[7,8,9]]
• Explanation: In this example, the elements are traversed diagonally starting from the top-left and moving towards the bottom-right, alternating between upward and downward diagonals.
Approach: The solution involves iterating over all diagonals and alternating the direction of traversal. Each element is added to the result array in the correct order.
Observations:
• The traversal alternates between upward and downward diagonals.
• We need to handle each diagonal separately and ensure the correct order of traversal.
• We can efficiently handle the diagonals by iterating over each one and managing the direction of traversal.
Steps:
• 1. Create a loop to iterate through diagonals, from the first element to the last.
• 2. For each diagonal, alternate the direction of traversal based on the diagonal's index.
• 3. Collect the elements from each diagonal in the result array.
Empty Inputs:
• The matrix will always contain at least one element.
Large Inputs:
• The solution must be efficient enough to handle large matrices (up to 10^4 elements).
Special Values:
• The matrix may contain negative and large values, but they should not affect the solution.
Constraints:
• Ensure the algorithm is optimized for large matrices, especially when m * n approaches 10^4.
vector<int> findDiagonalOrder(vector<vector<int>>& mat) {
int m = mat.size(), n = mat[0].size();
vector<int> res;
for(int s =0; s <= m + n -2; s++) {
for(int x =0; x <= s; x++) {
int i = x;
int j = s - i;
if(s%2==0) swap(i, j);
if(i >= m || j >= n) continue;
res.push_back(mat[i][j]);
}
}
return res;
}
