498. Diagonal Traverse LeetCode Solution

In this guide, you will get 498. Diagonal Traverse LeetCode Solution with the best time and space complexity. The solution to Diagonal Traverse problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Table of Contents

1. Problem Statement
2. Complexity Analysis
3. Diagonal Traverse solution in C++
4. Diagonal Traverse solution in Java
5. Diagonal Traverse solution in Python
6. Additional Resources

Problem Statement of Diagonal Traverse

Given an m x n matrix mat, return an array of all the elements of the array in a diagonal order.

Example 1:

Input: mat = [[1,2,3],[4,5,6],[7,8,9]]
Output: [1,2,4,7,5,3,6,8,9]

Example 2:

Input: mat = [[1,2],[3,4]]
Output: [1,2,3,4]

Constraints:

m == mat.length
n == mat[i].length
1 <= m, n <= 104
1 <= m * n <= 104
-105 <= mat[i][j] <= 105

Complexity Analysis

• Time Complexity: O(mn)
• Space Complexity: O(mn)

498. Diagonal Traverse LeetCode Solution in C++

class Solution {
 public:
  vector<int> findDiagonalOrder(vector<vector<int>>& matrix) {
    const int m = matrix.size();
    const int n = matrix[0].size();
    vector<int> ans(m * n);
    int d = 1;  // left-bottom -> right-top
    int row = 0;
    int col = 0;

    for (int i = 0; i < m * n; ++i) {
      ans[i] = matrix[row][col];
      row -= d;
      col += d;
      // out-of-bounds
      if (row == m)
        row = m - 1, col += 2, d = -d;
      if (col == n)
        col = n - 1, row += 2, d = -d;
      if (row < 0)
        row = 0, d = -d;
      if (col < 0)
        col = 0, d = -d;
    }

    return ans;
  }
};
/* code provided by PROGIEZ */

498. Diagonal Traverse LeetCode Solution in Java

class Solution {
  public int[] findDiagonalOrder(int[][] matrix) {
    final int m = matrix.length;
    final int n = matrix[0].length;
    int[] ans = new int[m * n];
    int d = 1; // left-bottom -> right-top
    int row = 0;
    int col = 0;

    for (int i = 0; i < m * n; ++i) {
      ans[i] = matrix[row][col];
      row -= d;
      col += d;
      // out-of-bounds
      if (row == m) {
        row = m - 1;
        col += 2;
        d = -d;
      }
      if (col == n) {
        col = n - 1;
        row += 2;
        d = -d;
      }
      if (row < 0) {
        row = 0;
        d = -d;
      }
      if (col < 0) {
        col = 0;
        d = -d;
      }
    }

    return ans;
  }
}
// code provided by PROGIEZ

498. Diagonal Traverse LeetCode Solution in Python

N/A
 # code by PROGIEZ

Additional Resources

• Explore all LeetCode problem solutions at Progiez here
• Explore all problems on LeetCode website here

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