Diagonal Traverse II

Problem Statement

The task is to return all elements of a given 2D integer array nums according to a specific diagonal traversal pattern. Diagonal traversal involves moving along the diagonals of the array matrix, starting from the top left corner, progressing diagonally downwards to the right, and then continuing the pattern until all elements are covered.

Examples

Example 1

Input:

nums = [[1,2,3],[4,5,6],[7,8,9]]

Output:

[1,4,2,7,5,3,8,6,9]

Example 2

Input:

nums = [[1,2,3,4,5],[6,7],[8],[9,10,11],[12,13,14,15,16]]

Output:

[1,6,2,8,7,3,9,4,12,10,5,13,11,14,15,16]

Constraints

• 1 <= nums.length <= 105
• 1 <= nums[i].length <= 105
• 1 <= sum(nums[i].length) <= 105
• 1 <= nums[i][j] <= 105

Approach and Intuition

Given the problem's requirements, our objective is to systematically traverse the 2D array in diagonals, starting from the element at the top-left. Here’s a detailed breakdown of the approach:

1. Initialize a dictionary or a list to gather elements by their corresponding diagonals. Each key in the dictionary can represent a diagonal, and the values are the elements in that diagonal.
2. Traverse through the array and place each element into the dictionary based on the sum of their index. Elements nums[i][j] will go into the diagonal labeled (i + j).
3. Each diagonal should be processed for the output - diagonals with an even key will keep the order as is, and diagonals with an odd key will reverse their order. This alternates the diagonals' direction as required.
4. Finally, concatenate all diagnostics sequentially based on their key values to get the final list.

Given the examples provided:

• For the first example, starting with the first element [0][0], we move diagonally to [1][0] and [0][1], and so on, finally the traversal fits perfectly into the requirements by the method of alternating the direction and using indexed placement.
• For the second example, the different lengths of the rows introduce variance in diagonal lengths, yet the approach using indexed summation (i+j) suitably categorizes each into the right diagonal sequence.

This method efficiently accumulates elements and achieves the diagonal order output using the constraints and properties established:

• The array dimensions are flexible (with given constraints on maximum size).
• The values are within the specified range and don’t affect the traversal order, but it's crucial to implement the solution to manage space complexity due to potentially large sizes.

Solutions

class Solution {
public:
    vector<int> traverseDiagonally(vector<vector<int>>& matrix) {
        queue<pair<int, int>> traversalQueue;
        traversalQueue.push({0, 0});
        vector<int> result;
            
        while (!traversalQueue.empty()) {
            auto [r, c] = traversalQueue.front();
            traversalQueue.pop();
            result.push_back(matrix[r][c]);
                
            if (c == 0 && r + 1 < matrix.size()) {
                traversalQueue.push({r + 1, c});
            }
                
            if (c + 1 < matrix[r].size()) {
                traversalQueue.push({r, c + 1});
            }
        }
            
        return result;
    }
};