<p>Given a 2D integer array <code>nums</code>, return <em>all elements of </em><code>nums</code><em> in diagonal order as shown in the below images</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/04/08/sample_1_1784.png" style="width: 158px; height: 143px;" />
<pre>
<strong>Input:</strong> nums = [[1,2,3],[4,5,6],[7,8,9]]
<strong>Output:</strong> [1,4,2,7,5,3,8,6,9]
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/04/08/sample_2_1784.png" style="width: 230px; height: 177px;" />
<pre>
<strong>Input:</strong> nums = [[1,2,3,4,5],[6,7],[8],[9,10,11],[12,13,14,15,16]]
<strong>Output:</strong> [1,6,2,8,7,3,9,4,12,10,5,13,11,14,15,16]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums[i].length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= sum(nums[i].length) &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums[i][j] &lt;= 10<sup>5</sup></code></li>
</ul>

class Solution {

public:
    vector<int> findDiagonalOrder(vector<vector<int>>& nums) {
        
        int n = nums.size(), m = 0; 
        map<int, vector<pair<int, int>>> store; 

        for(int i = 0; i < n; i++) {
            vector<int> row = nums[i]; 
            for(int j = 0; j < row.size(); j++) {
           //     cout << row[j] << "\t" << i << "\n"; 
                store[i + j].push_back({row[j], i});
            }
        }

        vector<int> elements; 
        for(auto it: store) {

            vector<pair<int, int>> pVec = it.second; 
            sort(pVec.begin(), pVec.end(), [](pair<int, int> &a, pair<int, int> &b) {
                return a.second > b.second; 
            }); 
            
            for(pair<int, int> p: pVec) {
                elements.push_back(p.first); 
            }
        }
        return elements; 
    }
};

// a[i][j] 
// while(inBoundary(i, j)) 
// cout << a[i - 1][j + 1] << // [3, 0] --> [2, 1] --> [1, 2] 
// 
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