Leetcode 1424: Diagonal Traverse II
You are given a 2D array
nums, where each sub-array represents a row in the grid. Your task is to return all the elements of nums in diagonal order, starting from the top-left to bottom-right diagonals, where the sum of the row and column indices is constant.Problem
Approach
Steps
Complexity
Input: The input consists of a 2D array `nums` where each sub-array represents a row in the grid.
Example: [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
Constraints:
• 1 <= nums.length <= 10^5
• 1 <= nums[i].length <= 10^5
• 1 <= sum(nums[i].length) <= 10^5
• 1 <= nums[i][j] <= 10^5
Output: The output is a list of integers representing the elements of `nums` traversed in diagonal order.
Example: [1, 4, 2, 7, 5, 3, 8, 6, 9]
Constraints:
• The output contains the elements of `nums` in the diagonal order.
Goal: The goal is to collect the elements of `nums` in diagonal order, where diagonals are formed by elements with the same sum of row and column indices.
Steps:
• Step 1: Group the elements of `nums` by the sum of their row and column indices.
• Step 2: Traverse the diagonals in reverse order to maintain the required diagonal pattern.
• Step 3: Collect the elements in the desired order and return them.
Goal: The constraints ensure the problem is solvable within a reasonable time and space limit.
Steps:
• 1 <= nums.length <= 10^5
• 1 <= nums[i].length <= 10^5
• 1 <= sum(nums[i].length) <= 10^5
• 1 <= nums[i][j] <= 10^5
Assumptions:
• The input grid is non-empty, and each sub-array represents a valid row of elements.
• Input: [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
• Explanation: The elements are collected diagonally starting from the top-left corner: 1, 4, 2, 7, 5, 3, 8, 6, 9.
• Input: [[1, 2, 3, 4, 5], [6, 7], [8], [9, 10, 11], [12, 13, 14, 15, 16]]
• Explanation: The elements are collected diagonally starting from the top-left: 1, 6, 2, 8, 7, 3, 9, 4, 12, 10, 5, 13, 11, 14, 15, 16.
Approach: The approach to solving this problem involves grouping the elements by their diagonal indices and then processing the diagonals from top-left to bottom-right.
Observations:
• Each diagonal consists of elements where the sum of their row and column indices is constant.
• To achieve the diagonal order, we can utilize a map to store the diagonals and process them accordingly.
• We can utilize a hashmap to group the diagonals and then traverse them in reverse order to achieve the desired output.
Steps:
• Step 1: Iterate through the grid and group elements by the sum of their row and column indices.
• Step 2: Traverse the diagonals in reverse order (starting from the highest diagonal sum) to maintain the diagonal traversal.
• Step 3: Append the elements from each diagonal in the correct order to the result list.
Empty Inputs:
• There are no empty inputs since the length of the grid is at least 1.
Large Inputs:
• For large inputs, ensure that the solution handles up to the maximum allowed grid size efficiently.
Special Values:
• Ensure the solution handles cases where the grid has rows of varying lengths.
Constraints:
• The space complexity should be managed to handle up to the maximum allowed input size.
vector<int> findDiagonalOrder(vector<vector<int>>& nums) {
int m = nums.size(), n = nums[0].size(), mx = 0;
unordered_map<int, vector<int>> mp;
for(int i = 0; i < nums.size(); i++)
for(int j = 0; j < nums[i].size(); j++) {
mp[i + j].push_back(nums[i][j]);
mx = max(mx, i + j);
}
vector<int> ans;
for(int i = 0; i <= mx; i++) {
for(auto x = mp[i].rbegin(); x != mp[i].rend(); x++)
ans.push_back(*x);
}
return ans;
}
};
/*
Primary attempt can redirect to two pointer simulation
But right attempt is, queueing over a map which shares same diagonals
1 : Function Definition
vector<int> findDiagonalOrder(vector<vector<int>>& nums) {
Defines the `findDiagonalOrder` function which takes a 2D vector of integers `nums` and returns a 1D vector of integers representing the diagonal order.
2 : Variable Initialization
int m = nums.size(), n = nums[0].size(), mx = 0;
Initializes variables `m` (number of rows), `n` (number of columns), and `mx` (to track the maximum diagonal index) for further processing.
3 : Data Structure Initialization
unordered_map<int, vector<int>> mp;
Declares an unordered map `mp` where each key corresponds to a diagonal index and the value is a vector of integers that stores the diagonal elements.
4 : Loop Constructs
for(int i = 0; i < nums.size(); i++)
Begins a loop that iterates over the rows of the `nums` vector.
5 : Loop Constructs
for(int j = 0; j < nums[i].size(); j++) {
Nested loop that iterates over the columns of the current row `i` of `nums`.
6 : Map Operations
mp[i + j].push_back(nums[i][j]);
Inserts the current element `nums[i][j]` into the map `mp` under the key `i + j`, which represents the diagonal index.
7 : Variable Update
mx = max(mx, i + j);
Updates the `mx` variable to track the maximum diagonal index encountered so far.
8 : Vector Initialization
vector<int> ans;
Declares a vector `ans` that will hold the final result of the diagonal traversal.
9 : Loop Constructs
for(int i = 0; i <= mx; i++) {
Begins a loop that iterates over each diagonal index from `0` to `mx`.
10 : Loop Constructs
for(auto x = mp[i].rbegin(); x != mp[i].rend(); x++)
Iterates over the elements of each diagonal in reverse order using a reverse iterator (`rbegin` to `rend`).
11 : Vector Operations
ans.push_back(*x);
Adds each element `*x` from the current diagonal to the `ans` vector.
12 : Return Statement
return ans;
Returns the `ans` vector, which contains the elements of the matrix in diagonal order.
Best Case: O(n), where n is the total number of elements in the grid.
Average Case: O(n), as we process each element once and traverse the diagonals.
Worst Case: O(n), as we process each element and diagonal once.
Description: The time complexity is linear in relation to the number of elements in the grid.
Best Case: O(n), since we store the elements in the hashmap based on their diagonal indices.
Worst Case: O(n), where n is the total number of elements in the grid.
Description: The space complexity is linear, as we store the elements based on their diagonals.
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