Diagonal Traverse LeetCode Solution

Problem – 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

Diagonal Traverse LeetCode Solution in Java

    public int[] findDiagonalOrder(int[][] matrix) {
        if (matrix.length == 0) return new int[0];
        int r = 0, c = 0, m = matrix.length, n = matrix[0].length, arr[] = new int[m * n];
        for (int i = 0; i < arr.length; i++) {
            arr[i] = matrix[r][c];
            if ((r + c) % 2 == 0) { // moving up
                if      (c == n - 1) { r++; }
                else if (r == 0)     { c++; }
                else            { r--; c++; }
            } else {                // moving down
                if      (r == m - 1) { c++; }
                else if (c == 0)     { r++; }
                else            { r++; c--; }
            }   
        }   
        return arr;
    }

Diagonal Traverse LeetCode Solution in C++

vector<int> findDiagonalOrder(vector<vector<int>>& matrix) {
        if(matrix.empty()) return {};
        
        const int N = matrix.size();
        const int M = matrix[0].size();
        
        vector<int> res;
        for(int s = 0; s <= N + M - 2; ++s)
        {
            // for all i + j = s
            for(int x = 0; x <= s; ++x) 
            {
                int i = x;
                int j = s - i;
                if(s % 2 == 0) swap(i, j);

                if(i >= N || j >= M) continue;
                
                res.push_back(matrix[i][j]);
            }
        }
        
        return res;
    }

Diagonal Traverse LeetCode Solution in Python

class Solution(object):
    def findDiagonalOrder(self, matrix):
        """
        :type matrix: List[List[int]]
        :rtype: List[int]
        """
        result = [ ]
        dd = collections.defaultdict(list)
        if not matrix: return result
        # Step 1: Numbers are grouped by the diagonals.
        # Numbers in same diagonal have same value of row+col
        for i in range(0, len(matrix)):
            for j in range(0, len(matrix[0])):
                dd[i+j+1].append(matrix[i][j]) # starting indices from 1, hence i+j+1.
        # Step 2: Place diagonals in the result list.
        # But remember to reverse numbers in odd diagonals.
        for k in sorted(dd.keys()):
            if k%2==1: dd[k].reverse()
            result += dd[k]
        return result

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