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Home Code Solutions

Diagonal Traverse – LeetCode Solution

Diagonal Traverse - LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.

BhautikBhalala by BhautikBhalala
September 10, 2022
Reading Time: 2 mins read
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Leetcode All Problems Solutions

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Table of Contents

  • Diagonal Traverse – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.
  • C++ Diagonal Traverse LeetCode Solution
  • Java Diagonal Traverse LeetCode Solution
  • Python 3 Diagonal Traverse LeetCode Solution
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Diagonal Traverse – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.

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

C++ Diagonal Traverse LeetCode Solution


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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;
    }

Java Diagonal Traverse LeetCode Solution


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 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;
    }

 



Python 3 Diagonal Traverse LeetCode Solution


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class Solution(object):
    def findDiagonalOrder(self, matrix):
        """
        :type matrix: List[List[int]]
        :rtype: List[int]
        """
        d={}
		#loop through matrix
        for i in range(len(matrix)):
            for j in range(len(matrix[i])):
				#if no entry in dictionary for sum of indices aka the diagonal, create one
                if i + j not in d:
                    d[i+j] = [matrix[i][j]]
                else:
				#If you've already passed over this diagonal, keep adding elements to it!
                    d[i+j].append(matrix[i][j])
		# we're done with the pass, let's build our answer array
        ans= []
		#look at the diagonal and each diagonal's elements
        for entry in d.items():
			#each entry looks like (diagonal level (sum of indices), [elem1, elem2, elem3, ...])
			#snake time, look at the diagonal level
            if entry[0] % 2 == 0:
				#Here we append in reverse order because its an even numbered level/diagonal. 
                [ans.append(x) for x in entry[1][::-1]]
            else:
                [ans.append(x) for x in entry[1]]
        return ans



 

Array-1180
String-562
Hash Table-412
Dynamic Programming-390
Math-368
Sorting-264


Greedy-257
Depth-First Search-256
Database-215
Breadth-First Search-200
Tree-195


Binary Search-191
Matrix-176
Binary Tree-160
Two Pointers-151
Bit Manipulation-140


Stack-133
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Design-116
Graph-108
Simulation-103
Prefix Sum-96


Backtracking-92
Counting-86
Sliding Window-73
Linked List-69
Union Find-66


Ordered Set-48
Monotonic Stack-47
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Trie-41
Binary Search Tree-40
Divide and Conquer-40


Enumeration-39
Bitmask-37
Queue-33
Memoization-32
Topological Sort-31


Geometry-30
Segment Tree-27
Game Theory-24
Hash Function-24
Binary Indexed Tree-21


Interactive-18
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Iterator-9
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Bucket Sort-6
Suffix Array-6
Minimum Spanning Tree-5
Counting Sort-5
Shell-4


Line Sweep-4
Reservoir Sampling-4
Eulerian Circuit-3
Radix Sort-3


Strongly Connected Componen-t2
Rejection Sampling-2
Biconnected Component-1

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Tags: Cc++14Diagonal Traversefull solutionGojavajava 15java 7java 8java8javascriptkotlinLeetCodeLeetCodeSolutionpypy 3Python 2python 3rubyrustSolution
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