solves reshape the matrix

Author: anishLearnsToCode

Diff for: README.md

@@ -149,7 +149,7 @@
 | 559 | [Maximum Depth of N-Ary Tree](https://leetcode.com/problems/maximum-depth-of-n-ary-tree) | Easy | [![Java](assets/java.png)](src/MaximumDepthOfNAryTree.java) [![Python](assets/python.png)](python/maximum_depth_of_n_ary_tree.py) |
 | 561 | [Array Partition I](https://leetcode.com/problems/array-partition-i) | Easy | [![Java](assets/java.png)](src/ArrayPartitionI.java) [![Python](assets/python.png)](python/array_partiton_I.py) |
 | 563 | [Binary Tree Tilt](https://leetcode.com/problems/binary-tree-tilt) | Easy | [![Java](assets/java.png)](src/BinaryTreeTilt.java) [![Python](assets/python.png)](python/binary_tree_tilt.py) |
-| 566 | [Reshape The Matrix](https://leetcode.com/problems/reshape-the-matrix) | Easy | |
+| 566 | [Reshape The Matrix](https://leetcode.com/problems/reshape-the-matrix) | Easy | [![Java](assets/java.png)](src/ReshapeTheMatrix.java) [![Python](assets/python.png)](python/reshape_the_matrix.py) |
 | 572 | [Subtree of Another Tree](https://leetcode.com/problems/subtree-of-another-tree) | Easy | |
 | 575 | [Distribute Candies](https://leetcode.com/problems/distribute-candies) | Easy | |
 | 581 | [Shortest Unsorted Continuous Subarray](https://leetcode.com/problems/shortest-unsorted-continuous-subarray) | Easy | |

Diff for: python/reshape_the_matrix.py

@@ -0,0 +1,13 @@
+from typing import List
+
+
+class Solution:
+    def matrixReshape(self, mat: List[List[int]], r: int, c: int) -> List[List[int]]:
+        if r * c != len(mat) * len(mat[0]) or (r == len(mat) and c == len(mat[0])): return mat
+        result = [[0] * c for _ in range(r)]
+        k = 0
+        for i in range(len(mat)):
+            for j in range(len(mat[0])):
+                result[k // c][k % c] = mat[i][j]
+                k += 1
+        return result

Diff for: src/ReshapeTheMatrix.java

@@ -0,0 +1,13 @@
+public class ReshapeTheMatrix {
+    public int[][] matrixReshape(int[][] mat, int r, int c) {
+        if (r * c != mat.length * mat[0].length || (mat.length == r && mat[0].length == c)) return mat;
+
+        int[][] result = new int[r][c];
+        for (int i = 0, k = 0 ; i < mat.length ; i++) {
+            for (int j = 0 ; j < mat[0].length ; j++, k++) {
+                result[k / c][k % c] = mat[i][j];
+            }
+        }
+        return result;
+    }
+}