## Algorithm

Problem Name: 566. Reshape the Matrix

In MATLAB, there is a handy function called `reshape` which can reshape an `m x n` matrix into a new one with a different size `r x c` keeping its original data.

You are given an `m x n` matrix `mat` and two integers `r` and `c` representing the number of rows and the number of columns of the wanted reshaped matrix.

The reshaped matrix should be filled with all the elements of the original matrix in the same row-traversing order as they were.

If the `reshape` operation with given parameters is possible and legal, output the new reshaped matrix; Otherwise, output the original matrix.

Example 1:

```Input: mat = [[1,2],[3,4]], r = 1, c = 4
Output: [[1,2,3,4]]
```

Example 2:

```Input: mat = [[1,2],[3,4]], r = 2, c = 4
Output: [[1,2],[3,4]]
```

Constraints:

• `m == mat.length`
• `n == mat[i].length`
• `1 <= m, n <= 100`
• `-1000 <= mat[i][j] <= 1000`
• `1 <= r, c <= 300`

## Code Examples

### #1 Code Example with C++ Programming

```Code - C++ Programming```

``````start coding..
class Solution {
public:
vector<vector<int>> matrixReshape(vector < vector<int>>& nums, int r, int c) {
int m = nums.size(), n = nums[0].size();
if(m * n != r * c) return nums;
vector<vector < int>>res(r, vector<int>(c));
int row = 0, col = 0;
for(int i = 0; i  <  r; i++)
for(int j = 0; j  <  c; j++){
res[i][j] = nums[row][col++];
if(col == n> col = 0, row++;
}
return res;
}
};
``````
Copy The Code &

Input

cmd
mat = [[1,2],[3,4]], r = 1, c = 4

Output

cmd
[[1,2,3,4]]

### #2 Code Example with Java Programming

```Code - Java Programming```

``````
class Solution {
public int[][] matrixReshape(int[][] nums, int r, int c) {
int numOfElem = nums.length*nums[0].length;
if (numOfElem != r*c) {
return nums;
}

int[] expandedArray = new int[numOfElem];
int k = 0;

for (int i=0;i < nums.length;i++) {
for (int j=0;j < nums[i].length;j++) {
expandedArray[k] = nums[i][j];
k++;
}
}

int[][] ans = new int[r][c];
int i=0;
int j=0;
for(k=0;k < numOfElem;k++) {
ans[i][j] = expandedArray[k];
j++;
if (j == c) {
j = 0;
i++;
}
}
return ans;
}
}
``````
Copy The Code &

Input

cmd
mat = [[1,2],[3,4]], r = 1, c = 4

Output

cmd
[[1,2,3,4]]

### #3 Code Example with Javascript Programming

```Code - Javascript Programming```

``````
const matrixReshape = function(nums, r, c) {
if (isValid(nums, r, c) === false) {
return nums;
}
const arr = [];
nums.forEach(el => arr.push(...el));
const res = [];
for (let start = 0; start  <  arr.length; start = start + c) {
res.push(arr.slice(start, start + c));
}
return res;
};

function isValid(matrix, r, c) {
if (matrix.length * matrix[0].length !== r * c) {
return false;
} else {
return true;
}
}

console.log(matrixReshape([[1, 2], [3, 4]], 1, 4));
console.log(matrixReshape([[1, 2], [3, 4]], 2, 4));
``````
Copy The Code &

Input

cmd
mat = [[1,2],[3,4]], r = 2, c = 4

Output

cmd
[[1,2],[3,4]]

### #4 Code Example with Python Programming

```Code - Python Programming```

``````start coding...
class Solution:
def matrixReshape(self, nums, r, c):
"""
:type nums: List[List[int]]
:type r: int
:type c: int
:rtype: List[List[int]]
"""
nums_ordered=[x for y in nums for x in y]
if r*c==len(nums)*len(nums[0]):
return [nums_ordered[c*i:c*(i+1)] for i in range(r)]
else:return nums

``````
Copy The Code &

Input

cmd
mat = [[1,2],[3,4]], r = 2, c = 4

Output

cmd
[[1,2],[3,4]]

### #5 Code Example with C# Programming

```Code - C# Programming```

``````
using System.Linq;

namespace LeetCode
{
public class _0567_PermutationInString
{
public bool CheckInclusion(string s1, string s2)
{
var s1Length = s1.Length;
var s2Length = s2.Length;

if (s1Length > s2Length) return false;

var s1Count = new int[26];
foreach (var ch in s1)
s1Count[ch - 'a']++;

var s2Count = new int[26];
for (int i = 0; i  <  s2Length; i++)
{
s2Count[s2[i] - 'a']++;
if (i >= s1Length)
s2Count[s2[i - s1Length] - 'a']--;

if (s2Count.SequenceEqual(s1Count))
return true;
}

return false;
}
}
}
``````
Copy The Code &

Input

cmd
mat = [[1,2],[3,4]], r = 1, c = 4

Output

cmd
[[1,2,3,4]]