notes/OJ notes/pages/Leetcode Reshape-The-Matrix.md

Leetcode Reshape-The-Matrix

2022-06-12

---

Data structures:

#DS #array #vector

Difficulty:

#leetcode #coding_problem #difficulty-easy

Related topics:

(tag:#array OR tag:#vector) -tag:#Kadane_s_algorithm 

• Binary Search Algorithm
• cpp_Range_based_for_loop
• Leetcode Binary-Search
• Leetcode First-Bad-Version
• Leetcode First-Unique-Character-In-a-String
• Leetcode Merge-Sorted-Array
• Leetcode Pascal's-Triangle
• Leetcode Ransom-Note
• Leetcode Search-a-2D-Matrix
• Leetcode Search-Insert-Position
• Leetcode Two-Sum
• Leetcode Valid-Anagram
• Leetcode Valid-Sudoku
• Two pointers approach

Links:

• Link to problem

---

Problem

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.

Examples

**Input:** mat = [[1,2],[3,4]], r = 1, 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

Thoughts

First is the O(mn) solution, I use nested for loop with two additional vars

Then, from the hint, I know we can transform any array to 1-D, then transform 1-D to any array, that is to said:

[!summary] For arr[m][n] and ans[r][c] and iterator i, with temp 1-D array temp[i]: temp[i] = arr[i / n][i % n] = ans[i / c][i % c]

Solution

O(nm) solution using two loops

class Solution {
public:
  vector<vector<int>> matrixReshape(vector<vector<int>> &mat, int r, int c) {
    // Initialize the answer vector
    vector<vector<int>> ans(r);
    for (int i = 0; i < ans.size(); i++) {
      ans[i] = vector<int>(c);
    }

    // if the vector is not possible, return original.
    if (mat[0].size() * mat.size() != r * c) {
      return mat;
    }

    int rCounter = 0;
    int cCounter = 0;
    for (int i = 0; i < mat.size(); i++) {
      for (int j = 0; j < mat[i].size(); j++) {
        if (cCounter >= c) {
          rCounter++;
          cCounter = cCounter % c;
        }

        ans[rCounter][cCounter++] = mat[i][j];
      }
    }
    return ans;
  }
};

O(mn) Solution using one loop

Tip

use vector<vector<int>> ans(r, vector<int>(c)); to initialize a 2-d vector

class Solution {
public:
  vector<vector<int>> matrixReshape(vector<vector<int>> &mat, int r, int c) {
    // Initialize the answer vector
    vector<vector<int>> ans(r, vector<int>(c));

    int m = mat.size();
    int n = mat[0].size();
    int total = m * n;

    // if the vector is not possible, return original.
    if (total != r * c) {
      return mat;
    }

    for (int i = 0; i < total; i++) {
      ans[i / c][i % c] = mat[i / n][i % n];
    }
    return ans;
  }
};