# 2D arrays 2

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# Agenda
- Revision
- Transpose
- Reverse every row in the given matrix
- Rotate by 90
- Intro to 2D ArrayList
- Syntax
- Functions
- Return even elements from everyrow.

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# Question

How do you declare an int 2D array in Java?

# Choices
- [x] int[][] mat = new int[rows][cols]
- [ ] int[][] mat = new int[cols][rows]
- [ ] int[][] mat = new int[rows][rows]
- [ ] int[][] mat = new int[cols][cols]

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# Question

How do you get the no. of rows in a 2D matrix mat?

# Choices
- [x] mat.length
- [ ] mat.length()
- [ ] mat.size
- [ ] mat.size()

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# Question

How do you get the number of columns in a 2D matrix for row index x?

# Choices
- [ ] mat[x].length()
- [x] mat[x].length
- [ ] mat[x].size
- [ ] mat[x].size()

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# Question

int[][] nums = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}};
1 2 3 
4 5 6
7 8 9
System.out.println(nums[1][2]);

# Choices
- [ ] 4
- [ ] 5
- [x] 6
- [ ] ArrayIndexOutOfBoundsException

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### Transpose

#### Given an rectangular matrix return the transpose of the matrix

> Rectangular matrix is matrix having number of rows not equal to number of columns

> Transpose of the matrix is new matrix in which the row of certain number in old matrix is converted to column of that particular number in new matrix like -
> >Row 1 of old matrix ---> column 1 of new matrix
> > Row 2 of old matrix ---> column 2 of new matrix
> > and so on...

#### Example 1

`mat[3][5]`

| 1 | 2 | 3 | 4 | 5 |
|:---:|:---:|:---:|:---:|:---:|
| 6 | 7 | 8 | 9 | 10 |
| 11 | 12 | 13 | 14 | 15 |

#### Explaination and solution

Intial matrix :-

`mat[3][5]`

| 1 | 2 | 3 | 4 | 5 |
|:---:|:---:|:---:|:---:|:---:|
| 6 | 7 | 8 | 9 | 10 |
| 11 | 12 | 13 | 14 | 15 |

Step 1 :- convert row 1 of intial to column 1

| 1 |
|:---:|
| 2 |
| 3 |
| 4 |
| 5 |

Step 2:- convert row 2 of intial to column 2
| 1 | 6 |
|:---:|:---:|
| 2 | 7 |
| 3 | 8 |
| 4 | 9 |
| 5 | 10 |

Step 3 :- convert row 3 of intial to column 3

| 1 | 6 | 11 |
|:---:|:---:|:---:|
| 2 | 7 | 12 |
| 3 | 8 | 13 |
| 4 | 9 | 14 |
| 5 | 10 | 15 |

Transpose of matrix is :-

| 1 | 6 | 11 |
|:---:|:---:|:---:|
| 2 | 7 | 12 |
| 3 | 8 | 13 |
| 4 | 9 | 14 |
| 5 | 10 | 15 |

#### Example 2

`mat[3][4]`

| 1 | 2 | 3 | 4 |
|:---:|:---:|:---:|:---:|
| 6 | 7 | 8 | 9 |
| 11 | 12 | 13 | 14 |

#### Explanation and solution
Transpose of matrix is :-

| 1 | 6 | 11 |
|:---:|:---:|:---:|
| 2 | 7 | 12 |
| 3 | 8 | 13 |
| 4 | 9 | 14 |

---

# Question

For a rectangular matrix, can we have the transpose in the same matrix?

# Choices
- [ ] Yes
- [x] No we need new matrix
- [ ] Maybe

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# Question

If dimensions of a matrix A is ( N x M ), and it is declared as int mat[][] = new int[N][M];
How will the transpose be declared?

# Choices
- [ ] int transpose[] = new int[N][M];
- [ ] int transpose[][] = new int[N][M];
- [x] int transpose[][] = new int[M][N];

---

# Question

What will be the transpose of this matrix?

10, 20, 30
14, 15, 18

# Choices
- [x] 10,14<br>20,15<br>30,18
- [ ] 10,20<br>30,14<br>15,18
- [ ] I am confused about what is transpose :(

---

#### Observations :-

* if we observe example 1
* Element at row 0 and column 1 in matrix mat becomes Element at column 0 and row 1 in transpose.
* similarly mat[2][3] ---> newMat[3][2]
* mat[1][4] ---> newMat[4][1]
* Is there any pattern between the position of element in intial matrix and tranpose matrix ?
* On observing we can say that :-

<div class="alert alert-block alert-warning">
Transpose[i][j] = Mat[j][i]
</div>

<div class="alert alert-block alert-warning">

**If dimensions of Mat are MxN then dimensions of transpose would be NxM**
</div>

#### Code

static int[][] transposeMatrix(int[][] Mat) {
    int m = Mat.length; 
    int n = Mat[0].length; 
    int[][] ans = new int[n][m]; 
    for (int i = 0; i < m; i++) {
        for (int j = 0; j < n; j++) {
            ans[j][i] = Mat[i][j]; 
        }
    }
    return ans;
}

---

### Reverse every row
Given a matrix reverse every row of matrix and return the same matrix

#### Example 1

`mat[3][5]`

| 3 | 2 | 6 | 1 | 9 |
|:---:|:---:|:---:|:---:|:---:|
| 14 | 18 | 2 | 4 | 10 |
| 5 | 6 | 3 | 9 | 8 |

#### Explanation and solution

Initial matrix :-

`mat[3][5]`

| 3 | 2 | 6 | 1 | 9 |
|:---:|:---:|:---:|:---:|:---:|
| 14 | 18 | 2 | 4 | 10 |
| 5 | 6 | 3 | 9 | 8 |

Step 1 :- Reverse row 1 of matrix

| 9 | 1 | 6 | 2 | 3 |
|:---:|:---:|:---:|:---:|:---:|
| 14 | 18 | 2 | 4 | 10 |
| 5 | 6 | 3 | 9 | 8 |

Step 2 :- Reverse row 2 of matrix

| 9 | 1 | 6 | 2 | 3 |
|:---:|:---:|:---:|:---:|:---:|
| 10 | 4 | 2 | 18 | 14 |
| 5 | 6 | 3 | 9 | 8 |

Step 4 :- Reverse row 3 of matrix

| 9 | 1 | 6 | 2 | 3 |
|:---:|:---:|:---:|:---:|:---:|
| 10 | 4 | 2 | 18 | 14 |
| 8 | 9 | 3 | 6 | 5 |

#### Example 2

`mat[3][4]`

| 1 | 2 | 3 | 4 |
|:---:|:---:|:---:|:---:|
| 6 | 7 | 8 | 9 |
| 11 | 12 | 13 | 14 |

#### Explanation and solution

| 4 | 3 | 2 | 1 |
|:---:|:---:|:---:|:---:|
| 9 | 8 | 7 | 6 |
| 14 | 13 | 12 | 11 |

---

# Question

What will be result if we reverse each row of this matrix?

10, 20, 30
14, 15, 18

# Choices
- [ ] 10, 20, 30<br>14, 15, 18

- [ ] 20, 10, 30<br>14, 15, 18

- [ ] 10, 20, 30<br>18, 15, 14

- [x] 30, 20, 10<br>18, 15, 14

---

#### Approach
* Our approach should be to traverse each row reverse it.
* But how to reverse a row ?
**Reversing a single row:-**
* First element of the row is swapped with the last element of the row. Similarly, the second element of the array is swapped with the second last element of the array and so on.
* 
* But if keep on swapping we would end up with intial configuartion again
* So **we swap till e>s**
This way at the end of traversal, we will have the entire row reversed.

### Code

 static int[][] reverseEachRow(int[][] Mat) {
    int m = Mat[0].length;
    int n = Mat.length;

    for (int i = 0; i < n; i++) {
        int s = 0;
        int e = m - 1;

        while (e > s) {
            // Swap elements in the current row
            int temp = Mat[i][s];
            Mat[i][s] = Mat[i][e];
            Mat[i][e] = temp;

            e--;
            s++;
        }
    }

    return Mat;
}

---

### Rotate by 90

Given a matrix rotate it by 90^o in clockwise direction ?
#### Testcase


#### Solution

#### Example


#### Approach
* **First we take transpose of matrix. On taking transpose:-**

* **Reverse each row of transpose to get the solution**.

**Code**

import java.util.*;

public class Main {
    static int[][] transposeMatrix(int Mat[][]) {
        int m = Mat.length; 
        int n = Mat[0].length; 
        int transposeMat[][] = new int[n][m]; 
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                transposeMat[j][i] = Mat[i][j]; 
            }
        }
        return transposeMat;
    }

    static int[][] reverseEachRow(int Mat[][]) {
        int m = Mat[0].length;
        int n = Mat.length;

        for (int i = 0; i < n; i++) {
            int s = 0;
            int e = m - 1;

            while (e > s) {
                int temp = Mat[i][s];
                Mat[i][s] = Mat[i][e];
                Mat[i][e] = temp;

                e--;
                s++;
            }
        }

        return Mat;
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int Mat[][] = new int[3][4]; 
        for (int i = 0; i < 3; i++) {
            for (int j = 0; j < 4; j++) {
                Mat[i][j] = sc.nextInt();
            }
        }
        int transpose[][] = transposeMatrix(Mat); 
        int matRotatedClockwise90Degree[][] = reverseEachRow(transpose); 
    }
}

---

### Syntax

ArrayList<Integer> l = new ArrayList<>();

Here each element in l is an integer.

### Properties

#### 1. add(element)
It is used to insert elements in ArrayList.

l.add(20);
l.add(30);
l.add(40);
l.add(35);

ArrayList :-

| 20 | 30 | 40 | 35 |
|:---:|:---:|:---:|:---:|

#### 2. set(element)
It is used to update values at particular index in ArrayList.

l.set(1, 80);
l.set(0, 90);

ArrayList :-

| 90 | 80 | 40 | 35 |
|:---:|:---:|:---:|:---:|

#### 3. get(index)
It is used to get values at particular index in ArrayList.

print(l.get(2));
print(l.get(3));

Output :

40
50

#### 4. remove(index)
It is used to remove value at particular index in ArrayList.

l.remove(2);

ArrayList :-

| 90 | 80 | 35 |
|:---:|:---:|:---:|

**Note:**

ArrayList<Integer> l = new ArrayList<>();

Each element in this ArrayList is of **Integer** type.

---

### 2D ArrayList

ArrayList of ArrayLists

### Syntax for 2D ArrayList

ArrayList<ArrayList< Datatype>> a = new ArrayList<>();

Each element in this 2D ArrayList is of **ArrayList< Datatype>**

### How to add elememts in 2D ArrayList

ArrayList<ArrayList<Integer>> arr = new ArrayList<>();

Here each Arraylist in arr is of type **ArrayList<Integer>**.

#### Pseudocode

ArrayList<ArrayList<Integer>> arr = new ArrayList<>();
ArrayList<Integer> d1 = new ArrayList<>();
d1.add(10);
d1.add(20);
d1.add(30);
d1.add(40);

ArrayList<Integer> d2 = new ArrayList<>();
d2.add(-1);
d2.add(4);
d2.add(8);

ArrayList<Integer> d3 = new ArrayList<>();
d1.add(50);
d1.add(60);
d1.add(70);
d1.add(80);

Output:

        {
            {10,20,30,40},
arr :       {-1,4,8},
            {50,60,70,80}
        }

### How to get elememts in 2D ArrayList

>Note: arr.get(i) = element at ith index.

#### Pseudocode

System.out.println(arr.get(1));
System.out.println(arr.get(2));

Output:

{-1,4,8}
{50,60,70,80}

### How to access element from ith ArrayList at jth index

>Note: arr.get(i).get(j) = element at ith ArrayList and jth index.

#### Pseudocode

System.out.println(arr.get(0).get(0));
System.out.println(arr.get(1).get(2));
System.out.println(arr.get(2).get(1));

Output:

10
8
60

### How to return no. of elements in ArrayList

#### Pseudocode

System.out.println(arr.size());
System.out.println(arr.get(0).size());
System.out.println(arr.get(1).size());

Output:

3
4
3

### How to modify elements in ArrayList

#### Pseudocode

System.out.println(arr.get(0).set(0,14));
System.out.println(arr.get(1).set(2,-9));
System.out.println(arr.get(2).set(0,20));

Output:

        {
            {14,20,30,40},
arr :       {-1,4,-9},
            {20,60,70,80}
        }

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### Problem 1

Print 2D ArrayList.

#### Pseudocode

void print(ArrayList< ArrayList< Integer>> arr) {
    
    int n = arr.size(); // Get the number of rows in the ArrayList

    // Iterate through each row
    for (int i = 0; i < n; i++) {
        // Get the number of columns in the current row 
        int m = arr.get(i).size();

        // Iterate through each element in the current row
        for (int j = 0; j < m; j++) {
            
            System.out.print(arr.get(i).get(j) + " ");
        }
        System.out.println();
    }
}

#### Dry run


---

### Even numbers
Given a 2D ArrayList, return a 2D ArrayList which contains even number from every row.

#### Example 1

arr = {
    {3,10,2},
    {2,7,6,9,4},
    {18,20,11,6}
}
    
Ans = {    
    {10, 2}
    {2,6,4}
    {18,20,6}
}   

#### Example 2

arr = {
    {3,6,2,9},
    {2,4,8,10},
    {3,9,7,15},
    8,3,2,14,19},
}
    
Ans = {    
    {6,2}
    {2,4,8,10}
    {}
    {8,2,14}
}   

#### Observation

We will traverse every element in ArrayList and insert even numbers in output.

#### Pseudocode

ArrayList<ArrayList<Integer>> even(ArrayList<>> arr){
    
    ArrayList<ArrayList<Integer>> ans = new ArrayList<>();
    int n = arr.size();
    for(int i = 0; i < n; i++) {
        ArrayList<Integer> l = new ArrayList<>();
        int m = arr[i].get(i).size();
        for(int j = 0; j < m; j++){
            if( arr.get(i).get(j) % 2 == 0){
               l.add(arr.get(i).get(j));            
            }
        }
        ans.add(l);
    }
    return ans;
}

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