# 1D Arrays 1

---
## Agenda

1. Introduction to Arrays
2. Reading Input
3. Indexing and Properties
4. Sum of all elements
5. Frequency of k in array
6. Max of all elements

---

### Example
Let's say we need to read four inputs for our programme. We can use the below approach.

#### Code

public static void main(String[] args) {
    int a, b, c, d;
    Scanner scanner = new Scanner(System.in);
    a = scanner.nextInt();
    b = scanner.nextInt();
    c = scanner.nextInt();
    d = scanner.nextInt();
}

Some one can suggest that we should use loop to get all four values like :-

#### Code

public static void main(){
    for(int i = 1; i <= 4;i ++ ){
        int a = sc.nextInt();
    }     
}

The above provided approach is wrong because what we are doing is updating the value of variable `a` in each iteration due this a would be set to the last input value provided.

---

### Concept of Arrays
* In above example what is instead of four there are hundreds of value to store. It would be manually infeasible to declare and set hundreds of variables.
* Therefore to overcome above problem we use **arrays**

#### Array
It is a data structure that can hold fixed number of values of same data type.

#### Syntax

datatype name[] = new datatype[size]

// example
float f[] = new float[10]
int arr[] = new int[10]

// Various ways
datatype[] name = new datatype[size]
datatype []name = new datatype[size]

---

# Question

Correct way to create an Array containing 5 int values in Java?

# Choices

- [x] int[] ar = new int[5]
- [ ] int[] ar = new int[4]
- [ ] int[] ar = int new[5]

---

## Explanation

Since size is 5 and datatype is int using above provided syntax rules:
int[] ar = new int[5]

---

### Indexing and Properties

* Indexing in array starts from **0**.

| index | 0 | 1 | 2 | 3 |
|:-----:|:---:|:---:|:---:|:---:|

* Accessing an element at **i^th** index in an array can be done as follows:-

 nameOfArray[i]

<img src = "https://d2beiqkhq929f0.cloudfront.net/public_assets/assets/000/045/869/original/upload_7feaa615de84a0773827cd5f5904fc43.png?1693761847"
height = "290" width = "500">

---

# Question

`int[] ar = new int[6];`

How can we access last position element ?

# Choices
- [x] ar[5]
- [ ] ar[6]
- [ ] ar[4]
- [ ] ar[7]

---

## Explanation

Since size is 6 indexing would be like :-

| index | 0 | 1 | 2 | 3 | 4 | 5 |
|:-----:|:---:|:---:|:---:|:---:|:---:|:---:|
| arr | 0 | 0 | 3 | 0 | 0 | 0 |

last element would be at index 5

---

# Question
`int[] ar = new int[10];`

How can we access last position element ?

# Choices
- [x] ar[9]
- [ ] ar[10]
- [ ] ar[7]
- [ ] ar[8]

---

## Explanation

Since size is 10 indexing would be like :-

| index | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
|:-----:|:---:|:---:|:---:|:---:|:---:|:---:|:---:|:---:|:---:|:---:|
| ar | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |

last element would be at index 9

---

# Question
Say int[] ar = new int[N]
How to access first element and last element ?

# Choices
- [x] ar[0] ar[N-1]
- [ ] ar[0] ar[N]
- [ ] ar[1] ar[N]

---

## Explanation
By observing previous questions we can generalize the idea that :-
* Last element in array 'arr' of size 'N' is accessed as `arr[N-1]`.
* First element in array 'arr' of size 'N' is accessed as `arr[0]`.

---

# Question
What is the Output of

public static void main(String args[]) {
    int[] arr = new int[10];
    int n = arr.length;
    System.out.println(n);
}

# Choices
- [x] 10
- [ ] 9
- [ ] 8

---

* By default values in an array of type `int` are intialized with **'0'**

---

# Question

What will be the output?

public static void main(String args[]) {
    int[] arr = new int[5];
    arr[0] = 10;
    arr[1] = 20;
    
    int sum = 0;
    for(int i = 0; i < 5;i ++ ) {
        sum += arr[i];
    }
    
    System.out.println(sum);
}

# Choices
- [x] 30
- [ ] error
- [ ] 20
- [ ] 43

---

## Explanation
By observing previous questions we can generalize the idea that :-
* Last element in array 'arr' of size 'N' is accessed as `arr[N-1]`.
* First element in array 'arr' of size 'N' is accessed as `arr[0]`.

#### Solution
<img src = "https://d2beiqkhq929f0.cloudfront.net/public_assets/assets/000/045/871/original/p17.png?1693762040"
height = "290" width = "500">

---

# Question

public static void main(String args[]) {
    int[] arr = new int[5];
    System.out.println(arr[0]);
}

# Choices
- [x] 0
- [ ] error
- [ ] random number
- [ ] 43

---

# Question

public static void main(String args[]) {
    int[] ar = new int[3];
    ar[0] = 10;
    ar[1] = 20;
    ar[2] = 30;
    System.out.print(ar[0] + ar[3]);
}

# Choices
- [x] error
- [ ] 0
- [ ] 40
- [ ] 60

---

# Question

public static void main(String args[]) {
    int[] ar = new int[3];
    ar[0] = 10;
    ar[1] = 20;
    ar[2] = 30;
    System.out.print(ar[0] + ar[2]);
}

# Choices
- [x] 40
- [ ] 0
- [ ] error
- [ ] 60

---

# Question

public static void main(String args[]) {
    int[] ar = new int[3];
    ar[0] = 10;
    ar[1] = 20;
    ar[2] = 30;
    System.out.print(ar[-1] + ar[3]);
}

# Choices
- [x] error
- [ ] 0
- [ ] 40
- [ ] 60

---

We can reassign an array to replace the previous value it was referencing.

**Code:**

public static void main(){
	int[] ar = new int[6];
	ar= new int[2];
	S.O.Pln(arr.length); 
}

**Output:**

2

---

* We can directly store elements into an array

**Code:**

int ar[] = {10,20,30};

---

### Creating and Reading an array

#### Create an array of size 4 and print sum of all it's element :-

* Let's create an array of size 4 and take input.

public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    
    int[] arr = new int[4]; 
    arr[0] = sc.nextInt();
    arr[1] = sc.nextInt();
    arr[2] = sc.nextInt();
    arr[3] = sc.nextInt();
}

* In above approach we have to take input for each index manually which is not a good idea.
* So **How can we take inputs efficiently ?**
* **Solution** :- We use a loop.
* But **how to apply loop to take array input ?**
* On observing above approach we will find that only index changes each time we take an input.
* **In each iteration we change the index number.**
* We iterate starting from 0 till last index i.e. array size -1.

public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    
    int[] arr = new int[4]; 
    for (int i = 0; i < 4; i ++ ) {
        arr[i] = sc.nextInt();
    }
}

---

### Passing Array to Functions

#### Create a Function Which Takes arr[] as A Parameter and Print the Array

* We need to declare a function which takes array as parameter to function.
* It can be done like :-
`Function nameOfFunction(dataType anyName[]){}`
* '`[]`' are important for distinguishing array type parameter from other variable type parameters.
* **How can we access the length of array from function ?**
* We use `array.length` for this purpose.
* We can pass array parameter to function call like:
`functionName(arrayName)`
* **We only need to pass array name.**

static void printArray(int[] ar) {
    int n = ar.length;
    for (int i = 0; i < n; i ++ ) {
        System.out.print(ar[i] + " ");
    }
    System.out.println(); 
}

public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);

    int[] arr = new int[4]; 
    for (int i = 0; i < 4; i ++ ) {
        arr[i] = sc.nextInt();
    }
    printArray(arr);
}

We take the sum of all elements of array as follows :-

public static void main(String[] args) {
        int[] arr = new int[4]; // creates an array of size 4
        Scanner scanner = new Scanner(System.in);

        for (int i = 0; i < 4; i ++ ) {
            arr[i] = scanner.nextInt();
        }
        int sum = 0;
        for (int i = 0; i < 4; i ++ ) {
            sum += arr[i]; // add element at ith index to sum variable
        }
        
        System.out.println(sum);
    }

---

### Problem 1

Given an array and k. Write a function to return the frequency of k in array?

#### Testcase

arr[7] = [3,6,7,6,11,6,14]

k = 6

#### solution

ans = 4

:::warning
Please take some time to think about the solution approach on your own before reading further.....
:::

#### Approach
* We need to create a function and pass array and k as parameters to the function.
* Inside the function :-
* Maintain a count variable which is intialised to 0.
* Iterate over the array:-
* If element at current index equals k increament count by 1.
* Return count.

#### Trace

![]()

#### Solution
<img src = "https://d2beiqkhq929f0.cloudfront.net/public_assets/assets/000/045/873/original/p22.png?1693762207"
height = "200" width = "500">

#### Pseudeocode

static int frequencyK(int[] ar, int k) {
    int n = ar.length;
    int count = 0;
    for (int i = 0; i < n; i ++ ) {
        if (ar[i] == k) {
            count ++ ;
        }
    }
    return count;
}

---

### Problem 2

Given an array . Write a function to return the maximum element present in array?

#### Testcase 1

arr[6] = [3,1,7,6,9,11]

#### solution

ans = 11

#### Testcase 2

arr[6] = [4,2,7,9,12,3]

#### solution

ans = 12

:::warning
Please take some time to think about the solution approach on your own before reading further.....
:::

#### Approach 1
* We need to create a function and pass array as parameters to the function.
* Inside the function :-
* Maintain a max variable which is intialised to 0.
* Iterate over the array:-
* If element at current index is greater than max then set max to current element.
* Return max.

#### Trace 1

<img src = "https://d2beiqkhq929f0.cloudfront.net/public_assets/assets/000/033/957/original/p23.png?1683759983"
height = "150" width = "500">

![]()

#### Trace 2

<img src = "https://d2beiqkhq929f0.cloudfront.net/public_assets/assets/000/033/958/original/p24.png?1683760023" height = "150" width = "500">

#### Code

 static int maxElement(int array[]) {
    int n = array.length;
    int max = Integer.MIN_VALUE; // Initialize max with the smallest possible integer value
    for (int i = 0; i < n; i ++ ) {
        if (array[i] > max) {
            max = array[i];
        }
    }
    return max;
}

**There is a flaw in above code.** Let's see it with help of an testcase.

#### Testcase 3

arr[4] = [ - 8, - 4, - 3, - 5]

#### solution

ans = - 3

* Let' apply approach 1 to testcase 3

<img src = "https://d2beiqkhq929f0.cloudfront.net/public_assets/assets/000/033/959/original/p25.png?1683760451" height = "150" width = "500" >

* In trace we get the answer as 0 whereas the correact answer is -3. **Why ?**

#### Issue

**Since max/ans variable is intialised to 0 which is already greater than all elements in array therefore max/ans is not updated.**

---
# Question

For taking sum of N numbers we initialise our sum variable with =

# Choices
- [x] 0
- [ ] 9
- [ ] 1

---
title: Quiz 13
description:
duration: 30
card_type : quiz_card
---
# Question

For taking product of N numbers we initialise our product variable with `=`

# Choices
- [ ] 0
- [ ] 9
- [x] 1

---

* Based upon observations from above questions we need to intialize max/ans in such a manner that it won't affect the answer.
* We intialize the ans/max variable to **- &#8734;**(negative infinity) so that it does not affect the final answer.
* We do this by **Integer.MIN_VALUE**

<img src = "https://d2beiqkhq929f0.cloudfront.net/public_assets/assets/000/033/961/original/p27.png?1683761781" height = "350" width = "500">

#### Code

static int maxElement(int[] ar) {
    int n = ar.length;
    int max = Integer.MIN_VALUE; 
    
    for (int i = 0; i < n; i ++ ) {
        if (ar[i] > max) {
            max = ar[i];
        }
    }
    return max;
}