1D Arrays 1¶

Agenda¶

Introduction to Arrays
Reading Input
Indexing and Properties
Sum of all elements
Frequency of k in array
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¶

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]
```

Question¶

int[] ar = new int[6];

How can we access last position element ?

Choices¶

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¶

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¶

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¶

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¶

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¶

Question¶

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

Choices¶

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¶

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¶

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¶

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:

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¶

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¶

Trace 2¶

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

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¶

0
9
1

title: Quiz 13

description:

duration: 30

Question¶

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

Choices¶

0
9
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 - ∞(negative infinity) so that it does not affect the final answer.
We do this by Integer.MIN_VALUE

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;
}