Refresher : While Loop¶

While Loop¶

Example¶

Say we need to print "Hello" 5 times.
We can do it as :-

System.out.print("Hello");
System.out.print("Hello");
System.out.print("Hello");
System.out.print("Hello");
System.out.print("Hello");

But what if we have to do the same 100 times or 1000 times ?
It would be not feasible.
Solution to above is to use while loop :-

int count = 0 ;

while(count < 5)
{
  System.out.print("Hello");
  count ++ ;   
}

Syntax¶

intialization

while(Condition)
{
    // loop work
    updation
}

Question¶

int i = 1;
i = i + 1;

What is the new value of i ?

Choices

2
1
0
3

Solution

i = 1
i = i + 1 => i = 1 + 1 = 2

Question 1¶

Question¶

Given an integer N as input. Print from 1 to N ?

Testcase 1¶

Input :
N = 4

Solution 1¶

Output : 1 2 3 4

Approach¶

Intialize the count with 1.
While loop will execute till count <= N.
Print count.
In loop update the count by increamenting 1.

Code¶

public static void main(){

    Scanner scn = new Scanner(System.in);
    int N = scn.nextInt();
    int count = 1;
    while (count <= N) {
        System.out.printLn(count + " ");
        count++ ;    
    }
}

Question 2¶

Question¶

Given an integer N as input. Print from N to 1 ?

Testcase 1¶

Input :
N = 4

Solution 1¶

Output : 4 3 2 1

Approach¶

Intialize the count with N.
While loop will execute till count >= 1.
Print count.
In loop update the count by decreamenting it by 1.

Code¶

public static void main() {

    Scanner scn = new Scanner(System.in);
    int N = scn.nextInt();
    int count = N;
    while (count >= 1) {
        System.out.print(count + " ");
        count--;
    }
}

Question 3¶

Question¶

Given an integer N as input. Print odd values from 1 to N ?

Testcase 1¶

Input : N = 8

Solution 1¶

Output : 1 3 5 7

Approach¶

Since odd numbers start from 1, intialize the count with 1.
While loop will execute till count <= N.
print count.
In loop update the count by increamenting it by 2 since adding 2 to previous odd will yeild next odd number.

Code¶

public static void main() {

    Scanner scn = new Scanner(System.in);
    int N = scn.nextInt();
    int count = 1;
    while (count <= N) {
        System.out.print(count + " ");
        count += 2;
    }
}

Question 4¶

Question¶

Given an integer N as input. Print odd values from 1 to N ?

Testcase 1¶

Input :
N = 8

Solution 1¶

Output : 1 3 5 7

Approach¶

Since odd numbers start from 1, intialize the count with 1.
While loop will execute till count <= N.
print count.
In loop update the count by increamenting it by 2 since adding 2 to previous odd will yeild next odd number.

Code¶

public static void main() {

    Scanner scn = new Scanner(System.in);
    int N = scn.nextInt();
    int count = 1;
    while (count <= N) {
        System.out.print(count + " ");
        count += 2;
    }
}

Question 5¶

Question¶

Given an integer N as input, print multiples Of 4 till N ?

Testcase 1¶

Input : N = 18

Solution 1¶

Output : 4 8 12 16

Approach¶

Since multiple of 4 numbers start from 4, intialize the count with 4.
While loop will execute till count <= N.
Print count.
In loop update the count by increamenting it by 4 since adding 4 to previous multiple will yeild the next multiple.

Code¶

public static void main() {

    Scanner scn = new Scanner(System.in);
    int N = scn.nextInt();
    int count = 4;
    while (count <= N) {
        System.out.print(count + " ");
        count += 4;
    }
}

Question¶

int i = 1;
while (i <= 10) {
    i = i * i;
    System.out.print(i + " ");
    i++;
}

What will be the output ?

Choices

1 4 25
Error
Infinite loop
100 12 34

Solution

|i = 1  | i <= 10 | i * i = 1 | i++ = 2 |---> iteration 1
|i = 2  | i <= 10 | i * i = 4 | i++ = 5 |---> iteration 2
|i = 5  | i <= 10 | i * i = 25 | i++ = 26 |---> iteration 3

Question¶

public static void main() {
    int i = 0;
    while (i <= 10) {
        System.out.print(i + " ");
        i = i * i;
    }
}

What will be the output ?

Choices

Loop will never end
1 4 9 16 25 36 49 64 81 100
1 2 3 4
0 0 0 0

Solution

Since i would always be less than 10 hence infinite loop

Question 6¶

Question¶

Q5 : Given an integer N as input, print perfect squares till N ?

Perfect square —> An integer whose square root is an integer

Testcase 1¶

Input : N = 30

Solution 1¶

Output : 1 4 9 16 25

Approach¶

Intialize count = 1
While loop will execute till count * count <= N.
Print count * count.
In loop update the count by increamenting it by 1.

Code¶

public static void main() {

    Scanner scn = new Scanner(System.in);
    int N = scn.nextInt();
    int count = 1;
    while (count * count <= N) {
        System.out.print(count * count + " ");
        count += 1;
    }
}

Question 7¶

Question¶

Given an integer N as input, print it's digits ?

Testcase 1¶

Input : N = 6531

Solution 1¶

Output : 1 3 5 6

Observations¶

N	N % 10
6531	1
653	3
65	5
6	6

We can see that answer to % 10 of numbers present in the table coincide with the answer.
We need to find a way to arrive at these numbers while iterating.
We can do this by dividing number in each iteration by 10.

N	N/10
6531	653
653	65
65	6
6	0

Approach¶

Input N.
While loop will execute till N > 0.
Print N % 10.
In loop update the N by dividing it by 10.

Code¶

public static void main() {

    Scanner scn = new Scanner(System.in);
    int N = scn.nextInt();
    while (N > 0) {
        System.out.print(N + " ");
        N /= 10;
    }
}

Edge Cases¶

TestCase 1

Input :-
N = 0

Solution 1

Output = 0

Approach to handle edge case -

Input N.
While loop will execute till N >= 0.
Print N % 10.
In loop update the N by dividing it by 10.

Code 1

public static void main() {

    Scanner scn = new Scanner(System.in);
    int N = scn.nextInt();
    while (N >= 0) {
        System.out.print(N + " ");
        N /= 10;
    }
}

* The above approach leads to infinite loop. * We need to handle above case seperately in if else.

TestCase 2

Input :-
N = - 6351 (i.e N < 0)

Solution 2

Output : 1 3 5 6

Since answer of N < 0 would be same as that of N > 0.
So we just multiple N with -1 in order to convert it to +ve number .

Code final¶

public static void main() {

    Scanner scn = new Scanner(System.in);
    int N = scn.nextInt();
    if (N == 0) {
        System.out.print(0);
    } else {

        if (N < 0) {
            N *= -1;
        }
        while (N > 0) {
            print(N + " ")
            N /= 10;
        }
    }
}

Question 8¶

Question¶

Given an integer N as input, print sum of it's digits ?

N > 0

Testcase 1¶

Input : N = 6531

Solution 1¶

Output : 15

Approach¶

Input N & Intialize sum = 0
While loop will execute till N > 0.
Add N % 10 to sum.
In loop update the N by dividing it by 10.

Pseudeocode¶

public static void main() {

    Scanner scn = new Scanner(System.in);
    int N = scn.nextInt();
    int Sum = 0;
    while (N > 0) {
        Sum += (N % 10);
        N /= 10;
    }
    System.out.print(Sum);
}

Question¶

Which of the following will add the digit d to the back of a number r

Choices

r * 10 + d
d * r
d + r
d * 10 + r

Solution

Let r = 13 & d = 4
We want to add d behind r i.e we need number 134
So r * 10 = 130
r * 10 + d => 130 + 4 = 134

Question 9¶

Question¶

Given an integer N as input, Reverse it ?

N > 0

Testcase 1¶

Input :
N = 6531

Solution 1¶

Output : 1356

Approach¶

Input N & Intialize reverse = 0
While loop will execute till N > 0.
Set reverse = reverse * 10 + N % 10.
In loop update the N by dividing it by 10.

Code¶

public static void main() {

    Scanner scn = new Scanner(System.in);
    int N = scn.nextInt();
    int reverse = 0;
    while (N > 0) {
        d = N % 10;
        reverse = reverse * 10 + d;
        N /= 10;
    }
    System.out.print(reverse);
}