Problems - Set 1

Easy

FizzBuzz

Given a positive integer A, return an array of strings with all the integers from 1 to N. But for multiples of 3 the array should have “Fizz” instead of the number. For the multiples of 5, the array should have “Buzz” instead of the number. For numbers which are multiple of 3 and 5 both, the array should have “FizzBuzz” instead of the number.

Example:

A = 5
Return: [1 2 Fizz 4 Buzz]

Method 1 - Using loop and if else

ArrayList<String> list = new ArrayList<>();

for (int i = 1; i <= A; i++) {
    if (i % 3 == 0 && i % 5 == 0) {
        list.add("FizzBuzz");
            } else if (i % 3 == 0) {
        list.add("Fizz");
            } else if (i % 5 == 0) {
        list.add("Buzz");
            } else {
        list.add(String.valueOf(i));
    }
}

Method 2 - Using StringBuilder for String Concatenation

Instead of performing multiple concatenations using + inside the if conditions, use a StringBuilder to construct the strings. This can improve performance when dealing with large datasets.

ArrayList<String> list = new ArrayList<>();

for (int i = 1; i <= A; i++) {
    StringBuilder sb = new StringBuilder();
    if (i % 3 == 0) sb.append("Fizz");
    if (i % 5 == 0) sb.append("Buzz");
    list.add(sb.length() > 0 ? sb.toString() : String.valueOf(i));
}

Method 3 - Using Streams (Java 8+)

If we want to use a more functional programming style, we can leverage Java Streams.

List<String> list = IntStream.rangeClosed(1, A)
    .mapToObj(i -> i % 3 == 0 && i % 5 == 0 ? "FizzBuzz" :
                 i % 3 == 0 ? "Fizz" :
                 i % 5 == 0 ? "Buzz" : String.valueOf(i))
    .collect(Collectors.toList());

Medium

Implement a method rand7() given rand5( )

Given a method that generates a random number between O and 4 (inclusive), write a method that generates a random number between O and 6 (inclusive).

We want to generate a uniform distribution over 7 numbers (0 to 6) using a generator that gives us 5 numbers (0 to 4).

The basic idea is:

1. Combine two calls to rand5() to generate a larger range that includes at least 7 values.

2. Reject values that are outside a clean multiple of 7 to preserve uniformity.

Use base-5 expansion:

• rand5() gives a random integer between 0 and 4 (i.e., 5 outcomes).

• rand7() should give a random integer between 0 and 6 (i.e., 7 outcomes).

But 5 can’t divide 7 evenly — so we need more randomness than a single call to rand5() can give.

By calling rand5() twice, we can generate more combinations.

Each rand5() has 5 values → so two calls give us:

Total combinations = 5 * 5 = 25

We want to generate a uniform integer from 0 to 24, which gives us 25 possible values.

Now, how do we combine the two rand5() calls to get those values?

Let’s take this:

int num = 5 * rand5() + rand5();

Assume:

• First call to rand5() → returns a (0 to 4)

• Second call to rand5() → returns b (0 to 4)

Then:

num = 5 * a + b

a	b	5 * a + b
0	0	0
0	1	1
0	2	2
0	3	3
0	4	4
1	0	5
1	1	6
...	...	...
4	4	24

So this gives us all numbers from 0 to 24, equally likely.

import java.util.Random;

public class Rand7FromRand5 {

    static Random random = new Random();

    // Provided rand5() function (returns 0–4)
    public static int rand5() {
        return random.nextInt(5);
    }

    public static int rand7() {
        while (true) {
            int num = 5 * rand5() + rand5(); // Generates number in range 0 to 24
            if (num < 21) {
                return num % 7; // Maps to 0 to 6 uniformly
            }
            // else, reject and retry
        }
    }

    public static void main(String[] args) {
        int[] count = new int[7];
        for (int i = 0; i < 70000; i++) {
            count[rand7()]++;
        }
        for (int i = 0; i < 7; i++) {
            System.out.println(i + " → " + count[i]);
        }
    }
}