# NAICHEF - Editorial

#1

Practice

Contest

Tester: Misha Chorniy

Editorialist: Bhuvnesh Jain

CAKEWALK

# Prerequisites

Probability, Looping

# Problem

You are given an N sided dice. You roll it twice and need to find the probability of getting A on the first throw and B on the second throw.

# Explanation

The probability of obtaining a number on the consecutive throw of a dice is independent of each other. For more details, you may refer here. The probability of getting a number X on throwing a N sided dice is given by:

ext{Probability} = \frac{ ext{Number of times X appears on the dice}}{N}

Thus, the overall probability of obtaining A on the first throw and B on the second throws is given by:

ext{Required Probability} = \frac{ ext{Number of times A appears on the dice}}{N} * \frac{ ext{Number of times B appears on the dice}}{N}

Thus, the problem reduces to finding the frequency of a number in an array. This can be easily done in O(1) space complexity and O(n) time complexity using a simple for loop as below


def count_frequency(array a, integer x):
count = 0
for number in a:
if number == x:
count += 1
return count



The constraints of the problem were such that all the operations can be done in integers only without any overflow issues.

# Time Complexity

O(n) per test case.

# Space Complexity

O(1)

### AUTHOR’S AND TESTER’S SOLUTIONS:

Tester’s solution can be found here.

Editorialist’s solution can be found here.

#2

Video solution (with a comparison of 3 different solutions): https://youtu.be/r9yvzd0tTJQ