What is the time complexity of Alice and Bob Playing Flower Game?

The optimal solution has a time complexity of O(1).

What data structures are used to solve Alice and Bob Playing Flower Game?

This problem uses Math. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Alice and Bob Playing Flower Game?

Insight 1: Alice wins if and only if the total number of flowers (x + y) is odd. This is because Alice can always force Bob into a position where the number of flowers remaining is even, ultimately leaving Bob with no flowers to pick when it's his turn if the initial sum is odd. Insight 2: We need to count the number of pairs (x, y) where 1 <= x <= n and 1 <= y <= m, and x + y is odd.

Which companies ask LeetCode 3021. Alice and Bob Playing Flower Game in interviews?

This problem is frequently asked at Rubrik.

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Alice and Bob Playing Flower Game - LeetCode 3021 Solution

Q: How to solve Alice and Bob Playing Flower Game?

The solution leverages the observation that Alice wins if x + y is odd. To determine how many pairs (x,y) satisfy this condition, we count the number of odd and even numbers within the ranges [1, n] and [1, m]. Then, we multiply the count of odd numbers in the [1, n] range by the count of even numbers in the [1, m] range, and vice-versa. Summing these two products yields the total number of (x, y) pairs where x + y is odd. The provided code directly calculates n * m / 2, because roughly half of the pairs will satisfy the condition when n and m are large enough, and the details of even/odd counts are handled implicitly within the division.

Q: Which companies ask LeetCode 3021. Alice and Bob Playing Flower Game in interviews?

This problem is frequently asked at Rubrik.

Alice and Bob Playing Flower Game - Complete Solution Guide

Alice and Bob Playing Flower Game is LeetCode problem 3021, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Alice and Bob are playing a turn-based game on a circular field surrounded by flowers. The circle represents the field, and there are x flowers in the clockwise direction between Alice and Bob, and y flowers in the anti-clockwise direction between them. The game proceeds as follows: Alice takes the first turn. In each turn, a player must choose either the clockwise or anti-clockwise direction and pick one flower from that side. At the end of the turn, if there are no flowers left at all, the cur

Detailed Explanation

Alice and Bob are playing a game where they pick flowers from a circle. There are 'x' flowers clockwise and 'y' flowers counter-clockwise between them. Alice goes first, and they alternate turns picking one flower from either direction. The player who picks the last flower wins. Given the ranges for 'x' (1 to n) and 'y' (1 to m), the problem asks to find the number of (x, y) pairs where Alice wins.

Solution Approach

The solution leverages the observation that Alice wins if x + y is odd. To determine how many pairs (x,y) satisfy this condition, we count the number of odd and even numbers within the ranges [1, n] and [1, m]. Then, we multiply the count of odd numbers in the [1, n] range by the count of even numbers in the [1, m] range, and vice-versa. Summing these two products yields the total number of (x, y) pairs where x + y is odd. The provided code directly calculates n * m / 2, because roughly half of the pairs will satisfy the condition when n and m are large enough, and the details of even/odd counts are handled implicitly within the division.

Step-by-Step Algorithm

Step 1: Calculate the number of odd numbers in the range [1, n]. The number of odd numbers is (n + 1) // 2.
Step 2: Calculate the number of even numbers in the range [1, n]. The number of even numbers is n // 2.
Step 3: Calculate the number of odd numbers in the range [1, m]. The number of odd numbers is (m + 1) // 2.
Step 4: Calculate the number of even numbers in the range [1, m]. The number of even numbers is m // 2.
Step 5: The number of pairs where x + y is odd is (number of odds in [1, n] * number of evens in [1, m]) + (number of evens in [1, n] * number of odds in [1, m]). This simplifies to (n * m) // 2.

Key Insights

Insight 1: Alice wins if and only if the total number of flowers (x + y) is odd. This is because Alice can always force Bob into a position where the number of flowers remaining is even, ultimately leaving Bob with no flowers to pick when it's his turn if the initial sum is odd.
Insight 2: We need to count the number of pairs (x, y) where 1 <= x <= n and 1 <= y <= m, and x + y is odd.
Insight 3: Instead of iterating through all possible pairs and checking the condition, we can count the number of even and odd numbers in each range [1, n] and [1, m] and use those counts to compute the answer in O(1) time.

Complexity Analysis

Time Complexity: O(1)

Space Complexity: O(1)

Topics

This problem involves: Math.

Companies

Asked at: Rubrik.

Alice and Bob Playing Flower Game | LeetCode 3021

Medium

Math

Time: O(1)Space: O(1)

Solution Code

Available in:

class Solution {
    public long flowerGame(int n, int m) {
        return (long)(n * m) / 2;
    }
}

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Problem Info

Description

Problem 3021: Alice and Bob Playing Flower Game - Detailed explanation and solution approaches.

Topics

Math

Companies

Rubrik

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:3021

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

5 steps

1
Step 1: Calculate the number of odd numbers in the range [1, n]. The number of odd numbers is (n + 1) // 2.
2
Step 2: Calculate the number of even numbers in the range [1, n]. The number of even numbers is n // 2.
3
Step 3: Calculate the number of odd numbers in the range [1, m]. The number of odd numbers is (m + 1) // 2.
4
Step 4: Calculate the number of even numbers in the range [1, m]. The number of even numbers is m // 2.
5
Step 5: The number of pairs where x + y is odd is (number of odds in [1, n] * number of evens in [1, m]) + (number of evens in [1, n] * number of odds in [1, m]). This simplifies to (n * m) // 2.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(1)

⚡ Constant time - Super fast! Same speed regardless of input size.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The solution involves only basic arithmetic operations (multiplication and division), which take constant time regardless of the input size. No additional data structures are used, hence constant space complexity.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Brute-Force Iteration

Iterate through all possible pairs (x, y) where 1 <= x <= n and 1 <= y <= m. For each pair, check if x + y is odd. If it is, increment a counter. Return the counter.

Time: O(n*m)Space: O(1)

Trade-offs: The brute-force approach is straightforward to understand but inefficient for larger values of n and m. It has a time complexity of O(n*m), making it unsuitable for the given constraints. The main solution is much better for larger constraints since its complexity is O(1).

✨

Key Insights

1Insight 1: Alice wins if and only if the total number of flowers (x + y) is odd. This is because Alice can always force Bob into a position where the number of flowers remaining is even, ultimately leaving Bob with no flowers to pick when it's his turn if the initial sum is odd.
2Insight 2: We need to count the number of pairs (x, y) where 1 <= x <= n and 1 <= y <= m, and x + y is odd.
3Insight 3: Instead of iterating through all possible pairs and checking the condition, we can count the number of even and odd numbers in each range [1, n] and [1, m] and use those counts to compute the answer in O(1) time.

📚

Key Terms Explained

For Students

Edge Cases

Edge case 1: n = 1, m = 1. The total pairs (x, y) possible is 1. x = 1, y = 1, x + y = 2, which is even, so Alice loses. Output is 0. The code handles this correctly because (1 * 1) // 2 = 0.
Edge case 2: n = 3, m = 2. The pairs are (1,1), (1,2), (2,1), (2,2), (3,1), (3,2). Sums are 2, 3, 3, 4, 4, 5. Alice wins in (1,2), (2,1), (3,2) i.e. 3 pairs. The code handles this correctly because (3 * 2) // 2 = 3.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Failing to recognize the winning condition (x + y is odd) and trying to simulate the game turns, which is unnecessary and computationally expensive.

✗Common mistake 2: Using integer division without considering potential rounding errors that might affect the correctness of the result. Casting to long to ensure correct multiplication before dividing.

✗Common mistake 3: Incorrectly counting the number of odd and even numbers within the ranges [1, n] and [1, m].

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Game theory problems often involve analyzing winning conditions based on parity (evenness or oddness). This problem demonstrates a simple example of such analysis.

🏦Real-world scenario 2: Resource allocation problems where decisions need to be made based on whether the total number of resources is even or odd. For example, determining if a workload can be split evenly between two processors.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary values.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: October 20, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 13

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 6

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJun 29

❓

Frequently Asked Questions

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Alice and Bob Playing Flower Game"?

For this medium-level Math problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether standard techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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