🧠 Solving LeetCode Until I Become Top 1% — Day `70`

This content originally appeared on DEV Community and was authored by Md. Rishat Talukder

Problem: 3021. Alice and Bob Playing Flower Game

Difficulty: #Medium
Tags: #Math, #Combinatorics

Problem Summary

Alice and Bob play a game with n flowers and m flowers. Each chooses one flower. Alice wins if the total number of petals is odd.

You need to count the number of winning pairs (x, y) where 1 ≤ x ≤ n and 1 ≤ y ≤ m.

My Thought Process

• Brute Force Idea:
  Loop through all pairs (x, y) and check if x + y is odd. That would be O(n * m), which is too slow for large inputs.

• Optimized Strategy:

  • The solution was unexpectedly simple once I realized the parity logic:
  • Alice wins only when one number is even and the other is odd.
  • Count how many odds and evens are in 1..n and 1..m.
  • odds_in_n = (n + 1) // 2
  • evens_in_n = n // 2
  • odds_in_m = (m + 1) // 2
  • evens_in_m = m // 2
  • Winning pairs = (odds_in_n * evens_in_m) + (evens_in_n * odds_in_m)
  • Simplifies to (n * m) // 2.

Code Implementation (Python)

class Solution:
    def flowerGame(self, n: int, m: int) -> int:
        return (n * m) // 2

Time & Space Complexity

• Time: O(1)
• Space: O(1)

Key Takeaways

• Learned how to reduce brute-force counting to simple parity logic.
• The trick is noticing that half of all pairs will have odd sums.
• In future, whenever I see problems about sums being odd/even, I’ll try counting odds and evens separately instead of brute force.

Reflection (Self-Check)

• [x] Could I solve this without help?
• [x] Did I write code from scratch?
• [x] Did I understand why it works?
• [x] Will I be able to recall this in a week? → Let’s see

Progress Tracker

Metric	Value
Day	70
Total Problems Solved	431
Confidence Today
Leetcode Rating	1530

This content originally appeared on DEV Community and was authored by Md. Rishat Talukder