How do you solve Alice and Bob Playing Flower Game on LeetCode?

Alice and Bob Playing Flower Game (LeetCode #3021) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(1) time complexity and O(1) space complexity. Key insight: Alice wins if x and y have different parities.

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

The optimal solution for Alice and Bob Playing Flower Game runs in O(1) time and uses O(1) space. This complexity is constant because we are performing a fixed number of arithmetic operations regardless of the input size.

What is the brute force solution for Alice and Bob Playing Flower Game?

The brute force approach for Alice and Bob Playing Flower Game is Brute Force, with O(n²) time complexity and O(1) space complexity. We can check every possible pair of (x, y) where x is in the range [1, n] and y is in the range [1, m]. For each pair, we determine if Alice can win based on the parity of x and y. The optimal Optimal Solution approach improves this to O(1).

What algorithm or data structure is used to solve Alice and Bob Playing Flower Game?

Alice and Bob Playing Flower Game uses the following concepts: Math. The recommended patterns to study are: Mathematical Counting, Parity Checking.

What are the interview tips for Alice and Bob Playing Flower Game?

Always check for edge cases, such as minimum values of n and m. Explain your thought process clearly while coding. Optimize your solution if you notice a pattern or mathematical relationship.

What are common mistakes when solving Alice and Bob Playing Flower Game?

Not considering the parity condition correctly. Overcomplicating the problem with unnecessary loops.

#3021

Alice and Bob Playing Flower Game

Medium

MathMathematical CountingParity Checking

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(1)
Space	O(1)	O(1)

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Intuition

Time O(1)Space O(1)

Instead of checking every pair, we can derive the count of valid pairs based on the parity of n and m. The key insight is that Alice wins if x and y have different parities.

⚙️

Algorithm

4 steps

1Step 1: Calculate the number of odd and even numbers in the range [1, n].
2Step 2: Calculate the number of odd and even numbers in the range [1, m].
3Step 3: The valid pairs are formed by combining odd x with even y and even x with odd y.
4Step 4: Return the sum of these combinations.

solution.py6 lines

1def countWinningPairs(n, m):
2    odd_n = (n + 1) // 2
3    even_n = n // 2
4    odd_m = (m + 1) // 2
5    even_m = m // 2
6    return odd_n * even_m + even_n * odd_m

ℹ

Complexity note: This complexity is constant because we are performing a fixed number of arithmetic operations regardless of the input size.

1Alice wins if x and y have different parities.
2Counting odds and evens allows for a quick calculation of valid pairs.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.