3021. Alice and Bob Playing Flower Game
Problem Description
Alice and Bob are playing a strategy game involving picking flowers arranged in a circle. The number of flowers between them in a clockwise direction is represented by x
, and in an anti-clockwise direction by y
. The game has specific rules:
- Alice will take the first turn.
- On each turn, the player can pick a flower from either the clockwise or anti-clockwise side.
- If there are no flowers left, the player who took the last turn wins by capturing their opponent.
We need to find the total number of unique starting configurations ((x, y)
pairs) that allow Alice to win the game guaranteeing these constraints:
x
is within the range[1, n]
.y
is within the range[1, m]
.
The crux of the problem lies in finding combinations where Alice has a winning strategy based on the rules set forth.
Intuition
The key to solving this problem is recognizing the importance of the parity (odd or even) of the sum of x
and y
. Since Alice goes first, for her to win, she needs to ensure the final move is hers. This can only be guaranteed if x + y
is odd since players alternate turns. In this case, if the sum is even, Bob will take the last turn and win. But if it's odd, Alice will.
Given this, we can split the possible ranges of x
and y
into odd and even counts:
-
When
x
is odd,y
must be even for the sum to be odd. The possible values forx
would be half ofn
, but rounded up, because in a range ofn
, we're looking for the number of odd numbers ((n + 1) // 2
). Form
being the range fory
, the formula to find even counts ism // 2
. -
Conversely, if
x
is even,y
must be odd to satisfy the odd sum condition. Here, the possible values forx
would be half ofn
, but rounded down this time (n // 2
), for even counts. Fory
, the similar logic of finding odd counts leads to(m + 1) // 2
.
Calculating both these scenarios gives us all the possible combinations where Alice will win:
- Count of (odd
x
, eveny
) pairs plus - Count of (even
x
, oddy
) pairs
The product of these counts for each scenario gives us the total winning configurations for Alice, which is the solution to the problem.
Learn more about Math patterns.
Solution Approach
The solution to this game problem uses a straightforward mathematical approach to count the number of valid (x, y)
pairs where Alice can win.
From our intuition, we know that Alice can only win if the total number of flowers (x + y
) is odd. To ensure this, we need to count the number of odd possibilities for x
and pair each with an even possibility for y
, and vice versa.
Here's how the implementation works, using the provided solution code as an example:
-
Calculate the number of odd possibilities for
x
, which amounts to half of the range ofn
rounded up:a1 = (n + 1) // 2
. The//
operator in Python performs integer division, which discards any fractional part, and adding 1 before division accounts for rounding up. -
Similarly, calculate the number of even possibilities for
y
, which is half of the range ofm
rounded down:b1 = (m + 1) // 2
. -
Next, calculate the number of even possibilities for
x
:a2 = n // 2
. -
And the number of odd possibilities for
y
:b2 = m // 2
. -
Finally, the total number of valid
(x, y)
pairs is the sum of the product of odds ofx
with evens ofy
and the product of evens ofx
with odds ofy
:return a1 * b2 + a2 * b1
.
This approach avoids the need for loops or complex data structures. It simply uses arithmetic operations to evaluate the necessary counts to satisfy the game's winning condition for Alice. By scaling the counts of odd and even possibilities against each other, the solution effectively maps out all permutations that lead to Alice's victory.
Note the clever use of integer division and rounding to quickly determine the counts of odd and even numbers within the specified ranges. This technique reduces a potentially complex combinatorial problem to a few arithmetic operations, showcasing the power of mathematical analysis in algorithm design.
Ready to land your dream job?
Unlock your dream job with a 2-minute evaluator for a personalized learning plan!
Start EvaluatorExample Walkthrough
Let's assume we have n = 3
and m = 4
. This means the range of x
(clockwise) can be [1, 2, 3]
, and the range of y
(anti-clockwise) can be [1, 2, 3, 4]
. We will illustrate how to find the total number of unique starting configurations (x, y)
pairs that allow Alice to win the game under these conditions.
-
Odd
x
, Eveny
Pairings:- The total number of odd values
x
can take in the range[1, 3]
is 2 (which are1
and3
). We calculate this usinga1 = (n + 1) // 2
, soa1 = (3 + 1) // 2
, which equals2
. - The range of
m
is[1, 2, 3, 4]
. The total number of even values thaty
can take is 2 (which are2
and4
). We calculate this usingb1 = m // 2
, sob1 = 4 // 2
, which equals2
. - Pairing odd
x
values with eveny
values gives us four configurations:(1, 2)
,(1, 4)
,(3, 2)
, and(3, 4)
.
- The total number of odd values
-
Even
x
, Oddy
Pairings:- The range of
n
is[1, 2, 3]
. The total number of even valuesx
can take is 1 (which is2
) calculated bya2 = n // 2
, soa2 = 3 // 2
, which equals1
. - For
y
, the total number of odd values in the range[1, 2, 3, 4]
is 2 (which are1
and3
). We calculate this usingb2 = (m + 1) // 2
, sob2 = (4 + 1) // 2
, which equals2
. - Pairing even
x
values with oddy
values gives us two configurations:(2, 1)
and(2, 3)
.
- The range of
Adding the counts together:
- We have
a1 * b1
configurations from the first scenario, which translates to2 * 2 = 4
. - We also have
a2 * b2
configurations from the second scenario, which translates to1 * 2 = 2
.
So, we add 4 + 2
to get a total of 6
unique starting configurations where Alice can win.
Therefore, for n = 3
and m = 4
, there are 6
unique game setups (x, y)
where Alice has a guaranteed winning strategy.
Solution Implementation
1class Solution:
2 def flowerGame(self, n: int, m: int) -> int:
3 # Calculate half of the garden, rounded up
4 half_up_n = (n + 1) // 2
5 half_up_m = (m + 1) // 2
6
7 # Calculate half of the garden, rounded down
8 half_down_n = n // 2
9 half_down_m = m // 2
10
11 # Calculate the result based on the game's logic, which seems to involve
12 # multiplying the halves of the two dimensions, with one dimension always rounded up
13 # and the other rounded down, then adding both results
14 result = half_up_n * half_down_m + half_down_n * half_up_m
15
16 return result
17
1class Solution {
2 // Method to compute the result of the flower game
3 public long flowerGame(int petals, int players) {
4 // a1 represents half the petals, rounded up
5 long upperHalfPetals = (petals + 1) / 2;
6 // b1 represents half the players, rounded up
7 long upperHalfPlayers = (players + 1) / 2;
8 // a2 represents half the petals, rounded down
9 long lowerHalfPetals = petals / 2;
10 // b2 represents half the players, rounded down
11 long lowerHalfPlayers = players / 2;
12
13 // Return the sum of cross-products of upper and lower halves of petals and players
14 // This probably follows some game rule logic based on the distribution of petals and players
15 return upperHalfPetals * lowerHalfPlayers + lowerHalfPetals * upperHalfPlayers;
16 }
17}
18
1class Solution {
2public:
3 // Function to play the flower game
4 // Parameters:
5 // n - number of flowers along the length of the garden
6 // m - number of flowers along the width of the garden
7 // Returns the total number of distinct pairs that can be picked
8 long long flowerGame(int n, int m) {
9 // Calculate half of the length of the garden rounded up
10 long long length_rounded_up = (n + 1) / 2;
11 // Calculate half of the width of the garden rounded up
12 long long width_rounded_up = (m + 1) / 2;
13
14 // Calculate half of the length of the garden rounded down
15 long long length_rounded_down = n / 2;
16 // Calculate half of the width of the garden rounded down
17 long long width_rounded_down = m / 2;
18
19 // Total number of distinct pairs is calculated by
20 // Adding the product of half the length (rounded up)
21 // and half the width (rounded down)
22 // with the product of half the length (rounded down)
23 // and half the width (rounded up)
24 // This is because a flower can be paired with any flower not on its row or column
25 // We round up and down to cover both even and odd numbers of n and m
26 long long total_pairs = length_rounded_up * width_rounded_down + length_rounded_down * width_rounded_up;
27
28 // Return the total number of pairs
29 return total_pairs;
30 }
31};
32
1// Defines a function that calculates a value based on the input parameters
2// related to a hypothetical 'flower game'.
3// @param playerCount - an integer representing the number of players in the game.
4// @param flowerCount - an integer representing the number of flowers in the game.
5// @returns an integer calculated based on a specific formula.
6function flowerGame(playerCount: number, flowerCount: number): number {
7 // Calculate the midpoints for the players and flowers, rounding down for even numbers
8 // and rounding up for odd numbers, which is effectively a ceiling division by 2.
9 const midpointPlayersCeil = Math.ceil(playerCount / 2);
10 const midpointFlowersCeil = Math.ceil(flowerCount / 2);
11
12 // Calculate the midpoints for the players and flowers, rounding down.
13 const midpointPlayersFloor = Math.floor(playerCount / 2);
14 const midpointFlowersFloor = Math.floor(flowerCount / 2);
15
16 // Returns the result of the game's specific formula calculation.
17 // The formula involves multiplying the opposing midpoints (ceiling midpoint of one group
18 // with the floor midpoint of the other group) and adding the two products together.
19 return (midpointPlayersCeil * midpointFlowersFloor) + (midpointPlayersFloor * midpointFlowersCeil);
20}
21
Time and Space Complexity
The time complexity of the code is O(1)
because all operations (+
, //
, *
) are constant time operations that do not depend on the size of the input. The calculations are done in a fixed number of steps regardless of the values of n
and m
.
The space complexity is also O(1)
as space usage does not scale with input size; only a fixed number of integer variables (a1
, b1
, a2
, b2
) are used, which occupy a constant amount of space.
Learn more about how to find time and space complexity quickly using problem constraints.
Which of these properties could exist for a graph but not a tree?
Recommended Readings
Math for Technical Interviews How much math do I need to know for technical interviews The short answer is about high school level math Computer science is often associated with math and some universities even place their computer science department under the math faculty However the reality is that you
LeetCode Patterns Your Personal Dijkstra's Algorithm to Landing Your Dream Job The goal of AlgoMonster is to help you get a job in the shortest amount of time possible in a data driven way We compiled datasets of tech interview problems and broke them down by patterns This way we
Recursion Recursion is one of the most important concepts in computer science Simply speaking recursion is the process of a function calling itself Using a real life analogy imagine a scenario where you invite your friends to lunch https algomonster s3 us east 2 amazonaws com recursion jpg You first
Got a question? Ask the Monster Assistant anything you don't understand.
Still not clear?  Submit the part you don't understand to our editors. Or join our Discord and ask the community.