Coin Game
Coin Game
You and a friend are playing a game with n
coins arranged in a straight line, where each coin has a distinct
value given by coins[i]
. Starting with you, both players take turns, picking one coin from either end of the
line and adding its value to their individual scores. Once a coin is picked up, it's removed from the line.
Given that your friend plays optimally to achieve the highest score, determine your maximum possible score.
Input
coins
: A list of the coins.
Output
Your maximum possible score, provided that you go first and your friend plays perfectly.
Examples
Example 1:
Input:
coins = [4, 4, 9, 4]
Output: 13
Explanation:
The coins start like this:
4, 4, 9, 4
You always go first, so you take the 4 from the left side:
4, 9, 4
Your friend takes any of the 4s (it doesn't matter)
9, 4
Now you take the 9, and your friend takes the remaining 4.
Your score in this case, is 4 + 9 = 13.
Constraints
1 <= len(coins) <= 1000
1 <= coins[i] <= 5 * 10^5
Try it yourself
Solution
Brute Force
A brute force solution would enumerate through all possibilities. For each of the n
turns, we either choose the left-most coin or the right-most coin and check which option maximizes our score.
The 2 cases mentioned above are described as follows:
- Case 1: We take coin
l
- Coins in the range
[l + 1, r]
are left - Since our opponent plays optimally, they will gain points equal to
maxScore(l + 1, r)
- Since we get all other coins, our score will be
sum(l, r) - maxScore(l + 1, r)
- Coins in the range
- Case 2: We take coin
r
- Coins in range
[l, r - 1]
are left - Since our opponent plays optimally, they will gain points equal to
maxScore(l, r - 1)
- Since we get all other coins, our score will be
sum(l, r) - maxScore(l, r - 1)
- Coins in range
Next, we choose the case that gives us the greatest score, or minimizes the opponent's score. Therefore, the solution is either:
maxScore(l, r) = max(sum(l, r) - maxScore(l + 1, r), sum(l, r) - maxscore(l, r - 1))
ormaxScore(l, r) = sum(l, r) - min(maxScore(l + 1, r), maxScore(l, r - 1))
Note that a common pitfall is to try to keep tracking of which player the current recursive call is for using an extra state variable. This is unnecessary because it doesn't really matter which user the recursive call is for because the current player always tries to minimize the other player's score (each user "plays perfectly in such a way that maximizes their score"). Each recursive call returns the best possible score the current player can get. The outermost call will return the best score for the first player.
Since there are n
turns, 2 possibilities each turn, and takes O(n)
to calculate the sum from l
to r
,
the final runtime is O(n * 2^n)
.
Here's a visual representation of the idea:
The following is an implementation of the idea above:
1from typing import List
2
3def max_score(coins: List[int], l: int, r: int) -> int:
4 # Base case: If only one coin is present, return its value
5 if l == r:
6 return coins[r]
7
8 # Calculate the sum of the coins in the range [l, r]
9 total = 0
10 for i in range(l, r + 1):
11 total += coins[i]
12
13 # Calculate the score when choosing the leftmost or rightmost coin
14 left_pick = max_score(coins, l + 1, r)
15 right_pick = max_score(coins, l, r - 1)
16
17 # Return the best score after choosing the optimal coin
18 return total - min(left_pick, right_pick)
19
20def coin_game(coins: List[int]) -> int:
21 n = len(coins)
22 return max_score(coins, 0, n - 1)
23
24if __name__ == "__main__":
25 coins = [int(x) for x in input().split()]
26 res = coin_game(coins)
27 print(res)
28
1import java.util.Arrays;
2import java.util.List;
3import java.util.Scanner;
4import java.util.stream.Collectors;
5
6class Solution {
7 public static int maxScore(List<Integer> coins, int l, int r) {
8 // Base case: If only one coin is present, return its value
9 if (l == r) {
10 return coins.get(r);
11 }
12
13 // Calculate the sum of the coins in the range [l, r]
14 int sum = 0;
15 for (int i = l; i <= r; i++) {
16 sum += coins.get(i);
17 }
18
19 // Calculate the score when choosing the leftmost or rightmost coin
20 int leftPick = maxScore(coins, l + 1, r);
21 int rightPick = maxScore(coins, l, r - 1);
22
23 // Return the best score after choosing the optimal coin
24 return sum - Math.min(leftPick, rightPick);
25 }
26
27 public static int coinGame(List<Integer> coins) {
28 int n = coins.size();
29 return maxScore(coins, 0, n - 1);
30 }
31
32 public static List<String> splitWords(String s) {
33 return s.isEmpty() ? List.of() : Arrays.asList(s.split(" "));
34 }
35
36 public static void main(String[] args) {
37 Scanner scanner = new Scanner(System.in);
38 List<Integer> coins = splitWords(scanner.nextLine()).stream().map(Integer::parseInt).collect(Collectors.toList());
39 scanner.close();
40 int res = coinGame(coins);
41 System.out.println(res);
42 }
43}
44
1"use strict";
2
3function maxScore(coins, l, r) {
4 // Base case: If only one coin is present, return its value
5 if (l === r) {
6 return coins[r];
7 }
8
9 // Calculate the sum of the coins in the range [l, r]
10 let sum = 0;
11 for (let i = l; i <= r; i++) {
12 sum += coins[i];
13 }
14
15 // Calculate the score when choosing the leftmost or rightmost coin
16 const leftPick = maxScore(coins, l + 1, r);
17 const rightPick = maxScore(coins, l, r - 1);
18
19 // Return the best score after choosing the optimal coin
20 return sum - Math.min(leftPick, rightPick);
21}
22
23function coinGame(coins) {
24 const n = coins.length;
25 return maxScore(coins, 0, n - 1);
26}
27
28function splitWords(s) {
29 return s === "" ? [] : s.split(" ");
30}
31
32function* main() {
33 const coins = splitWords(yield).map((v) => parseInt(v));
34 const res = coinGame(coins);
35 console.log(res);
36}
37
38class EOFError extends Error {}
39{
40 const gen = main();
41 const next = (line) => gen.next(line).done && process.exit();
42 let buf = "";
43 next();
44 process.stdin.setEncoding("utf8");
45 process.stdin.on("data", (data) => {
46 const lines = (buf + data).split("\n");
47 buf = lines.pop();
48 lines.forEach(next);
49 });
50 process.stdin.on("end", () => {
51 buf && next(buf);
52 gen.throw(new EOFError());
53 });
54}
55
1#include <algorithm>
2#include <iostream>
3#include <iterator>
4#include <sstream>
5#include <string>
6#include <vector>
7
8int max_score(std::vector<int>& coins, int l, int r) {
9 // Base case: If only one coin is present, return its value
10 if (l == r) {
11 return coins[r];
12 }
13
14 // Calculate the sum of the coins in the range [l, r]
15 int sum = 0;
16 for (int i = l; i <= r; i++) {
17 sum += coins[i];
18 }
19
20 // Calculate the score when choosing the leftmost or rightmost coin
21 int left_pick = max_score(coins, l + 1, r);
22 int right_pick = max_score(coins, l, r - 1);
23
24 // Return the best score after choosing the optimal coin
25 return sum - std::min(left_pick, right_pick);
26}
27
28int coin_game(std::vector<int>& coins) {
29 int n = coins.size();
30 return max_score(coins, 0, n - 1);
31}
32
33template<typename T>
34std::vector<T> get_words() {
35 std::string line;
36 std::getline(std::cin, line);
37 std::istringstream ss{line};
38 ss >> std::boolalpha;
39 std::vector<T> v;
40 std::copy(std::istream_iterator<T>{ss}, std::istream_iterator<T>{}, std::back_inserter(v));
41 return v;
42}
43
44int main() {
45 std::vector<int> coins = get_words<int>();
46 int res = coin_game(coins);
47 std::cout << res << '\n';
48}
49
Slight Optimization
So far, for each recursive call we are spending O(n)
time to calculate the sum between the range [l, r]
. However, instead of using O(n)
every time we need the sum we can use a prefix sum array to get the range sum in O(1)
time and a single O(n)
pre-computation.
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Top-down DP
In essence, the DP top-down solution is the brute force solution with memoization. You may have noticed maxScore(l, r)
can be be memoized.
Let’s formalize our idea above by specifying our DP state (i.e. what’s important that can lead us to our answer). In this case, let dp(l, r)
be the maximum score we can achieve if coins in the range [l, r]
are the only ones present and we go first.
Furthermore, our base case is: dp(l, r) = v_r
if l = r
. That is, since the range contains only one coin, we simply take that coin.
Now we deal with the transition. Exactly like our brute force solution, we consider two cases of picking either the left or right coin.
Next, we choose the case that gives us the greatest score or minimizes the opponent's score. Therefore, the solution is either:
dp(l, r) = max(sum(l, r) - dp(l + 1, r), sum(l, r) - dp(l, r - 1))
ordp(l, r) = sum(l, r) - min(dp(l + 1, r), dp(l, r - 1))
The following is a top-down solution to the problem:
1import sys
2from typing import List
3
4sys.setrecursionlimit(1500)
5
6def max_score(dp: List[List[int]], prefix_sum: List[int], l: int, r: int) -> int:
7 # Return memoized result
8 if dp[l][r] != 0:
9 return dp[l][r]
10
11 # Calculate total coins value from l to r
12 total = prefix_sum[r] - prefix_sum[l - 1]
13
14 # If only one coin is present, return its value
15 if l == r:
16 dp[l][r] = total
17 else:
18 # Consider left and right coin and choose the better option
19 dp[l][r] = total - min(
20 max_score(dp, prefix_sum, l + 1, r),
21 max_score(dp, prefix_sum, l, r - 1),
22 )
23
24 return dp[l][r]
25
26def coin_game(coins: List[int]) -> int:
27 n = len(coins)
28 # Initialize the DP table with zeroes
29 dp = [[0 for i in range(n + 1)] for j in range