# Interval Dynamic Programming | Coin Game

Interval DP is another sub-type of the dynamic programming technique that deals with ranges or intervals. In general, the final answer to interval DP problems will be the answer to the entire range `[1, n]`, where subproblems are computed by finding the answer to all possible ranges, `[l, r]` where `l <= r`. Alternate names for interval DP are left-right DP or L-R DP.

Interval DP is one of the more challenging types of dynamic programming problems. They might be too difficult for the real interviews. We are including them here for completeness. Don't sweat if you can't get it the firs time.

## Coin Game

There are `n` coins in a straight line. The `i`th coin has a value of `coins[i]`. You play this game with a friend alternating turns, starting with you, remove one coin from one end of the line and add that coin's value to your score.

If your friend plays perfectly in such a way that maximizes their score, what is 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:

``1coins = [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`

## 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)`
• 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)`

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))` or
• `maxScore(l, r) = sum(l, r) - min(maxScore(l + 1, r), maxScore(l, r - 1))`

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:

``````1def max_score(coins, l, r):
2  if l == r:
3    return coins[r]
4
5  sum = 0
6  for i in range(l, r + 1):
7    sum += coins[i]
8
9  left_pick = max_score(coins, l + 1, r)
10  right_pick = max_score(coins, l, r - 1)
11  return sum - min(left_pick, right_pick)
12
13def coin_game(coins):
14  n = len(coins)
15  return max_score(coins, 0, n - 1)
16``````

### 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.

 `Expand 1 lines ...` `2` `2` ``1 if l == r:`` `3` `3` ``1 return coins[r]`` `4` `4` `5` `-` ``1 sum = 0`` `5` `+` ``1 sum = coins[r] - coins[l - 1] # query sum from [l, r] in O(1)`` `6` `-` ``1 for i in range(l, r + 1):`` `7` `-` ``1 sum += coins[i]`` `8` `-` ``1 `` `9` `6` ``1 left_pick = max_score(coins, l + 1, r)`` `10` `7` ``1 right_pick = max_score(coins, l, r - 1)`` `11` `8` ``1 return sum - min(left_pick, right_pick)`` `12` `9` `13` `10` ``1def coin_game(coins):`` `14` `11` ``1 n = len(coins)`` `15` `-` ``1 return max_score(coins, 0, n - 1)`` `12` `+` ``1 prefix_sum = [0 for i in range(n + 1)] # precompute prefix sum`` `13` `+` ``1 for i in range(1, n + 1):`` `14` `+` ``1 prefix_sum[i] = prefix_sum[i - 1] + coins[i - 1]`` `15` `+` `16` `+` ``1 return max_score(prefix_sum, 0, n - 1)``

### 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))` or
• `dp(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
2sys.setrecursionlimit(1500)
3
4def max_score(dp, prefix_sum, l, r):
5  if dp[l][r] != 0:
6    return dp[l][r]
7
8  sum = prefix_sum[r] - prefix_sum[l - 1]
9  if l == r:
10    dp[l][r] = sum
11  else:
12    dp[l][r] = sum - min(max_score(dp, prefix_sum, l + 1, r), max_score(dp, prefix_sum, l, r - 1))
13
14  return dp[l][r]
15
16def coin_game(coins):
17  n = len(coins)
18  dp = [[0 for i in range(n + 1)] for j in range(n + 1)]
19  prefix_sum = [0 for i in range(n + 1)]
20  for i in range(1, n + 1):
21    prefix_sum[i] = prefix_sum[i - 1] + coins[i - 1]
22  return max_score(dp, prefix_sum, 1, n)
23``````

### Bottom-up DP

The iterative version is a bit trickier. The idea is that we loop through all the possible lengths starting from 1 to `n`. Then for each size, we consider all possible left starting positions and calculate the respective right ending position. We do it in this order because we are building solutions from the smallest case and building the solutions up. That is, first considering each item as an interval of length 1 then merging their solutions together to form larger and larger interval lengths until we get to an interval of size `n`.

Note: There are slight tweaks in the implementation of this idea such as how we loop through the lengths.

Here's a visualization of the table generated: And here's the bottom-up iterative code:

``````1from typing import List
2
3def coin_game(coins: List[int]) -> int:
4  n = len(coins)
5  prefix_sum = [0 for i in range(n + 1)]
6  for i in range(1, n + 1):
7    prefix_sum[i] = prefix_sum[i - 1] + coins[i - 1]
8
9  dp = [[0 for i in range(n + 1)] for j in range(n + 1)]
10  for size in range(0, n):
11    for l in range(1, n - size + 1):
12      r = l + size
13      if l == r:
14        dp[l][r] = prefix_sum[r] - prefix_sum[l - 1]
15      else:
16        dp[l][r] = prefix_sum[r] - prefix_sum[l - 1] - min(dp[l + 1][r], dp[l][r - 1])
17  return dp[n]
18
19if __name__ == '__main__':
20    coins = [int(x) for x in input().split()]
21    res = coin_game(coins)
22    print(res)
23``````