Leetcode #1563: Stone Game V
In this guide, we solve Leetcode #1563 Stone Game V in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
There are several stones arranged in a row, and each stone has an associated value which is an integer given in the array stoneValue. In each round of the game, Alice divides the row into two non-empty rows (i.e.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Array, Math, Dynamic Programming, Game Theory
Intuition
The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.
A carefully chosen DP state captures exactly what we need to build the final answer.
Approach
Define the DP state and recurrence, then compute states in the correct order.
Optionally compress space once the recurrence is clear.
Steps:
- Choose a DP state definition.
- Write the recurrence and base cases.
- Compute states in the correct order.
Example
Input: stoneValue = [6,2,3,4,5,5]
Output: 18
Explanation: In the first round, Alice divides the row to [6,2,3], [4,5,5]. The left row has the value 11 and the right row has value 14. Bob throws away the right row and Alice's score is now 11.
In the second round Alice divides the row to [6], [2,3]. This time Bob throws away the left row and Alice's score becomes 16 (11 + 5).
The last round Alice has only one choice to divide the row which is [2], [3]. Bob throws away the right row and Alice's score is now 18 (16 + 2). The game ends because only one stone is remaining in the row.
Python Solution
def max(a: int, b: int) -> int:
return a if a > b else b
class Solution:
def stoneGameV(self, stoneValue: List[int]) -> int:
def dfs(i: int, j: int) -> int:
if i >= j:
return 0
ans = l = 0
r = s[j + 1] - s[i]
for k in range(i, j):
l += stoneValue[k]
r -= stoneValue[k]
if l < r:
if ans >= l * 2:
continue
ans = max(ans, l + dfs(i, k))
elif l > r:
if ans >= r * 2:
break
ans = max(ans, r + dfs(k + 1, j))
else:
ans = max(ans, max(l + dfs(i, k), r + dfs(k + 1, j)))
return ans
s = list(accumulate(stoneValue, initial=0))
return dfs(0, len(stoneValue) - 1)
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.