877. Stone Game #

Problem #

Alex and Lee play a game with piles of stones.  There are an even number of piles arranged in a row, and each pile has a positive integer number of stones piles[i].

The objective of the game is to end with the most stones.  The total number of stones is odd, so there are no ties.

Alex and Lee take turns, with Alex starting first.  Each turn, a player takes the entire pile of stones from either the beginning or the end of the row.  This continues until there are no more piles left, at which point the person with the most stones wins.

Assuming Alex and Lee play optimally, return True if and only if Alex wins the game.

Example 1:

Input: piles = [5,3,4,5]
Output: true
Explanation:
Alex starts first, and can only take the first 5 or the last 5.
Say he takes the first 5, so that the row becomes [3, 4, 5].
If Lee takes 3, then the board is [4, 5], and Alex takes 5 to win with 10 points.
If Lee takes the last 5, then the board is [3, 4], and Alex takes 4 to win with 9 points.
This demonstrated that taking the first 5 was a winning move for Alex, so we return true.

Constraints:

• 2 <= piles.length <= 500
• piles.length is even.
• 1 <= piles[i] <= 500
• sum(piles) is odd.

Problem Summary #

Alex and Lee are playing a game with several piles of stones. An even number of piles of stones are arranged in a row, and each pile has a positive integer number of stones piles[i] . The game is decided by who has the most stones in hand. The total number of stones is odd, so there are no ties. Alex and Lee take turns, with Alex starting first. Each turn, a player takes the entire pile of stones from either the beginning or the end of the row. This continues until there are no more piles left, at which point the player with the most stones in hand wins. Assuming Alex and Lee both play optimally, return true when Alex wins the game, and return false when Lee wins the game.

Solution Approach #

• When encountering a stone problem, it is easy to think about whether it is related to parity. This problem specifies that the number of stone piles must be even. So try using this as the breakthrough point. Alex takes first, either taking the stones at the beginning of the row with index 0, or the stones at the end of the row with index n-1. Suppose he takes the stones at index 0; the remaining stone pile indices are from 1 ~ n-1, meaning Lee can only take a stone pile with an odd index, 1 or n-1. Suppose Alex first takes the stones at index n-1; the remaining stone pile indices are from 0 ~ n-2, meaning Lee can only take a stone pile with an even index. Therefore, Alex’s winning strategy is, in each round of taking stones, to choose the larger of the total number of stones in the odd-indexed piles and the total number of stones in the even-indexed piles for that round. Then in the next round, Lee can only take from the set of piles with the relatively smaller total number of stones, and the parity of the indices of the piles Lee takes is completely controlled by Alex’s previous move. So as long as Alex moves first, he can suppress Lee in every round and thus is guaranteed to win.

Code #

package leetcode

func stoneGame(piles []int) bool {
	return true
}