< Previous
Next >
You are given an array of integers stones
where stones[i]
is the weight of the ith
stone.
We are playing a game with the stones. On each turn, we choose any two stones and smash them together. Suppose the stones have weights x
and y
with x <= y
. The result of this smash is:
- If
x == y
, both stones are destroyed, and
- If
x != y
, the stone of weight x
is destroyed, and the stone of weight y
has new weight y - x
.
At the end of the game, there is at most one stone left.
Return the smallest possible weight of the left stone. If there are no stones left, return 0
.
Example 1:
Input: stones = [2,7,4,1,8,1]
Output: 1
Explanation:
We can combine 2 and 4 to get 2, so the array converts to [2,7,1,8,1] then,
we can combine 7 and 8 to get 1, so the array converts to [2,1,1,1] then,
we can combine 2 and 1 to get 1, so the array converts to [1,1,1] then,
we can combine 1 and 1 to get 0, so the array converts to [1], then that's the optimal value.
Example 2:
Input: stones = [31,26,33,21,40]
Output: 5
Example 3:
Input: stones = [1,2]
Output: 1
Constraints:
1 <= stones.length <= 30
1 <= stones[i] <= 100
[Dynamic Programming]
Hints
Hint 1
Think of the final answer as a sum of weights with + or - sign symbols infront of each weight. Actually, all sums with 1 of each sign symbol are possible.
Hint 2
Use dynamic programming: for every possible sum with N stones, those sums +x or -x is possible with N+1 stones, where x is the value of the newest stone. (This overcounts sums that are all positive or all negative, but those don't matter.)