Last Stone Weight in Python
Suppose we have some rocks, each rock has a positive integer weight. In each turn, we will take two heaviest rocks and smash them together. consider the stones have weights x and y and x <= y. The result of this smash is can be of two types.
- If x = y, then both stones are totally destroyed;
- Otherwise when x != y, the stone of weight x is totally destroyed, and the stone of weight y has new weight y-x.
Finally, there is at most 1 stone left. We have to find the weight of this stone (0 when there are no stones left.)
So if the stone weights are [2,7,4,1,8,1], then the result will be 1. At first select 7 and 8, then get 1, so array will be [2,4,1,1,1], secondly take 2 and 4. The array will be [2,1,1,1], after that select 2 and 1, array will be [1,1,1], then select two stones with weight 1, then both will be destroyed, so array will be [1]. This is the answer
To solve this, we will follow these steps −
- If stone weight array W has no element, then return 0
- If W has only one element then return the W[0]
- while W has more than one elements −
- sort W
- s1 := last element of W, s2 := second last element of W
- if s1 = s2, then remove s1 and s2 from W
- otherwise s1 := |s1 – s2|, remove last element from W, then set s1 as last element of W
- if W has one element, then return that element, otherwise return 0
Example
Let us see the following implementation to get better understanding −
Live Demo
class Solution(object):
def lastStoneWeight(self, stones):
"""
:type stones: List[int]
:rtype: int
"""
if len(stones) ==0:
return 0
if len(stones) == 1:
return 1
while len(stones)>1:
stones.sort()
s1,s2=stones[-1],stones[-2]
if s1==s2:
stones.pop(-1)
stones.pop(-1)
else:
s1 = abs(s1-s2)
stones.pop(-1)
stones[-1] = s1
if len(stones):
return stones[-1]
return 0
ob1 = Solution()
print(ob1.lastStoneWeight([2,7,4,1,6,1]))
Input
[2,7,4,1,6,1]
Output
1
Published on 16-Jan-2020 16:39:11
