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Copy path0673-number-of-longest-increasing-subsequence.py
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0673-number-of-longest-increasing-subsequence.py
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classSolution:
deffindNumberOfLIS(self, nums: List[int]) ->int:
# 1. O(n^2) Recursive solution with Caching
dp= {} # key = index, value = [length of LIS, count]
lenLIS, res=0, 0# length of LIS, count of LIS
defdfs(i):
ifiindp:
returndp[i]
maxLen, maxCnt=1, 1# length and count of LIS
forjinrange(i+1, len(nums)):
ifnums[j] >nums[i]: # make sure increasing order
length, count=dfs(j)
iflength+1>maxLen:
maxLen, maxCnt=length+1, count
eliflength+1==maxLen:
maxCnt+=count
nonlocallenLIS, res
ifmaxLen>lenLIS:
lenLIS, res=maxLen, maxCnt
elifmaxLen==lenLIS:
res+=maxCnt
dp[i] = [maxLen, maxCnt]
returndp[i]
foriinrange(len(nums)):
dfs(i)
returnres
# 2. O(n^2) Dynamic Programming
dp= {} # key = index, value = [length of LIS, count]
lenLIS, res=0, 0# length of LIS, count of LIS
# i = start of subseq
foriinrange(len(nums) -1, -1, -1):
maxLen, maxCnt=1, 1# len, cnt of LIS start from i
forjinrange(i+1, len(nums)):
ifnums[j] >nums[i]:
length, count=dp[j] # len, cnt of LIS start from j
iflength+1>maxLen:
maxLen, maxCnt=length+1, count
eliflength+1==maxLen:
maxCnt+=count
ifmaxLen>lenLIS:
lenLIS, res=maxLen, maxCnt
elifmaxLen==lenLIS:
res+=maxCnt
dp[i] = [maxLen, maxCnt]
returnres
