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Longest Increasing Subsequence - Hard


Problem - Given an array of integers. Find the longest increasing subsequence in the array. Eg. [42, 91, 46, 73, 75, 91, 95]. The LIS for this array is 6 which is [42,46,73,75,91,95]. Note: Subsequence does not require the elements to be contiguous

As always, we begin with our naive solution which is a DFS/Bruteforce search trying every possible combination until we get the longest sequence. This is simply exhaustive and will explode on bigger arrays. Time Complexity : O(2^n) ~ Exponential

def longest_increasing_subsequence_dfs(lst):
lis_seq = []
sz = len(lst)
def lis(index,seq):
if len(seq) + (sz - index) < len(lis_seq): return #prune the rest.
for i in xrange(index,sz):
if seq[-1] < lst[i]:
seq.append(lst[i])
lis(i+1,seq)
seq.pop()
else:
if len(seq) > len(lis_seq):
l = lis_seq
l[:] = seq
for i in xrange(sz):
lis(i + 1,[lst[i]])
return lis_seq
view raw Longest Increasing Sequence DFS hosted with ❤ by GitHub
Now, If you notice we repeatedly try a prefix subsequence repeatedly. This tells us, the problem is a good candidate for a dynamic programming algorithm.

We define lis[0...L], where L is the length of the array. We define lis[i] as the length of longest increasing subsequence that ends at i. 
  • lis[0] = 1 { trivial since, the longest increasing sequence that ends at the start of the array contains just the first element, also true for the arr[i] where arr[i] is the smaller than preceding elements}
  • lis[i] = 1 + max( lis[j = 0..i-1] where A[i] > A[j] ) 
To calculate lis[i], we have a new element A[i] which can be part of a new LIS by appending to an existing LIS ending at j where j varies from 0 to i - 1 if and only if, A[i] > A[j]. If there is no such element, ie A[i] is smaller than A[0..i-1] therefore, lis[i] = 1
To build back the LIS, we also store a prev[] array, which holds the preceding index of the previous member of the LIS. The prev for the first element will be marked as negative to indicate termination. We build the LIS backwards and finally reverse the sequence.

Time Complexity : O(n^2)

def longest_increasing_subsequence_dp(lst):
if not lst: return None
lis_len = []
prev = []
for i,v in enumerate(lst):
mx = 0
mxi = -1
for j in xrange(i):
if v > lst[j] and mx < lis_len[j]: #can v extend the current longest increasing subsequence?
mx = lis_len[j]
mxi = j
lis_len.append(1 + mx)
prev.append(mxi)
index = max(xrange(len(lst)),key=lis_len.__getitem__)
lis = []
while index >= 0: # go backwards till index < 0
lis.append(lst[index])
index = prev[index]
lis.reverse()
return lis
view raw Longest Increasing Subsequence DP hosted with ❤ by GitHub

I'll end this post here, but I will follow up with a O(nlogn) method which can be used to solve the problem as well. I will also describe the applications and some problems you can solve using LIS. Until next time.

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