# Longest Common Subsequence

This problem is very practical. In Unix utility `diff`

, a line of one file is compared w/ that of another. In `git diff`

, a line of newer version is compared w/ that of older version. In this problem, we simply compute the common chars of 2 strings.

## # Solution

Let `text1`

and `text2`

, respectively.

## Initial Wrong Solution

Following is a wrong solution because I didn't consider the case that a subsequence later seen can be longer than a subsequence earlier seen.

This test case failed b/c `p`

is before `qr`

in shorter `text1`

while after `qr`

in `text2`

.

```
text1 = "oxcpqrsvwf"
text2 = "shmtulqrypy"
```

```
def longestCommonSubsequence(self, text1: str, text2: str) -> int:
lcs = 0
l = 0
shorter = text1 if len(text1) <= len(text2) else text2
longer = text1 if len(text1) > len(text2) else text2
for i in range(0, len(shorter)):
# find the shorter ith elt in longer[l:]
j = longer[l:].find(shorter[i])
if j!=-1:
l=j+1
lcs+=1
return lcs
```

Thus, this problem should be solved by **DP** (iterative/recursive w/ memoization).

### # Iterative DP (squared space)

As explained in this video, I construct a matrix to store the *length of longest common subsequence* seen so far. If the two chars match, `M[i][j]`

is `M[i-1][j-1]+1`

; o.w., it's the max of `M[i-1][j]`

and `M[i][j-1]`

. At the end of for loop, return `M[-1][-1]`

.

Complexity:

- Time:
- Space:

```
def longestCommonSubsequence(self, text1: str, text2: str) -> int:
n,m = len(text1),len(text2)
M = [[0] * (m + 1) for _ in range(n + 1)]
for i in range(1, n+1):
for j in range(1, m+1):
if text1[i-1]==text2[j-1]:
M[i][j] = M[i-1][j-1]+1
else:
M[i][j] = max(M[i-1][j], M[i][j-1])
return M[-1][-1]
```

### # Iterative DP (linear space)

This is similar to iterative DP, but has less space complexity: smaller matrix **only needs previous row** to calculate current row, so only 2 rows are needed.

Please note that the commented line `M[0],M[1] = M[1],M[0]`

doesn't work because Python list is copy **by reference**, not **by value**.

Complexity:

- Time:
- Space:

```
def longestCommonSubsequence(self, text1: str, text2: str) -> int:
n,m = len(text1),len(text2)
# find the shorter string
s = text1 if n<=m else text2
l = text1 if n>m else text2
M = [[0]*(len(s)+1) for i in range(2)]
for i in range(1,len(l)+1):
for j in range(1,len(s)+1):
if s[j-1]==l[i-1]:
M[i%2][j] = M[1-i%2][j-1] + 1
else:
M[i%2][j] = max(M[1-i%2][j],M[i%2][j-1])
# M[0],M[1] = M[1],M[0]
return M[-1][-1]
```

### # Recursive DP (REDO)

https://leetcode.com/problems/longest-common-subsequence/discuss/436719/Python-very-detailed-solution-with-explanation-and-walkthrough-step-by-step.

https://leetcode.com/problems/longest-common-subsequence/discuss/398711/ALL-4-ways-Recursion-greater-Top-down-greaterBottom-Up-greater-Efficient-Solution-O(N)-including-VIDEO-TUTORIAL