Gary - 15 days ago 4

Python Question

I am reviewing the GeeksforGeeks code for the Longest Increasing Subsequence and I am trying to figure out how to print the actual longest increasing subsequence, and not just the length.

I know that I somehow need to store the indices of the maximum, and then print out the actual numbers from that array... I'm just having trouble implementing it.

The code from the G4G website is as follows:

`# Dynamic programming Python implementation of LIS problem`

# lis returns length of the longest increasing subsequence

# in arr of size n

def lis(arr):

n = len(arr)

# Declare the list (array) for LIS and initialize LIS

# values for all indexes

lis = [1]*n

# Compute optimized LIS values in bottom up manner

for i in range (1 , n):

for j in range(0 , i):

if arr[i] > arr[j] and lis[i]< lis[j] + 1 :

lis[i] = lis[j]+1

# Initialize maximum to 0 to get the maximum of all

# LIS

maximum = 0

# Pick maximum of all LIS values

for i in range(n):

maximum = max(maximum , lis[i])

return maximum

# end of lis function

# Driver program to test above function

arr = [10, 22, 9, 33, 21, 50, 41, 60]

print "Length of lis is", lis(arr)

# This code is contributed by Nikhil Kumar Singh

Answer

As you have pointed out, the indices of the LIS need to be stored as well.

This can be done by introducing an additional array `prev`

. In the original code, `lis(n)`

represents the length of the LIS ending at `n`

, and we have `prev(n)`

representing the index number immediately before `n`

in the LIS ending at `n`

(i.e. the second last index number of the LIS), if the LIS ending at `n`

is of length `1`

, we simply define `prev(n) = n`

.

Whenever you update the value of `lis(n)`

, `prev(n)`

also needs to be updated accordingly. Please find the enhanced code below for your reference:

```
# Dynamic programming Python implementation of LIS problem
# lis returns length of the longest increasing subsequence
# in arr of size n
def lis(arr):
n = len(arr)
# Declare the list (array) for LIS and initialize LIS
# values for all indexes
lis = [1]*n
prev = [0]*n
for i in range(0, n):
prev[i] = i
# Compute optimized LIS values in bottom up manner
for i in range (1 , n):
for j in range(0 , i):
if arr[i] > arr[j] and lis[i]< lis[j] + 1:
lis[i] = lis[j]+1
prev[i] = j
# Initialize maximum to 0 to get the maximum of all
# LIS
maximum = 0
idx = 0
# Pick maximum of all LIS values
for i in range(n):
if maximum < lis[i]:
maximum = lis[i]
idx = i
seq = [arr[idx]]
while idx != prev[idx]:
idx = prev[idx]
seq.append(arr[idx])
return (maximum, reversed(seq))
# end of lis function
# Driver program to test above function
arr = [10, 22, 9, 33, 21, 50, 41, 60]
ans = lis(arr)
print "Length of lis is", ans[0]
print "The longest sequence is", ", ".join(str(x) for x in ans[1])
```