# Find the longest subsequence of an array having LCM at most K in Python

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Suppose we have an array A of n different numbers and another positive integer K, we have to find the longest sub-sequence in the array having Least Common Multiple (LCM) at most K. After than return the LCM and the length of the sub-sequence, following the indexes (starting from 0) of the elements of the obtained sub-sequence. Otherwise, return -1.

So, if the input is like A = [3, 4, 5, 6], K = 20, then the output will be LCM = 12, Length = 3, Indexes = [0,1,3]

To solve this, we will follow these steps −

• n := size of A

• my_dict := a map

• for i in range 0 to n, do

• my_dict[A[i]] := my_dict[A[i]] + 1

• count := an array of size (k + 1) fill with 0

• for each key in my_dict, do

• if key <= k, then

• i := 1

• while key * i <= k, do

• count[key * i] := count[key * i] + my_dict[key]

• i := i + 1

• otherwise,

• come out from the loop

• lcm := 0, size := 0

• for i in range 1 to k + 1, do

• if count[i] > size, then

• size := count[i]

• lcm := i

• if lcm is same as 0, then

• return -1

• otherwise,

• display lcm and size

• for i in range 0 to n, do

• if lcm mod A[i] is same as 0, then

• display i

## Example

Let us see the following implementation to get better understanding −

Live Demo

from collections import defaultdict
def get_seq(A,k):
n = len(A)
my_dict = defaultdict(lambda:0)
for i in range(0, n):
my_dict[A[i]] += 1
count = [0] * (k + 1)
for key in my_dict:
if key <= k:
i = 1
while key * i <= k:
count[key * i] += my_dict[key]
i += 1
else:
break
lcm = 0
size = 0
for i in range(1, k + 1):
if count[i] > size:
size = count[i]
lcm = i
if lcm == 0:
print(-1)
else:
print("LCM = {0}, Length = {1}".format(lcm, size))
print("Index values: ", end = "")
for i in range(0, n):
if lcm % A[i] == 0:
print(i, end = " ")

k = 20
A = [3, 4, 5, 6]
get_seq(A,k)

## Input

[3, 4, 5, 6] , 20

## Output

LCM = 12, Length = 3 Index values: 0 1 3
Updated on 20-Aug-2020 07:49:33