Find a sorted subsequence of size 3 in linear time in Python

Suppose we have an array with N numbers, we have to check whether the 3 elements such that b[i]< b[j] < b[k] and i < j < k in linear (O(n)) time. If there are multiple such triplets, then print any one of them.

So, if the input is like [13, 12, 11, 6, 7, 3, 31], then the output will be [6,7,31]

To solve this, we will follow these steps −

• n := size of A
• maximum := n-1, minimum := 0
• smaller := an array of size 1000, and fill with 0
• smaller[0] := -1
• for i in range 1 to n, do
  • if A[i] <= A[minimum], then
    • minimum := i
    • smaller[i] := -1
  • otherwise,
    • smaller[i] := minimum
• greater := an array of size 1000, and fill with 0
• greater[n-1] := -1
• for i in range n-2 to -1, decrease by 1, do
  • if A[i] >= A[maximum], then
    • maximum := i
    • greater[i] := -1
  • otherwise,
    • greater[i] := maximum
• for i in range 0 to n, do
  • if smaller[i] is not same as -1 and greater[i] is not same as -1, then
    • return A[smaller[i]], A[i], A[greater[i]]
• return "Nothing"

Example

Let us see the following implementation to get better understanding −

def find_inc_seq(A):
   n = len(A)
   maximum = n-1
   minimum = 0
   smaller = [0]*10000
   smaller[0] = -1
   for i in range(1, n):
      if (A[i] <= A[minimum]):
         minimum = i
         smaller[i] = -1
      else:
         smaller[i] = minimum
   greater = [0]*10000
   greater[n-1] = -1
   for i in range(n-2, -1, -1):
      if (A[i] >= A[maximum]):
         maximum = i
         greater[i] = -1
      else:
         greater[i] = maximum
   for i in range(0, n):
      if smaller[i] != -1 and greater[i] != -1:
         return A[smaller[i]], A[i], A[greater[i]]
   return "Nothing"
arr = [13, 12, 11, 6, 7, 3, 31]
print(find_inc_seq(arr) )

Input

[13, 12, 11, 6, 7, 3, 31]

Output

(6, 7, 31)