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Find longest bitonic sequence such that increasing and decreasing parts are from two different arrays in Python
Suppose we have two arrays; we have to find the longest possible bitonic sequence so that the increasing part should be from first array and should be a subsequence of first array. similarly decreasing part of must be from second array and a subsequence of the second one.
So, if the input is like A = [2, 6, 3, 5, 4, 6], B = [9, 7, 5, 8, 4, 3], then the output will be [2, 3, 4, 6, 9, 7, 5, 4, 3]
To solve this, we will follow these steps −
Define a function index_ceiling() . This will take arr, T, left, right, key
while right - left > 1, do
mid := left +(right - left) / 2;
if arr[T[mid]] >= key, then
right := mid
otherwise,
left := mid
return right
Define a function long_inc_seq() . This will take A
n := size of A
tails_idx := an array of size n, fill with 0
prev_idx := an array of size n, fill with -1
length := 1
for i in range 1 to n, do
if A[i] < A[tails_idx[0]], then
tails_idx[0] := i
otherwise when A[i] > A[tails_idx[length - 1]], then
prev_idx[i] := tails_idx[length - 1]
tails_idx[length] := i
length := length + 1
otherwise,
pos := index_ceiling(A, tails_idx, -1, length - 1, A[i])
prev_idx[i] := tails_idx[pos - 1]
tails_idx[pos] := i
i := tails_idx[length - 1]
while i >= 0, do
insert A[i] at the end of answer
i := prev_idx[i]
From the main method, do the following −
n1 := size of A, n2 := size of B
long_inc_seq(A)
answer := reverse answer
B := reverse B
long_inc_seq(B)
return answer
Example
Let us see the following implementation to get better understanding −
answer = [] def index_ceiling(arr,T, left,right, key): while (right - left > 1): mid = left + (right - left) // 2; if (arr[T[mid]] >= key): right = mid else: left = mid return right def long_inc_seq(A): n = len(A) tails_idx = [0]*(n) prev_idx = [-1]*(n) length = 1 for i in range(1, n): if (A[i] < A[tails_idx[0]]): tails_idx[0] = i elif (A[i] > A[tails_idx[length - 1]]): prev_idx[i] = tails_idx[length - 1] tails_idx[length] = i length += 1 else: pos = index_ceiling(A, tails_idx, -1, length - 1, A[i]) prev_idx[i] = tails_idx[pos - 1] tails_idx[pos] = i i = tails_idx[length - 1] while(i >= 0): answer.append(A[i]) i = prev_idx[i] def long_bitonic(A,B): n1 = len(A) n2 = len(B) global answer long_inc_seq(A) answer = answer[::-1] B = B[::-1] long_inc_seq(B) A = [2, 6, 3, 5, 4, 6] B = [9, 7, 5, 8, 4, 3] long_bitonic(A,B) print(answer)
Input
[2, 6, 3, 5, 4, 6], [9, 7, 5, 8, 4, 3]
Output
[2, 3, 4, 6, 9, 7, 5, 4, 3]
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