Suppose we have a sorted doubly linked list of unique positive numbers; we have to find pairs in the doubly linked list whose product is same as a given value x. We have to keep in mind that, this will be solved without consuming any extra space.
So, if the input is like L = 1 ⇔ 2 ⇔ 4 ⇔ 5 ⇔ 6 ⇔ 8 ⇔ 9 and x = 8, then the output will be (1, 8), (2, 4)
To solve this, we will follow these steps −
curr := head, nxt := head
while nxt.next is not None is non-zero, do
nxt := nxt.next
found := False
while curr and nxt are not null and curr and nxt are different and nxt.next is not curr, do
if (curr.data * nxt.data) is same as x, then
found := True
display a pair curr.data, nxt.data
curr := curr.next
nxt := nxt.prev
otherwise,
if (curr.data * nxt.data) < x, then
curr := curr.next
otherwise,
nxt := nxt.prev
if found is False, then
display "Not found"
Example
Let us see the following implementation to get better understanding −
class ListNode:
def __init__(self, data):
self.data = data
self.prev = None
self.next = None
def insert(head, data):
node = ListNode(0)
node.data = data
node.next = node.prev = None
if (head == None):
(head) = node
else :
node.next = head
head.prev = node
head = node
return head
def get_pair_prod(head, x):
curr = head
nxt = head
while (nxt.next != None):
nxt = nxt.next
found = False
while (curr != None and nxt != None and curr != nxt and nxt.next != curr) :
if ((curr.data * nxt.data) == x) :
found = True
print("(", curr.data, ", ", nxt.data, ")")
curr = curr.next
nxt = nxt.prev
else :
if ((curr.data * nxt.data) < x):
curr = curr.next
else:
nxt = nxt.prev
if (found == False):
print( "Not found")
head = None
head = insert(head, 9)
head = insert(head, 8)
head = insert(head, 6)
head = insert(head, 5)
head = insert(head, 4)
head = insert(head, 2)
head = insert(head, 1)
x = 8
get_pair_prod(head, x)Input
head = None head = insert(head, 9) head = insert(head, 8) head = insert(head, 6) head = insert(head, 5) head = insert(head, 4) head = insert(head, 2) head = insert(head, 1) x = 8
Output
( 1 , 8 ) ( 2 , 4 )