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21 changes: 21 additions & 0 deletions Problem1.py
Original file line number Diff line number Diff line change
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# Reverse Linked List
# Time complexity On
# SPace complexity -> O1
# Logic -> Maintain prev and curr pointer, eveery loop update next of curre to prev, and move both 1 step ahead while maintaing
# the pointer to the next element of the current node using tmp
class ListNode:
def __init__(self, val=0, next=None):
self.val = val
self.next = next

class Solution:
def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:
prev = None
curr = head

while curr !=None:
tmp = curr.next
curr.next = prev
prev = curr
curr = tmp
return prev
30 changes: 30 additions & 0 deletions Problem2.py
Original file line number Diff line number Diff line change
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# Remove Nth Node From End of List

# Time complexity -> O(2n) -> O(n)
# Space complexity -> O1
# Logic -> iterate once find the length, in 2nd iteration stop at the desired position drop the node

# Definition for singly-linked list.
class ListNode:
def __init__(self, val=0, next=None):
self.val = val
self.next = next


class Solution:
def removeNthFromEnd(self, head: Optional[ListNode], n: int) -> Optional[ListNode]:
length = 0

curr = head
while curr!=None:
length+=1
curr=curr.next
dummyNode = ListNode(-1,head)
curr=dummyNode
while length > n:
length-=1
curr=curr.next
print(curr)
curr.next=curr.next.next

return dummyNode.next
37 changes: 37 additions & 0 deletions Problem3.py
Original file line number Diff line number Diff line change
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#Linked List Cycle II

# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None

# Time -> On
# Space -> O1
# Logic -> Take 2 pointers, slow and fast, fast is of double speed. If they meeet, means cycle else No cycle
# if cycle move fast to begining now move at same speed they'll meet at the junction


class Solution:
def detectCycle(self, head: Optional[ListNode]) -> Optional[ListNode]:
slow = head
fast = head

isCyclic = False
while fast != None and fast.next!=None:
slow = slow.next
fast = fast.next.next
if slow == fast:
isCyclic = True
break

if not isCyclic:
return None


fast = head
while fast != slow:
slow=slow.next
fast=fast.next

return fast