3 snippets across 2 stacks - Python, Interview Prep
Also written as list reversal
items[start:stop:step]nums = [10, 20, 30, 40, 50, 60]
print(nums[1:4])
print(nums[::2])
print(nums[::-1])
nums[1:3] = [200, 300]
print(nums)[20, 30, 40]
[10, 30, 50]
[60, 50, 40, 30, 20, 10]
[10, 200, 300, 40, 50, 60]Note Slice assignment can replace a range of elements. The replacement does not need to be the same length as the slice being replaced.
list.sort(key=None, reverse=False)
sorted(iterable, key=None, reverse=False)prices = [29.99, 5.50, 12.00, 89.99]
prices.sort()
print(prices)
names = ["Charlie", "alice", "Bob"]
ordered = sorted(names, key=str.lower)
print(ordered)[5.5, 12.0, 29.99, 89.99]
['alice', 'Bob', 'Charlie']Note .sort() modifies the list in place and returns None. sorted() returns a new list. Use key= for custom ordering logic.
Iterative: maintain prev, current, next pointers.
At each step: save next, point current.next to prev, advance both.// JavaScript - Iterative
function reverseList(head) {
let prev = null, current = head;
while (current !== null) {
const next = current.next;
current.next = prev;
prev = current;
current = next;
}
return prev;
}
// Recursive
function reverseListRecursive(head) {
if (!head || !head.next) return head;
const newHead = reverseListRecursive(head.next);
head.next.next = head;
head.next = null;
return newHead;
}
# Python - Iterative
def reverse_list(head):
prev, current = None, head
while current:
nxt = current.next
current.next = prev
prev = current
current = nxt
return prev
# Recursive
def reverse_list_recursive(head):
if not head or not head.next:
return head
new_head = reverse_list_recursive(head.next)
head.next.next = head
head.next = None
return new_head1→2→3→4 becomes 4→3→2→1Note Iterative: O(n) time, O(1) space. Recursive: O(n) time, O(n) stack space. This is the single most common linked list question. Draw the pointer changes on paper. Follow-up: reverse a sublist from position m to n.