Auckland Programming Algorithms and Performance Meetup about Stacks & LeetCode Problems

1. Stacks
Rodrigo Villarreal && Paulo Almeida
Auckland Data Structures and
Algorithms

2. Agenda
● Introduction to Stacks
○ Design
○ Real-world applications
● Practice
○ Examples
○ Let’s get our hands dirty!
● What’s next?

3. What is a Stack ?
A stack is an abstract data type that serves as a collection of
elements, with two principal operations:
1 1 2 3 5 6bottom top

4. What is a Stack ?
A stack is an abstract data type that serves as a collection of
elements, with two principal operations:
1 1 2 3 5
6
pop()bottom

5. What is a Stack ?
A stack is an abstract data type that serves as a collection of
elements, with two principal operations:
1 1 2 3 5
8
push(x)
bottom

6. What is a Stack ?
A stack is an abstract data type that serves as a collection of
elements, with two principal operations:
1 1 2 3 5 8
Both operations change the top of the stack.
bottom top

7. What is a Stack ?
This behaviour is sometimes referred as LIFO (Last in, First
out).

8. What is a Stack ?
This behaviour is sometimes referred to as LIFO (Last in, First
out).
1
push(x)
bottom top

9. What is a Stack ?
This behaviour is sometimes referred to as LIFO (Last in, First
out).
1
push(x)
1
bottom

10. What is a Stack ?
This behaviour is sometimes referred to as LIFO (Last in, First
out).
1
push(x)
2
1bottom

11. What is a Stack ?
This behaviour is sometimes referred to as LIFO (Last in, First
out).
1
push(x)
3
1 2bottom

12. What is a Stack ?
This behaviour is sometimes referred to as LIFO (Last in, First
out).
1 31 2bottom top

13. What is a Stack ?
This behaviour is sometimes referred to as LIFO (Last in, First
out).
1 pop()
3
1 2bottom

14. What is a Stack ?
This behaviour is sometimes referred to as LIFO (Last in, First
out).
1 pop()
2
1bottom

15. What is a Stack ?
This behaviour is sometimes referred to as LIFO (Last in, First
out).
1 pop()
1
bottom

16. What is a Stack ?
This behaviour is sometimes referred to as LIFO (Last in, First
out).
1
pop()
bottom top

17. What is a Stack ?
Like in a stack of books. One can
only add or remove from the
top of the stack.
….
(Or push/pop respectively)

18. Big O
Time Complexity (Avg/Worst)
Search: O(n)
Access: O(n)
Insertion: O(1)
Deletion: O(1)
Space Complexity (Worst): O(n)

19. Where would I use Stacks in the real world ?
I’m glad you asked :)

20. Real life examples
Parsing mathematical expressions
String expr = ( 9 + ( 4 / 2 ) ) * 2

21. Real life examples
Parsing mathematical expressions
String expr =
(
push(x)
bottom top
( 9 + ( 4 / 2 ) ) * 2

22. String expr =
(bottom top
( 9 + ( 4 / 2 ) ) * 2
Real life examples
Parsing mathematical expressions

23. String expr =
(bottom
push(x)
9
( 9 + ( 4 / 2 ) ) * 2
Real life examples
Parsing mathematical expressions

24. String expr =
(bottom 9
( 9 + ( 4 / 2 ) ) * 2
Real life examples
Parsing mathematical expressions

25. String expr =
(bottom 9
push(x)
+
( 9 + ( 4 / 2 ) ) * 2
Real life examples
Parsing mathematical expressions

26. String expr =
(bottom 9 +
( 9 + ( 4 / 2 ) ) * 2
Real life examples
Parsing mathematical expressions

27. String expr =
(bottom 9 +
push(x)
(
( 9 + ( 4 / 2 ) ) * 2
Real life examples
Parsing mathematical expressions

28. String expr =
(bottom 9 + (
( 9 + ( 4 / 2 ) ) * 2
Real life examples
Parsing mathematical expressions

29. String expr =
(bottom 9 + (
push(x)
4
( 9 + ( 4 / 2 ) ) * 2
Real life examples
Parsing mathematical expressions

30. String expr =
(bottom 9 + ( 4
( 9 + ( 4 / 2 ) ) * 2
Real life examples
Parsing mathematical expressions

31. String expr =
(bottom 9 + ( 4
( 9 + ( 4 / 2 ) ) * 2
push(x)
/
Real life examples
Parsing mathematical expressions

32. String expr =
(bottom 9 + ( 4
( 9 + ( 4 / 2 ) ) * 2
/
Real life examples
Parsing mathematical expressions

33. String expr =
(bottom 9 + ( 4
( 9 + ( 4 / 2 ) ) * 2
/
push(x)
2
Real life examples
Parsing mathematical expressions

34. String expr =
(bottom 9 + ( 4
( 9 + ( 4 / 2 ) ) * 2
/ 2
Real life examples
Parsing mathematical expressions

35. String expr =
(bottom 9 + ( 4
( 9 + ( 4 / 2 ) ) * 2
/ 2
pop()pop()pop()pop()
Real life examples
Parsing mathematical expressions

36. String expr =
(bottom 9 +
( 9 + ( 4 / 2 ) ) * 2
push(x)
2
Real life examples
Parsing mathematical expressions

37. String expr =
(bottom 9 +
( 9 + ( 4 / 2 ) ) * 2
2
Real life examples
Parsing mathematical expressions

38. String expr =
(bottom 9 +
( 9 + ( 4 / 2 ) ) * 2
2
pop()pop()pop()pop()
Real life examples
Parsing mathematical expressions

39. String expr =
bottom
( 9 + ( 4 / 2 ) ) * 2
11
push(x)
top
Real life examples
Parsing mathematical expressions

40. String expr =
bottom
( 9 + ( 4 / 2 ) ) * 2
11 top
Real life examples
Parsing mathematical expressions

41. String expr =
bottom
( 9 + ( 4 / 2 ) ) * 2
11
push(x)
*
Real life examples
Parsing mathematical expressions

42. String expr =
bottom
( 9 + ( 4 / 2 ) ) * 2
11 *
Real life examples
Parsing mathematical expressions

43. String expr =
bottom
( 9 + ( 4 / 2 ) ) * 2
11 *
push(x)
2
Real life examples
Parsing mathematical expressions

44. String expr =
bottom
( 9 + ( 4 / 2 ) ) * 2
11 * 2
Real life examples
Parsing mathematical expressions

45. String expr =
bottom
( 9 + ( 4 / 2 ) ) * 2
11 * 2
pop()pop()pop()
Real life examples
Parsing mathematical expressions

46. String expr = ( 9 + ( 4 / 2 ) ) * 2
bottom 22
push(x)
top
Real life examples
Parsing mathematical expressions

47. Using Stacks in Java
Declare stack:
Stack<Type> stack = new Stack<>();
Push element:
stack.push(element);
Pop element:
stack.pop();

48. Valid Parentheses (L #20)
Given a string containing just the characters
determine if the input string is valid.
An input string is valid if:
● Open brackets must be closed by the same type of brackets.
● Open brackets must be closed in the correct order.
( ) { } [ ]

49. Valid Parentheses (L #20)
Test cases
Example Output
“()” true
“()[]{}” true
“(]” false
“([)]” false
“{[]}” true

50. DIVISORIA

51. DIVISORIA

52. Valid Parentheses (L #20)
Time Complexity = O(n)
Space Complexity = O(n)

53. Still not convinced… where else would I use
stacks?
Can you relate to this example?

54. Real life examples
Back / Forward Browser Button
Back Stack Forward Stack

55. Real life examples
Back / Forward Browser Button
Back Stack Forward Stack
www.google.comcurrent

56. Real life examples
Back / Forward Browser Button
Back Stack Forward Stack
www.google.com
www.facebook.comcurrent

57. Real life examples
Back / Forward Browser Button
Back Stack Forward Stack
www.google.com
www.facebook.com
current www.twitter.com

58. Real life examples
Back / Forward Browser Button
Back Stack Forward Stack
www.google.com
www.facebook.com
current
www.twitter.com
www.meetup.com

59. Real life examples
Back / Forward Browser Button
Back Stack Forward Stack
www.google.com
www.facebook.com
current
www.twitter.com
www.meetup.com
www.linkedin.com

60. Real life examples
Back / Forward Browser Button
Back Stack Forward Stack
www.google.com
www.facebook.com
current
www.twitter.com
www.meetup.com
www.linkedin.com

61. Real life examples
Back / Forward Browser Button
Back Stack Forward Stack
www.google.com
www.facebook.com
current www.twitter.com
www.meetup.com
www.linkedin.com

62. Real life examples
Back / Forward Browser Button

63. Real life examples
Back / Forward Browser Button
What happens if I write a new URL while I am in Twitter?
Back Stack Forward Stack
www.google.com
www.facebook.com
current www.twitter.com
www.meetup.com
www.linkedin.com

64. Real life examples
Back / Forward Browser Button
Forward Stack gets cleared
Back Stack Forward Stack
www.google.com
www.facebook.com
current
www.twitter.com
www.gmail.com

65. Remove All Adjacent Duplicates In String (L
#1047)
Given a string S of lowercase letters, a duplicate removal consists of choosing two adjacent and equal
letters, and removing them.
We repeatedly make duplicate removals on S until we no longer can.
Return the final string after all such duplicate removals have been made. It is guaranteed the answer is
unique.
Input: "abbaca"
Output: "ca"
Explanation: For example, in "abbaca" we could remove "bb" since the letters are adjacent and equal,
and this is the only possible move. The result of this move is that the string is "aaca", of which only "aa" is
possible, so the final string is "ca".

66. Remove All Adjacent Duplicates In String (L
#1047)
Test cases
Input: "abbaca"
Output: "ca"
Input: "abcdeedcba"
Output: ""
Input: "abcdefg"
Output: "abcdefg"

67. Remove All Adjacent Duplicates In String (L
#1047)
Approach #1
Iterate the string and remove the adjacents. If we find a pair, we set a flag to true indicating we
need to iterate again (since we have created a new string)
Input: "abcdeedcba"
Iteration #1: "abcddcba" Iteration #5: "abba"
Iteration #2: "abccba" Iteration #6: "aa"
Iteration #3: "abccba" Iteration #7 : ""
Iteration #4: "abccba"

68. Remove All Adjacent Duplicates In String (L
#1047)
Approach #1
Time Complexity = O(n^2)
Space Complexity = O(n)

69. Remove All Adjacent Duplicates In String (L
#1047)
Approach #2
We’ll use a stack this time as our support tool to solve this problem. We’ll iterate the string and
we’ll check if the stack has elements and if top of the stack (the previous char) is equal to the
element we’re currently checking.
If the previous statement is false, add the char to the stack
If the previous statement is true, pop the previous char and don’t add the current to the stack

70. Remove All Adjacent Duplicates In String (L
#1047)
a b b a c a
Input
Stack

71. Remove All Adjacent Duplicates In String (L
#1047)
a b b a c a
Input
Stack

72. Remove All Adjacent Duplicates In String (L
#1047)
a b b a c a
Input
Stack
a

73. Remove All Adjacent Duplicates In String (L
#1047)
a b b a c a
Input
Stack
a

74. Remove All Adjacent Duplicates In String (L
#1047)
a b b a c a
Input
Stack
a b

75. Remove All Adjacent Duplicates In String (L
#1047)
a b b a c a
Input
Stack
a b

76. Remove All Adjacent Duplicates In String (L
#1047)
a b b a c a
Input
Stack
a

77. Remove All Adjacent Duplicates In String (L
#1047)
a b b a c a
Input
Stack
a

78. Remove All Adjacent Duplicates In String (L
#1047)
a b b a c a
Input
Stack

79. Remove All Adjacent Duplicates In String (L
#1047)
a b b a c a
Input
Stack

80. Remove All Adjacent Duplicates In String (L
#1047)
a b b a c a
Input
Stack
c

81. Remove All Adjacent Duplicates In String (L
#1047)
a b b a c a
Input
Stack
c

82. Remove All Adjacent Duplicates In String (L
#1047)
a b b a c a
Input
Stack
c a

83. Remove All Adjacent Duplicates In String (L
#1047)
Final Result
c a

84. Remove All Adjacent Duplicates In String (L
#1047)
Approach #1
Time Complexity = O(n^2)
Space Complexity = O(n)
Approach #2
Time Complexity = O(n)
Space Complexity = O(n)

85. Remove All Adjacent Duplicates In String (L
#1047)
Approach #1
Time Complexity = O(n^2)
Space Complexity = O(n)
Approach #2 ← Winner!
Time Complexity = O(n)
Space Complexity = O(n)

87. Leetcode Problems
Easy
Remove All Adjacent Duplicates In String (Leetcode #1047)
Valid Parentheses (Leetcode #20)
Min Stack (Leetcode #155)
Decode String (Leetcode #394)
Medium
Design Browser History (Leetcode #1472)
Remove K Digits (Leetcode #402)
Online Stock Span (Leetcode #901)
Hard
Basic Calculator (Leetcode #224)

88. What next?
The next session will be about Trees!