The trick here is to name the state correctly, then let the implementation follow. For 232. Implement Queue using Stacks, the solution in this repository is mainly a data structure design solution.
Guide
What?
We want to turn the problem statement into a smaller set of decisions the computer can repeat safely. Instead of trying to be clever immediately, read the code as a sequence of questions:
- What state are we keeping?
- How do we move from one state to the next?
- When do we know the answer is already determined?
For this file, the main tools are: data structure design, stack.
Guide
When?
This is the kind of solution you want when the problem has structure hiding inside a messy-looking input. The accepted code reduces that pressure by storing exactly the information that remains useful later.
The important function names to track are push, pop, peek, empty.
Guide
Why?
The solution works because it narrows the problem until every update has a clear reason to exist.
- The stack stores unfinished context, which is usually the cleanest way to handle nested or monotonic structure.
- The final return is not magic; it is the invariant after the loops or recursion have finished doing their accounting.
Guide
How?
Walk through the solution in this order:
- Read the setup variables first.
- Follow the main loop or recursive helper next.
- Watch where invalid states get skipped.
- Check which value survives to the return statement.
The most important competitive-programming habit here is to trust the invariant. Once the invariant is right, the loops become much less scary.
Guide
Complexity
- Time: O(n) to O(n log n), depending on the dominant loop or data structure operation
- Space: O(n) in the usual case for auxiliary containers or recursion
Guide
C++ Solution
Your submission
The accepted solution
01//Runtime: 4 ms, faster than 100.00% of C++ online submissions for Implement Queue using Stacks.
02//Memory Usage: 8.8 MB, less than 97.79% of C++ online submissions for Implement Queue using Stacks.
03
04/**
05push: O(n), O(n)
06pop: O(1), O(1)
07peek: O(1), O(1)
08empty: O(1), O(1)
09**/
10
11class MyQueue {
12public:
13 stack<int> stk;
14
15 /** Initialize your data structure here. */
16 MyQueue() {
17 }
18
19 /** Push element x to the back of queue. */
20 void push(int x) {
21 stack<int> tmp;
22 while(!stk.empty()){
23 tmp.push(stk.top());
24 stk.pop();
25 }
26 stk.push(x);
27 while(!tmp.empty()){
28 stk.push(tmp.top());
29 tmp.pop();
30 }
31 }
32
33 /** Removes the element from in front of queue and returns that element. */
34 int pop() {
35 int tmp = stk.top();
36 stk.pop();
37 return tmp;
38 }
39
40 /** Get the front element. */
41 int peek() {
42 return stk.top();
43 }
44
45 /** Returns whether the queue is empty. */
46 bool empty() {
47 return stk.empty();
48 }
49};
50
51/**
52 * Your MyQueue object will be instantiated and called as such:
53 * MyQueue* obj = new MyQueue();
54 * obj->push(x);
55 * int param_2 = obj->pop();
56 * int param_3 = obj->peek();
57 * bool param_4 = obj->empty();
58 */
59
60 /**
61 Approach #2 (Two Stacks) Push - O(1)O(1) per operation, Pop - Amortized O(1)O(1) per operation.
62 **/
63
64 /**
65push: O(1), O(n)
66pop: amortized O(1), worst O(n), O(1)
67peek: O(1), O(1)
68empty: O(1), O(1)
69**/
70
71/**
72class MyQueue {
73public:
74 stack<int> s1, s2;
75 int front;
76
77 /** Initialize your data structure here. */
78 MyQueue() {
79 }
80
81 /** Push element x to the back of queue. */
82 void push(int x) {
83 if(s1.empty()) front = x;
84 s1.push(x);
85 }
86
87 /** Removes the element from in front of queue and returns that element. */
88 int pop() {
89 if(s2.empty()){
90 while(!s1.empty()){
91 s2.push(s1.top());
92 s1.pop();
93 }
94 }
95 int tmp = s2.top();
96 s2.pop();
97 return tmp;
98 }
99
100 /** Get the front element. */
101 int peek() {
102 if(!s2.empty()){
103 return s2.top();
104 }
105 return front;
106 }
107
108 /** Returns whether the queue is empty. */
109 bool empty() {
110 return s1.empty() && s2.empty();
111 }
112};
113**/Cost