895. Maximum Frequency Stack

This problem looks busy at first, but the accepted solution is built around one steady invariant. For 895. Maximum Frequency Stack, the solution in this repository is mainly a data structure design solution.

Guide

What?

Before optimizing anything, pin down what information is still useful after each move. 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.

The notes already sitting in the source point us in the right direction:

TLE
30 / 37 test cases passed.

Guide

When?

Use this approach when the hard part is not syntax, but deciding what must stay true after every update. The accepted code reduces that pressure by storing exactly the information that remains useful later.

The important function names to track are push, pop.

Guide

Why?

The code is doing bookkeeping so your brain does not have to keep the entire search space open at once.

A map keeps the lookup side cheap; the code pays a little memory to avoid repeated searching.
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:

Start from the smallest reliable state.
Expand one legal move at a time.
Cache, count, or merge information as soon as it becomes settled.
Let the final stored value answer the original question.

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

solution.cpp

01//TLE
02//30 / 37 test cases passed.
03class FreqStack {
04public:
05    map<int, int> counter;
06    vector<int> vec;
07    
08    FreqStack() {
09        
10    }
11    
12    void push(int x) {
13        vec.push_back(x);
14        counter[x]++;
15    }
16    
17    int pop() {
18        map<int, int>::iterator p = max_element(counter.begin(), counter.end(), 
19            [] (const pair<int, int> & p1, const pair<int, int> & p2) {
20                return p1.second < p2.second;
21            }
22        );
23        int mode = p->second;
24        int key;
25        for(int i = vec.size()-1; i >= 0; i--){
26            key = vec[i];
27            if(counter[key] == mode){
28                vec[i] = -1;
29                counter[key]--;
30                if(counter[key] == 0){
31                    counter.erase(key);
32                }
33                break;
34            }
35        }
36        return key;
37    }
38};
39
40/**
41 * Your FreqStack object will be instantiated and called as such:
42 * FreqStack* obj = new FreqStack();
43 * obj->push(x);
44 * int param_2 = obj->pop();
45 */
46 
47 //Runtime error
48 class FreqStack {
49public:
50    map<int, int> counter;
51    vector<int> vec;
52    
53    FreqStack() {
54        
55    }
56    
57    void push(int x) {
58        vec.push_back(x);
59        counter[x]++;
60    }
61    
62    int pop() {
63        map<int, int>::iterator p = max_element(counter.begin(), counter.end(), 
64            [] (const pair<int, int> & p1, const pair<int, int> & p2) {
65                return p1.second < p2.second;
66            }
67        );
68        int mode = p->second;
69        int key;
70        auto rit = find_if(vec.rbegin(), vec.rend(), 
71               [this, &mode](const int& key){
72                   return this->counter[key] == mode;
73               });
74        int idx = vec.rend() - rit - 1;
75        vec[idx] = -1;
76        counter[key]--;
77        
78        return key;
79    }
80};
81
82/**
83 * Your FreqStack object will be instantiated and called as such:
84 * FreqStack* obj = new FreqStack();
85 * obj->push(x);
86 * int param_2 = obj->pop();
87 */
88
89//Approach 1: Stack of Stacks
90//Runtime: 408 ms, faster than 10.44% of C++ online submissions for Maximum Frequency Stack.
91//Memory Usage: 73.1 MB, less than 91.67% of C++ online submissions for Maximum Frequency Stack.
92//time: push and pop, O(1), space: O(n)
93class FreqStack {
94public:
95    map<int, int> freq;
96    map<int, stack<int>> group;
97    int maxfreq;
98    
99    FreqStack() {
100        maxfreq = 0;
101    }
102    
103    void push(int x) {
104        int f = freq[x] + 1;
105        freq[x] = f;
106        maxfreq = max(maxfreq, f);
107        
108        // don't need to check whether group[f] is initialized?
109        group[f].push(x);
110        
111        /*
112        if x is pushed multiple times, 
113        then group[1], group[2], group[3], ... will all have x 
114        */
115    }
116    
117    int pop() {
118        int x = group[maxfreq].top(); group[maxfreq].pop();
119        freq[x]--;
120        if(group[maxfreq].size() == 0){
121            /*
122            In this circumstance, maxfreq is consecutive
123            */
124            maxfreq--;
125        }
126        return x;
127    }
128};
129
130/**
131 * Your FreqStack object will be instantiated and called as such:
132 * FreqStack* obj = new FreqStack();
133 * obj->push(x);
134 * int param_2 = obj->pop();
135 */

Cost

Complexity

Time

O(n) to O(n log n), depending on the dominant loop or data structure operation

Dominated by the main traversal, recursion, or data-structure operations in the code.

Space

O(n) in the usual case for auxiliary containers or recursion

Auxiliary state plus the answer structure where the problem requires one.