[LeetCode] Maximum Frequency Stack

Problem: Design a stack-like data structure to push elements to the stack and pop the most frequent element from the stack.

Implement the FreqStack class:

• FreqStack() constructs an empty frequency stack.
• void push(int val) pushes an integer val onto the top of the stack.
• int pop() removes and returns the most frequent element in the stack.
• If there is a tie for the most frequent element, the element closest to the stack's top is removed and returned.

Example:

Input
["FreqStack", "push", "push", "push", "push", "push", "push", "pop", "pop", "pop", "pop"]
[[], [5], [7], [5], [7], [4], [5], [], [], [], []]
Output
[null, null, null, null, null, null, null, 5, 7, 5, 4]

Explanation
FreqStack freqStack = new FreqStack();
freqStack.push(5); // The stack is [5]
freqStack.push(7); // The stack is [5,7]
freqStack.push(5); // The stack is [5,7,5]
freqStack.push(7); // The stack is [5,7,5,7]
freqStack.push(4); // The stack is [5,7,5,7,4]
freqStack.push(5); // The stack is [5,7,5,7,4,5]
freqStack.pop();   // return 5, as 5 is the most frequent. The stack becomes [5,7,5,7,4].
freqStack.pop();   // return 7, as 5 and 7 is the most frequent, but 7 is closest to the top. The stack becomes [5,7,5,4].
freqStack.pop();   // return 5, as 5 is the most frequent. The stack becomes [5,7,4].
freqStack.pop();   // return 4, as 4, 5 and 7 is the most frequent, but 4 is closest to the top. The stack becomes [5,7].

Constraints:

• 0 <= val <= 109
• At most 2 * 104 calls will be made to push and pop.
• It is guaranteed that there will be at least one element in the stack before calling pop.

Approach: We can use two hash maps to solve this problem:

1. FrequencyMap: This map will have frequency as key and corresponding list of elements as values.
2. ElementsMap: This map will have elements as key and it's frequency as value.

Once we have two maps in our class along with a member integer say max_frequncy, We can apply following operations at push and pop:

Push: 

    ElementsMap[val] = ElementsMap[val] + 1

    curr_freq = ElementsMap[val]

    FrequencyMap[curr_freq].AddAtLast(val)

    max_freq = MAX(curr_freq, max_freq)

Pop:

    val = FrequencyMap[max_freq].GetLast()

    FrequencyMap[max_freq].RemoveLast()

    IF ISEMPTY(FrequencyMap[max_freq])):

        FrequencyMap.Remove(max_freq)

        --max_freq

    ElementsMap[val] = ElementsMap[val] - 1

    RETURN val

That's all!

Implemetation in C#:

public class FreqStack

{

    public FreqStack()

    {

        this.freqMap = new Dictionary<int, List<int>>();

        this.elementsMap = new Dictionary<int, int>();

        this.maxFreq = 0;

    }

   

    public void Push(int val)

    {

        if (!elementsMap.ContainsKey(val))

        {

            elementsMap[val] = 0;

        }

        ++elementsMap[val];

        int currFreq = elementsMap[val];

        if (!freqMap.ContainsKey(currFreq))

        {

            freqMap[currFreq] = new List<int>();

        }

        freqMap[currFreq].Add(val);

        if (currFreq > this.maxFreq)

        {

            this.maxFreq = currFreq;

        }

    }

   

    public int Pop()

    {

        int maxFreqListCount = this.freqMap[this.maxFreq].Count;

        int val = this.freqMap[this.maxFreq][maxFreqListCount - 1];

        this.freqMap[this.maxFreq].RemoveAt(maxFreqListCount - 1);

        if (this.freqMap[this.maxFreq].Count == 0)

        {

            this.freqMap.Remove(this.maxFreq);

            --this.maxFreq;

        }

        --this.elementsMap[val];

        if (this.elementsMap[val] == 0)

        {

            this.elementsMap.Remove(val);

        }

        return val;

    }

    private Dictionary<int, List<int>> freqMap;

    private Dictionary<int, int> elementsMap;

    int maxFreq;

}

Complexiy: O(1) for push and pop.