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895. Maximum Frequency Stack

Question link

(https://leetcode.com/problems/maximum-frequency-stack/)

Question Description

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 1:

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.

Tags

  • stack

Code Implementation

class FreqStack {
private:
    unordered_map<int, int> freq;  //key, freq
    unordered_map<int, stack<int>> m; //freq, stack intem
    int maxFreq=0; //max freq
public:    
    void push(int x) {
        freq[x]++;
        maxFreq = max(maxFreq, freq[x]);
        m[freq[x]].push(x);
        return;
    }
    
    int pop() {
        int ret =  m[maxFreq].top();
        m[maxFreq].pop();
        if(m[maxFreq].size()==0) maxFreq--;
        freq[ret]--;
        return ret;
    }
};

Time Complexity Analysis