# 895. Maximum Frequency Stack LeetCode Solution | Easy Approach Share:

Maximum Frequency Stack 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"]
[[], , , , , , , [], [], [], []]
Output
```
[null, null, null, null, null, null, null, 5, 7, 5, 4]

Explanation FreqStack freqStack = new FreqStack(); freqStack.push(5); // The stack is  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`.

Time: O(n)
Space: O(n)

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