LeetCode 895. Maximum Frequency Stack

 895. 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"]
[[], [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.

	class FreqStack {
	public:
	// keep each element's frequent
	unordered_map<int, int> freq;
	// keep each frequent's element as an order
	unordered_map<int, stack<int>> group;
	// keap max frequent
	int maxfreq;
	FreqStack() {
	maxfreq = 0;
	}
	void push(int val) {
	// update frequent of value
	int f = freq[val] + 1;
	freq[val] = f;
	// update max frequent if needed
	if(f > maxfreq)
	maxfreq = f;
	// freq: [v1, v2, v3] as stack
	group[f].push(val);
	}
	int pop() {
	// maxfreq's top val
	int x = group[maxfreq].top(); group[maxfreq].pop();
	// decrease frequent
	freq[x] = freq[x] - 1;
	if(group[maxfreq].size() == 0)
	maxfreq--;
	return x;
	}
	};
	/**
	* Your FreqStack object will be instantiated and called as such:
	* FreqStack* obj = new FreqStack();
	* obj->push(val);
	* int param_2 = obj->pop();
	*/

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