Middle Operable Deque - Amazon Top Interview Questions


Problem Statement :


Implement a data structure with the following methods:

void appendLeft(int val) when appends val to the left of the deque.

int popLeft() which pops the leftmost value of the deque. If it's empty, return -1.

void append(int val) when appends val to the end of the deque.

int pop() which pops the last value of the deque. If it's empty, return -1.

void appendMiddle(int val) when appends val to the middle of the deque.

int popMiddle() which pops the middle value of the deque. If it's empty, return -1.

Middle is defined to be floor(k / 2) where k is the length of the deque.

Constraints

0 ≤ n ≤ 100,000 where n is the number of calls to any method.

Example 1

Input

methods = ["constructor", "append", "appendLeft", "appendMiddle", "appendLeft", "popMiddle", "pop", "popLeft"]
arguments = [[], [1], [3], [2], [0], [], [], []]`

Output

[None, None, None, None, None, 3, 1, 0]



Solution :



title-img




                        Solution in C++ :

class MiddleOperableDeque {
    deque<int> q1, q2;

    public:
    MiddleOperableDeque() {
    }

    void appendLeft(int val) {
        q1.push_front(val);
        if (q1.size() > q2.size() + 1) q2.push_front(q1.back()), q1.pop_back();
    }

    int popLeft() {
        if (q1.empty()) return -1;
        int ret = q1.front();
        q1.pop_front();
        if (q1.size() < q2.size()) q1.push_back(q2.front()), q2.pop_front();
        return ret;
    }

    void append(int val) {
        q2.push_back(val);
        if (q1.size() < q2.size()) q1.push_back(q2.front()), q2.pop_front();
    }

    int pop() {
        if (q2.empty()) {
            if (q1.empty()) return -1;
            int ret = q1.back();
            q1.pop_back();
            return ret;
        }
        int ret = q2.back();
        q2.pop_back();
        if (q1.size() > q2.size() + 1) q2.push_front(q1.back()), q1.pop_back();
        return ret;
    }

    void appendMiddle(int val) {
        if (q1.size() == q2.size())
            q1.push_back(val);
        else
            q2.push_front(q1.back()), q1.back() = val;
    }

    int popMiddle() {
        if (q1.empty()) return -1;
        int ret = q1.back();
        q1.pop_back();
        if (q1.size() < q2.size()) q1.push_back(q2.front()), q2.pop_front();
        return ret;
    }
};
                    




                        Solution in Python : 
                            
class MiddleOperableDeque:
    def __init__(self):
        # Represent the queue as the concatenation of two deques.
        # We will make sure the gap between the deques corresponds to the "middle".
        self.left = deque()
        self.right = deque()

    def _rebalance(self):
        # Ensure that len(self.left) == floor(n / 2) and len(self.right) == ceil(n / 2)
        if len(self.left) > len(self.right):
            self.right.appendleft(self.left.pop())
        elif len(self.right) > len(self.left) + 1:
            self.left.append(self.right.popleft())

    def appendLeft(self, val):
        self.left.appendleft(val)
        self._rebalance()

    def popLeft(self):
        if self.left:
            val = self.left.popleft()
        elif self.right:
            # Special case: When there is only one element, it's in `self.right`
            val = self.right.popleft()
        else:
            return -1
        self._rebalance()
        return val

    def append(self, val):
        self.right.append(val)
        self._rebalance()

    def pop(self):
        if self.right:
            val = self.right.pop()
        else:
            # No special case - if self.right is empty then the whole queue is empty
            return -1
        self._rebalance()
        return val

    def appendMiddle(self, val):
        # If the overall length is odd, tiebreak by putting the element to the left of the middle element
        self.left.append(val)
        self._rebalance()

    def popMiddle(self):
        if not self.left and not self.right:
            return -1
        elif len(self.left) == len(self.right):
            # If the overall length is even, tiebreak by taking the element on the left
            val = self.left.pop()
        else:
            val = self.right.popleft()
        self._rebalance()
        return val
                    


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