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 :
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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