Stack of Stacks - Amazon Top Interview Questions
Problem Statement :
Implement a data structure with the following methods: StackOfStacks(int capacity) which instantiates an instance that represents an infinite number of stacks, each with size capacity. void push(int val) which pushes the value val to the leftmost stack that's not full. int pop() which pops the value of the top element of the rightmost non-empty stack. If every stack is empty, return -1. int popStack(int idx) which pops the value of the top element of the idx (0-indexed) stack. If the stack at index idx is empty, return -1. Constraints 0 ≤ n ≤ 100,000 where n is the number of calls to push, pop and popStack Example 1 Input methods = ["constructor", "push", "push", "push", "popStack", "pop", "popStack"] arguments = [[2], [1], [2], [3], [0], [], [1]]` Output [None, None, None, None, 2, 3, -1] Explanation s = StackOfStacks(2) s.push(1) s.push(2) s.push(3) s.popStack(0) == 2 s.pop() == 3 s.popStack(1) == -1
Solution :
Solution in C++ :
class StackOfStacks {
int cap;
vector<stack<int>> stacks;
set<int> avails, idxes;
public:
StackOfStacks(int capacity) : cap(capacity) {
}
void push(int val) {
if (cap < 1) return;
if (avails.empty()) {
stacks.emplace_back();
stacks.back().push(val);
// cout << "push " << val << " to " << stacks.size() - 1 << endl;
if (cap > 1 && stacks.back().size() == 1) avails.insert(stacks.size() - 1);
idxes.insert(stacks.size() - 1);
} else {
auto it = avails.begin();
stacks[*it].push(val);
// cout << "push " << val << " to " << *it << endl;
if (stacks[*it].size() == 1) idxes.insert(*it);
if (stacks[*it].size() >= cap) avails.erase(it);
}
}
int pop() {
if (idxes.empty()) return -1;
auto it = idxes.end();
--it;
int ret = stacks[*it].top();
stacks[*it].pop();
if (stacks[*it].size() == cap - 1) avails.insert(*it);
if (stacks[*it].empty()) idxes.erase(it);
return ret;
}
int popStack(int idx) {
if (idx >= stacks.size() || stacks[idx].empty()) return -1;
int ret = stacks[idx].top();
stacks[idx].pop();
if (stacks[idx].size() == 0) idxes.erase(idx);
if (stacks[idx].size() == cap - 1) avails.insert(idx);
return ret;
}
};
Solution in Java :
import java.util.*;
class StackOfStacks {
private final TreeSet<Integer> avails = new TreeSet<>();
private final TreeSet<Integer> occups = new TreeSet<>();
private final List<Stack<Integer>> stacks = new ArrayList<>();
private final int N;
public StackOfStacks(int capacity) {
N = capacity;
}
public void push(int val) {
int idx = -1;
Stack<Integer> stack = null;
if (avails.isEmpty()) {
idx = stacks.size();
stacks.add(stack = new Stack<>());
avails.add(idx);
occups.add(idx);
} else {
idx = avails.first().intValue();
stack = stacks.get(idx);
}
stack.push(val);
if (stack.size() == N)
avails.remove(idx);
if (stack.size() == 1)
occups.add(idx);
}
public int pop() {
if (occups.isEmpty())
return -1;
final int idx = occups.last().intValue();
Stack<Integer> stack = stacks.get(idx);
final int res = stack.pop();
if (stack.size() + 1 == N)
avails.add(idx);
if (stack.size() == 0)
occups.remove(idx);
return res;
}
public int popStack(int idx) {
if (idx >= stacks.size())
return -1;
Stack<Integer> stack = stacks.get(idx);
if (stack.isEmpty())
return -1;
final int res = stack.pop();
if (stack.size() + 1 == N)
avails.add(idx);
if (stack.size() == 0)
occups.remove(idx);
return res;
}
}
Solution in Python :
class StackOfStacks:
def __init__(self, cap):
self.cap = cap
self.s = defaultdict(list)
self.c = Counter() # size of stack at idx
self.sl = []
self.ne = SortedList()
heapify(self.sl)
self.full = set()
self.stack_cnt = 0
def push(self, val):
if len(self.sl) == 0:
heappush(self.sl, self.stack_cnt)
self.stack_cnt += 1
idx = heappop(self.sl)
self.c[idx] += 1
self.s[idx].append(val)
if self.c[idx] == self.cap:
self.full.add(idx)
else:
heappush(self.sl, idx)
if self.c[idx] == 1:
self.ne.add(idx)
def pop(self):
if len(self.ne) == 0:
return -1
idx = self.ne[-1]
ret = self.s[idx].pop()
self.c[idx] -= 1
if self.c[idx] == 0:
self.ne.pop()
if self.c[idx] == self.cap - 1:
self.full.remove(idx)
heappush(self.sl, idx)
return ret
def popStack(self, idx):
if self.c[idx] == 0:
return -1
self.c[idx] -= 1
ret = self.s[idx].pop()
if self.c[idx] == self.cap - 1:
self.full.remove(idx)
heappush(self.sl, idx)
if self.c[idx] == 0:
self.ne.remove(idx)
return ret
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