Problema
A generic Stack<T> works for any type. Here you extend the idea: a stack of integers that can also report its minimum in constant time.
Process the operations in order:
push <x>: push x.pop: remove the top value (ignored when the stack is empty).min: report the smallest value currently in the stack, or -1 if it is empty.
Implement min_stack_operations(operations).
Input
operations: The operations.
Output
Return the answers of the min operations, in order.
Restricciones
1 ≤ number of operations ≤ 200,000
0 ≤ x ≤ 10^9
Ejemplos
Entrada:
operations = ["push 5", "push 3", "min", "pop", "min"]
Salida esperada:
expected = [3, 5]Entrada:
std::vector<std::string> operations = {"push 5", "push 3", "min", "pop", "min"};
Salida esperada:
std::vector<long long> expected = {3, 5};Entrada:
string[] operations = new string[] {"push 5", "push 3", "min", "pop", "min"};
Salida esperada:
long[] expected = new long[] {3, 5};Explicación
With 5 and 3 on the stack the minimum is 3. After popping 3, only 5 is left: [3, 5].
Entrada:
operations = ["min", "pop", "push 7", "min"]
Salida esperada:
expected = [-1, 7]Entrada:
std::vector<std::string> operations = {"min", "pop", "push 7", "min"};
Salida esperada:
std::vector<long long> expected = {-1, 7};Entrada:
string[] operations = new string[] {"min", "pop", "push 7", "min"};
Salida esperada:
long[] expected = new long[] {-1, 7};Explicación
The empty stack reports -1 and the pop is ignored; then the minimum is 7: [-1, 7].
Cómo abordarlo
Inténtalo primero por tu cuenta y luego abre las pistas una a una.
Pista 1
Store, next to each value, the minimum of the stack at that point.
Pista 2
The current minimum is then simply the top of that second stack.
Pista 3
Ignore pop on an empty stack, and answer -1 for min when it is empty.
Complejidad esperada
O(number of operations) time, O(number of operations) extra space
Referencias
Este reto está adaptado para CodeUtility. Sigue el enunciado y las restricciones indicadas arriba.