Min Stack - Beauty of OOP
Problem Statement:
Design a stack that supports push, pop, top, and retrieving the minimum element in constant time.
Implement the MinStack
class:
MinStack()
initializes the stack object.void push(int val)
pushes the elementval
onto the stack.void pop()
removes the element on the top of the stack.int top()
gets the top element of the stack.int getMin()
retrieves the minimum element in the stack.
You must implement a solution with O(1)
time complexity for each function.
Example 1:
Input ["MinStack","push","push","push","getMin","pop","top","getMin"] [[],[-2],[0],[-3],[],[],[],[]] Output [null,null,null,null,-3,null,0,-2] Explanation MinStack minStack = new MinStack(); minStack.push(-2); minStack.push(0); minStack.push(-3); minStack.getMin(); // return -3 minStack.pop(); minStack.top(); // return 0 minStack.getMin(); // return -2
Constraints:
-231 <= val <= 231 - 1
- Methods
pop
,top
andgetMin
operations will always be called on non-empty stacks. - At most
3 * 104
calls will be made topush
,pop
,top
, andgetMin
.
Solution:
We use pre-built stack class but with our own class object instead of int which will have information on minimum value at any point on stack.
class Item
{
public:
int val;
int minVal;
Item(int v, int mv): val(v), minVal(mv){}
};
class MinStack {
stack<Item> stk;
public:
MinStack()
{}
void push(int val)
{
int m = (stk.empty()) ? val : min(stk.top().minVal, val);
stk.push(Item(val, m));
}
void pop()
{
stk.pop();
}
int top()
{
return stk.top().val;
}
int getMin()
{
return stk.top().minVal;
}
};