LeetCode in Python

155. Min Stack

Easy

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 element val 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.

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 and getMin operations will always be called on non-empty stacks.
• At most 3 * 104 calls will be made to push, pop, top, and getMin.

To solve the problem and implement the MinStack class that supports push, pop, top, and getMin operations in constant time, we can use two stacks:

1. Main Stack: This stack stores all the elements.
2. Min Stack: This stack keeps track of the minimum elements. Whenever a new element is pushed onto the main stack, it is also pushed onto the min stack if it is less than or equal to the current minimum. When an element is popped from the main stack, it is also popped from the min stack if it is the current minimum.

Steps:

1. Initialization:
   • Create two stacks: main_stack and min_stack.
2. Push Operation:
   • Push the value onto main_stack.
   • If min_stack is empty or the value is less than or equal to the top of min_stack, push the value onto min_stack.
3. Pop Operation:
   • Pop the value from main_stack.
   • If the popped value is equal to the top of min_stack, pop it from min_stack.
4. Top Operation:
   • Return the top value of main_stack.
5. GetMin Operation:
   • Return the top value of min_stack.

Implementation:

class MinStack:
    def __init__(self):
        self.main_stack = []
        self.min_stack = []

    def push(self, val: int) -> None:
        self.main_stack.append(val)
        if not self.min_stack or val <= self.min_stack[-1]:
            self.min_stack.append(val)

    def pop(self) -> None:
        if self.main_stack:
            if self.main_stack[-1] == self.min_stack[-1]:
                self.min_stack.pop()
            self.main_stack.pop()

    def top(self) -> int:
        if self.main_stack:
            return self.main_stack[-1]

    def getMin(self) -> int:
        if self.min_stack:
            return self.min_stack[-1]

Explanation:

1. Initialization:
   • The __init__ method initializes main_stack and min_stack as empty lists.
2. Push Operation:
   • push(val: int) -> None:
     • Adds val to main_stack.
     • If min_stack is empty or val is less than or equal to the current minimum (top of min_stack), it adds val to min_stack.
3. Pop Operation:
   • pop() -> None:
     • Removes the top element from main_stack.
     • If this element is the same as the top element of min_stack, it removes the top element from min_stack as well.
4. Top Operation:
   • top() -> int:
     • Returns the top element of main_stack.
5. GetMin Operation:
   • getMin() -> int:
     • Returns the top element of min_stack, which is the current minimum in the stack.

This implementation ensures that all operations (push, pop, top, and getMin) are performed in constant time O(1).

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