URL: LeetCode Problem
Problem Description
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.
- You must implement a solution with O(1) time complexity for each function.
Examples:
- 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.
Answer
Intuition
While the straightforward approach is to implement the stack using a class with the required methods, a more efficient alternative exists. This approach involves maintaining two arrays: one for storing all pushed elements (array), and another (min_array) for storing the minimum value at each level of the stack.
Approach
push
The push method appends the given element to array. It also appends a value to min_array — but only if min_array is empty or if the given value is smaller than the last element in min_array. In the latter case, the given value is appended to min_array.
pop
The pop method removes the last element from both array and min_array.
top
Either of the last element can be returned.
getMin
Returns the last element of min_array, which is the minimum value currently in the stack.
Complexity
Time complexity: O(n)
Space complexity: O(n)
Code
class MinStack:
array = []
min_array = []
def __init__(self):
self.array = []
self.min_array = []
def push(self, val: int) -> None:
self.array.append(val)
self.min_array.append(val if not self.min_array else min(val, self.min_array[-1]))
def pop(self) -> None:
self.array.pop()
self.min_array.pop()
def top(self) -> int:
return self.array[-1]
def getMin(self) -> int:
return self.min_array[-1]