This approach utilizes two stacks. One stack will hold the main elements, while the other stack will keep track of the minimum elements. When you push a value, it's added to the main stack, and if it's less than or equal to the current minimum, it's also pushed onto the min stack. During a pop operation, if the value being popped is the current minimum, it's also popped from the min stack.
Time Complexity: O(1) for each operation.
Space Complexity: O(n), where n is the number of elements in the stack.
1#include <stack>
2#include <climits>
3class MinStack {
4 std::stack<int> s;
5 std::stack<int> minStack;
6public:
7 MinStack() {
8 }
9
10 void push(int val) {
11 s.push(val);
12 if (minStack.empty() || val <= minStack.top()) {
13 minStack.push(val);
14 }
15 }
16
17 void pop() {
18 if (s.top() == minStack.top()) {
19 minStack.pop();
20 }
21 s.pop();
22 }
23
24 int top() {
25 return s.top();
26 }
27
28 int getMin() {
29 return minStack.top();
30 }
31};
32The C++ implementation uses two standard library stacks. The second stack, minStack, only stores the minimum values so far. Any time a value is pushed that's smaller or equal to the top of minStack, it's pushed onto minStack as well. When removing a value, if it's the same as the top of minStack, it's also popped from minStack.
This approach reduces space usage by storing a pair (value, minimum so far) in a single stack. Each element maintains the current minimum alongside its actual value, providing quick minimum retrieval by just examining the top of the stack.
Time Complexity: O(1) for each operation.
Space Complexity: O(n), each element only holds a pair so space is efficient relative to the number of elements.
1class MinStack:
2 def __init__(self):
3 self.stack = []
4
5 def push(self, val: int) -> None:
6 if not self.stack:
7 self.stack.append((val, val))
8 else:
9 self.stack.append((val, min(val, self.stack[-1][1])))
10
11 def pop(self) -> None:
12 self.stack.pop()
13
14 def top(self) -> int:
15 return self.stack[-1][0]
16
17 def getMin(self) -> int:
18 return self.stack[-1][1]
19Python takes advantage of tuples to store both the value and current minimum together. When popping or retrieving the top, the value is easily accessible, and minimum retrieval likewise by referring to the second element of the tuple.