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 <stdio.h>
2#include <limits.h>
3#define MAX_SIZE 30000
4
5typedef struct {
6 int data[MAX_SIZE];
7 int minData[MAX_SIZE];
8 int top;
9 int minTop;
10} MinStack;
11
12MinStack* minStackCreate() {
13 MinStack* stack = (MinStack*)malloc(sizeof(MinStack));
14 stack->top = -1;
15 stack->minTop = -1;
16 return stack;
17}
18
19void minStackPush(MinStack* obj, int val) {
20 obj->data[++obj->top] = val;
21 if (obj->minTop == -1 || val <= obj->minData[obj->minTop]) {
22 obj->minData[++obj->minTop] = val;
23 }
24}
25
26void minStackPop(MinStack* obj) {
27 if (obj->top == -1) return;
28 if (obj->data[obj->top] == obj->minData[obj->minTop]) {
29 obj->minTop--;
30 }
31 obj->top--;
32}
33
34int minStackTop(MinStack* obj) {
35 return obj->data[obj->top];
36}
37
38int minStackGetMin(MinStack* obj) {
39 return obj->minData[obj->minTop];
40}
41
42void minStackFree(MinStack* obj) {
43 free(obj);
44}
45The C solution uses simple arrays to maintain the main stack and the minimum stack. We use indices to track the current position in both stacks. The main operations are very similar to basic stack operations but with an additional stack handling minimums.
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.