LeetCode - Min Stack
Problem description
Design a stack that supports push, pop, top, and retrieving the minimum element in constant time.
- push(x) – Push element x onto stack.
- pop() – Removes the element on top of the stack.
- top() – Get the top element.
- getMin() – Retrieve the minimum element in the stack.
Example 1:
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
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:
- Methods pop, top and getMin operations will always be called on non-empty stacks.
Analysis
The first thought is to use another stack to save the min value.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
class MinStack {
Stack<Integer> vals;
Stack<Integer> mins;
/** initialize your data structure here. */
public MinStack() {
vals = new Stack<>();
mins = new Stack<>();
}
public void push(int x) {
vals.push(x);
if (mins.size() == 0){
mins.push(x);
}
else {
mins.push(Math.min(x, mins.peek()));
}
}
public void pop() {
vals.pop();
mins.pop();
}
public int top() {
return vals.peek();
}
public int getMin() {
return mins.peek();
}
}
/**
* Your MinStack object will be instantiated and called as such:
* MinStack obj = new MinStack();
* obj.push(x);
* obj.pop();
* int param_3 = obj.top();
* int param_4 = obj.getMin();
*/
A better way is to save the min only if the min needs to be updated.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
class MinStack {
private Stack<Integer> stack = new Stack<>();
private Stack<Integer> minStack = new Stack<>();
public MinStack() { }
public void push(int x) {
stack.push(x);
if (minStack.isEmpty() || x <= minStack.peek()) {
minStack.push(x);
}
}
public void pop() {
if (stack.peek().equals(minStack.peek())) {
minStack.pop();
}
stack.pop();
}
public int top() {
return stack.peek();
}
public int getMin() {
return minStack.peek();
}
}
Another improvement is to save the repeat times for each min.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
class MinStack {
private Stack<Integer> stack = new Stack<>();
private Stack<int[]> minStack = new Stack<>();
public MinStack() { }
public void push(int x) {
// We always put the number onto the main stack.
stack.push(x);
// If the min stack is empty, or this number is smaller than
// the top of the min stack, put it on with a count of 1.
if (minStack.isEmpty() || x < minStack.peek()[0]) {
minStack.push(new int[]{x, 1});
}
// Else if this number is equal to what's currently at the top
// of the min stack, then increment the count at the top by 1.
else if (x == minStack.peek()[0]) {
minStack.peek()[1]++;
}
}
public void pop() {
// If the top of min stack is the same as the top of stack
// then we need to decrement the count at the top by 1.
if (stack.peek().equals(minStack.peek()[0])) {
minStack.peek()[1]--;
}
// If the count at the top of min stack is now 0, then remove
// that value as we're done with it.
if (minStack.peek()[1] == 0) {
minStack.pop();
}
// And like before, pop the top of the main stack.
stack.pop();
}
public int top() {
return stack.peek();
}
public int getMin() {
return minStack.peek()[0];
}
}