Min Stack
Problem Statement :
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. 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.
Solution :
Solution in C :
typedef struct Stack {
int dataArray[10000];
int top;
} MinStack;
void minStackGetMinForPush(MinStack*);
int min = 0; // For Minimum Element in the Stack
/** initialize your data structure here. */
MinStack* minStackCreate() {
static MinStack stack;
stack.top=-1;
return &stack;
}
void minStackPush(MinStack* obj, int x) {
if(obj->top == 10000){
return;
}
else{
obj->top++;
obj->dataArray[obj->top] = x;
minStackGetMinForPush(obj);
}
}
void minStackPop(MinStack* obj) {
if(obj->top == -1){
return;
}
else{
obj->top--;
}
if(obj->top == -1){
return;
}
int minTemp = obj->dataArray[obj->top];
for(int i=0;i<obj->top;i++){
if(minTemp>obj->dataArray[i]){
minTemp = obj->dataArray[i];
}
}
min = minTemp;
}
int minStackTop(MinStack* obj) {
return obj->dataArray[obj->top];
}
int minStackGetMin(MinStack* obj) {
return min;
}
void minStackGetMinForPush(MinStack* obj){
// if(obj->top == -1){
// return;
// }
if(obj->top==0){
min = obj->dataArray[obj->top];
}
else if(obj->top>0){
if(min > obj->dataArray[obj->top]){
min = obj->dataArray[obj->top];
}
}
}
void minStackFree(MinStack* obj) {
obj->top = -1;
}
Solution in C++ :
class MinStack {
public:
typedef struct node{
int v;
int minUntilNow;
node* next;
}node;
MinStack() : topN(nullptr){
}
void push(int val) {
node* n = new node;
n->v = n->minUntilNow = val;
n->next = nullptr;
if(topN == nullptr){
topN = n;
}
else{
n->minUntilNow = min(n->v,topN->minUntilNow);
n->next = topN;
topN = n;
}
}
void pop() {
topN = topN->next;
}
int top() {
return topN->v;
}
int getMin() {
return topN->minUntilNow;
}
private:
node* topN;
};
Solution in Java :
class MinStack {
LinkedList<TplusMin> stack;
private class TplusMin {
int val;
int min;
public TplusMin(int val, int min) {
this.val = val;
this.min = min;
}
}
public MinStack() {
stack = new LinkedList<>();
}
public void push(int val) {
int newMin;
if (stack.size() == 0){
newMin = val;
}
else {
int currentMin = stack.getFirst().min;
newMin = val < currentMin ? val : currentMin;
}
stack.addFirst(new TplusMin(val, newMin));
}
public void pop() {
stack.removeFirst();
}
public int top() {
return stack.peekFirst().val;
}
public int getMin() {
return stack.peekFirst().min;
}
}
Solution in Python :
class MinStack:
def __init__(self):
self.stack = []
self.minStack = []
def push(self, val: int) -> None:
self.stack.append(val)
if self.minStack:
val = min(self.minStack[-1],val)
self.minStack.append(val)
def pop(self) -> None:
self.stack.pop()
self.minStack.pop()
def top(self) -> int:
return self.stack[-1]
def getMin(self) -> int:
return self.minStack[-1]
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